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13.3: New Page - Mathematics


13.3: New Page - Mathematics

13.3: New Page - Mathematics

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Mathematics Departments of CUNY

Here are the locations and contact information for CUNY’s mathematics departments. See also CUNY’s Colleges page and map for general campus information.


Alphabetical Listing

Bernard M. Baruch College
24th St & Lexington Ave
One Bernard Baruch Way
Rm 6-230
New York, NY 10010
(phone) (646) 312-4110
(fax) (646) 312-4111

Borough of Manhattan Community College
199 Chambers Street
Rm N-520
New York, NY 10007
(phone) (212) 220-1335
(fax) (212) 748-7459

Bronx Community College
University Ave and W 181st St
Rm CP 315
Bronx, NY 10453
(phone) (718) 289-5411
(fax) (718) 289-6056

Brooklyn College
2900 Bedford Ave
1156 Ingersoll Hall
Brooklyn, NY 11210
(phone) (718) 951-5246
(fax) (718) 951-4674

City College of New York
Convent Ave & 138th Street
Rm NAC 8/133
New York, NY 10031
(phone) (212) 650-5346
(fax) (212) 862-0004
(email) [email protected]

College of Staten Island
2800 Victory Blvd
Rm 1S-215
Staten Island, NY 10314
(phone) (718) 982-3600

Graduate Center
365 5th Ave
Rm 4208
New York, NY 10016
(phone) (212) 817-8530
(fax) (212) 817-1527
(email) [email protected]

Hostos Community College
500 Grand Concourse
Rm B-408
Bronx, NY 10451
(phone) (718) 518-6615

Hunter College
695 Park Ave
Rm 919/944 E
New York, NY 10065
(phone) (212) 772-5300

John Jay College of Criminal Justice
444 West 56th Street
Rm 4221N
New York, NY 10019
(phone) (212) 237-8925

Kingsborough Community College
2001 Oriental Boulevard
Rm F309
Brooklyn, NY 11235
(phone) (718) 368-5931
(fax) (718) 368-4868
(email) [email protected]

Lehman College
250 Bedford Park Blvd W
Rm GI-211
Bronx, NY 10468
(phone) (718) 960-8117
(fax) (718) 960-8969

LaGuardia Community College
31-10 Thomson Avenue
Rm E218
Long Island City, NY 11101
(phone) (718) 482-5710

Medgar Evers College
1650 Bedford Ave
Brooklyn, NY 11225
(phone) (718) 270-6416

New York City College of Technology
300 Jay Street
Rm N-711
Brooklyn, NY 11201
(phone) (718) 260-5380
(fax) (718) 254-8537

Queens College
65-30 Kissena Blvd
Kiely 237
Flushing, NY 11367
(phone) (718) 997-5800

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CUNYMath Blog

Hypatia, a Greek scholar, is one of the first well-known women in mathematics.


Welcome!

The main department telephone lines at (212) 650-5346,7 are being transferred to a personal phone outside the college and will be answered roughly between 11am and 7pm on Monday and Wednesday, and 9am to 5pm on Tuesday, Thursday, and Friday, when the college is not closed for a holiday or snow.

Preferably, email your concerns and questions to [email protected], which is also being monitored.

Staff members and administrators are either forwarding their telephone lines or monitoring their campus voicemail on a regular basis.

The Assistant Chair, Prof. Bak, and the Department Advisors for non-majors, Mr. Park and Mr. Turner, are holding regular online advising hours during which you can contact them in real time. The details can found on our Administration page which can be accessed via the left side bar or you can click here.

Additionally, Mr. Turner, who is also the Artino Lab Director, is available during his aforementioned posted office hours to answer programming questions from students in Math 328, 377, and 412/A1200.

Instructors can be contacted via email or their office hours. You can find a listing of instructors by clicking here or you can click here to view office hours for the Summer 2021 term.

We hope everyone stays safe and healthy, and we look forward to seeing each other in person once again.

Meet the Platonic Solids: Octahedron

The Platonic solids have been known since antiquity, and they play a prominent role in Plato's description of the physical world. The planar faces of each solid are identical polygons. Only equilateral triangles, squares and regular pentagons appear.

Although the platonic solids seem to be purely geometric objects, they embody a number of deep algebraic features. Their symmetries, for example, relate to the solution of polynomial equations of low degree.

If you would like to learn more about Platonic solids, you can start here.


Common Core Grade 3 Math (Worksheets, Homework, Lesson Plans)

Looking for video lessons that will help you in your Common Core Grade 3 Math classwork or homework?
Looking for Common Core Math Worksheets and Lesson Plans that will help you prepare lessons for Grade 3 students?

The following lesson plans and worksheets are from the New York State Education Department Common Core-aligned educational resources. The Lesson Plans and Worksheets are divided into seven modules.

Grade 3 Homework, Lesson Plans And Worksheets

Lesson 1: Understand equal groups of as multiplication. (Video)

Lesson 4: Understand the meaning of the unknown as the size of the group in division. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 5: Understand the meaning of the unknown as the number of groups in division. (Video)

Lesson 7, Lesson 8: Demonstrate the commutativity of multiplication and practice related facts by skip-counting objects in array models. (Video)

Lesson 9: Find related multiplication facts by adding and subtracting equal groups in array models. (Video)

Lesson 11: Model division as the unknown factor in multiplication using arrays and tape diagrams. (Video)

Lesson 12: Interpret the quotient as the number of groups or the number of objects in each group using units of 2. (Video)

Lesson 14: Skip-count objects in models to build fluency with multiplication facts using units of 4. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 15: Relate arrays to tape diagrams to model the commutative property of multiplication. (Video)

Lesson 16: Use the distributive property as a strategy to find related multiplication facts. (Video)

Lesson 20: Solve two-step word problems involving multiplication and division and assess the reasonableness of answers. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 1: Explore time as a continuous measurement using a stopwatch. (Video)

Lesson 2: Relate skip-counting by 5 on the clock and telling time to a continuous measurement model, the number line. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 3: Count by fives and ones on the number line as a strategy to tell time to the nearest minute on the clock. (Video)

Lesson 4: Solve word problems involving time intervals within 1 hour by counting backward and forward using the number line and clock. (Video)

Lesson 6: Build and decompose a kilogram to reason about the size and weight of 1 kilogram, 100 grams, 10 grams, and 1 gram. (Video)

Lesson 7: Develop estimation strategies by reasoning about the weight in kilograms of a series of familiar objects to establish mental benchmark measures. (Video)

Lesson 8: Solve one-step word problems involving metric weights within 100 and estimate to reason about solutions. (Video)

Lesson 9: Decompose a liter to reason about the size of 1 liter, 100 milliliters, 10 milliliters, and 1 milliliter. (Video)

Lesson 10: Estimate and measure liquid volume in liters and milliliters using the vertical number line. (Video)

Lesson 12: Round two-digit measurements to the nearest ten on the vertical number line. (Video)

Lesson 13: Round two- and three-digit numbers to the nearest ten on the vertical number line. (Video)

Lesson 15: Add measurements using the standard algorithm to compose larger units once. (Video)

Lesson 16: Add measurements using the standard algorithm to compose larger units twice. (Video)

Lesson 18: Decompose once to subtract measurements including three-digit minuends with zeros in the tens or ones place. (Video)

Lesson 19: Decompose twice to subtract measurements including three-digit minuends with zeros in the tens and ones places. (Video)

Lesson 20: Estimate differences by rounding and apply to solve measurement word problems. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 1: Study commutativity to find known facts of 6, 7, 8, and 9. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 2: Apply the distributive and commutative properties to relate multiplication facts 5 × n + n to 6 × n and n × 6 where n is the size of the unit. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 4: Count by units of 6 to multiply and divide using number bonds to decompose. (Video)

Lesson 5: Count by units of 7 to multiply and divide using number bonds to decompose. (Video)

Lesson 6: Use the distributive property as a strategy to multiply and divide using units of 6 and 7. (Video)

Lesson 8: Understand the function of parentheses and apply to solving problems. (Video)

Lesson 9: Model the associative property as a strategy to multiply. (Video)

Lesson 10: Use the distributive property as a strategy to multiply and divide. (Video)

Lesson 12: Apply the distributive property and the fact 9 = 10 - 1 as a strategy to multiply. (Video)

Lesson 16: Reason about and explain arithmetic patterns using units of 0 and 1 as they relate to multiplication and division. (Video) (Worksheet Sprint A) (Worksheet Sprint B)

Lesson 17: Identify patterns in multiplication and division facts using the multiplication table. (Video)

Lesson 19: Multiply by multiples of 10 using the place value chart. (Video)

Lesson 20: Use place value strategies and the associative property n × (m × 10) = (n × m) × 10 (where n and m are less than 10) to multiply by multiples of 10. (Video)

Lesson 1: Understand area as an attribute of plane figures. (Video)

Lesson 2: Decompose and recompose shapes to compare areas. (Video)

Lesson 3: Model tiling with centimeter and inch unit squares as a strategy to measure area. (Video)

Lesson 5: Form rectangles by tiling with unit squares to make arrays. (Video)

Lesson 6: Draw rows and columns to determine the area of a rectangle, given an incomplete array. (Video)

Lesson 7: Interpret area models to form rectangular arrays. (Video)

Lesson 9: Analyze different rectangles and reason about their area. (Video)

Lesson 10: Apply the distributive property as a strategy to find the total area of a large rectangle by adding two products. (Video)

Lesson 12: Solve word problems involving area. (Video)

Lesson 13, Lesson 14: Find areas by decomposing into rectangles or completing composite figures to form rectangles. (Video)

Lesson 1: Specify and partition a whole into equal parts, identifying and counting unit fractions using concrete models. (Video)

Lesson 2: Specify and partition a whole into equal parts, identifying and counting unit fractions by folding fraction strips. (Video)

Lesson 3: Specify and partition a whole into equal parts, identifying and counting unit fractions by drawing pictorial area models. (Video)

Lesson 5: Partition a whole into equal parts and define the equal parts to identify the unit fraction numerically. (Video)

Lesson 6: Build non-unit fractions less than one whole from unit fractions. (Video)

Lesson 7: Identify and represent shaded and non-shaded parts of one whole as fractions. (Video)

Lesson 8: Represent parts of one whole as fractions with number bonds. (Video)

Lesson 10: Compare unit fractions by reasoning about their size using fraction strips. (Video)

Lesson 11: Compare unit fractions with different sized models representing the whole. (Video)

Lesson 12: Specify the corresponding whole when presented with one equal part. (Video)

Lesson 13: Identify a shaded fractional part in different ways depending on the designation of the whole. (Video)

Lesson 14: Place unit fractions on a number line with endpoints 0 and 1. (Video)

Lesson 15: Place any fraction on a number line with endpoints 0 and 1. (Video)

Lesson 16: Place whole number fractions and unit fractions between whole numbers on the number line. (Video)

Lesson 17: Practice placing various fractions on the number line. (Video)

Lesson 18: Compare fractions and whole numbers on the number line by reasoning about their distance from 0. (Video)

Lesson 19: Understand distance and position on the number line as strategies for comparing fractions. (Optional) (Video)

Lesson 20: Recognize and show that equivalent fractions have the same size, though not necessarily the same shape. (Video)

Lesson 21: Recognize and show that equivalent fractions refer to the same point on the number line. (Video)

Lesson 22, Lesson 23: Generate simple equivalent fractions by using visual fraction models and the number line. (Videos)

Lesson 24: Express whole numbers as fractions and recognize equivalence with different units. (Video)

Lesson 25: Express whole number fractions on the number line when the unit interval is 1. (Video)

Lesson 26: Decompose whole number fractions greater than 1 using whole number equivalence with various models. (Video)

Lesson 28: Compare fractions with the same numerator pictorially. (Video)

Lesson 29: Compare fractions with the same numerator using <, >, or = and use a model to reason about their size. (Video)

Lesson 1: Generate and organize data. (Video)

Lesson 2: Rotate tape diagrams vertically. (Video)

Lesson 5: Create ruler with 1-inch, 1/2-inch, and 1/4-inch intervals and generate measurement data. (Video)

Lesson 6: Interpret measurement data from various line plots. (Video)

Lesson 7, Lesson 8: Represent measurement data with line plots. (Video)

Lesson 1, Lesson 2: Solve word problems in varied contexts using a letter to represent the unknown. (Video)

Lesson 4: Compare and classify quadrilaterals. (Video)

Lesson 5: Compare and classify other polygons. (Video)

Lesson 6: Draw polygons with specified attributes to solve problems. (Video)

Lesson 7: Reason about composing and decomposing polygons using tetrominoes. (Video)

Lesson 8: Create a tangram puzzle and observe relationships among the shapes. (Video)

Lesson 10: Decompose quadrilaterals to understand perimeter as the boundary of a shape. (Video)

Lesson 11: Tessellate to understand perimeter as the boundary of a shape. (Optional) (Video)

Lesson 12: Measure side lengths in whole number units to determine the perimeter of polygons. (Video)

Lesson 13: Explore perimeter as an attribute of plane figures and solve problems. (Video)

Lesson 14: Determine the perimeter of regular polygons and rectangles when whole number measurements are missing. (Video)

Lesson 15: Solve word problems to determine perimeter with given side lengths. (Video)

Lesson 16: Use string to measure the perimeter of various circles to the nearest quarter inch. (Video)

Lesson 18: Construct rectangles from a given number of unit squares and determine the perimeters. (Video)

Lesson 19: Use a line plot to record the number of rectangles constructed from a given number of unit squares. (Video)

Lesson 20, Lesson 21: Construct rectangles with a given perimeter using unit squares and determine their areas. (Video)

Lesson 23: Solve a variety of word problems with perimeter. (Video)

Lesson 24, Lesson 25, Lesson 26, Lesson 27: Use rectangles to draw a robot with specified perimeter measurements, and reason about the different areas that may be produced. (Video)

Lesson 28, Lesson 29: Solve a variety of word problems involving area and perimeter using all four operations. (Video)

Lesson 31, Lesson 32: Explore and create unconventional representations of one-half. (Video)

Lesson 33: Solidify fluency with Grade 3 skills.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

We welcome your feedback, comments and questions about this site or page. Please submit your feedback or enquiries via our Feedback page.


Majoring/Minoring in Math

Is a math major right for me?

Do you like thinking about mathematics problems? Do you enjoy discussing math with your friends? Are you curious as to how mathematics is used in our ever more complex world? Do you want to study a subject that will enhance your analytical and thinking skills? If you answered yes to any of these questions, you may want to consider majoring in mathematics.

What can I do with a math degree after college?

There are lots of things you can do: Start a career in the actuarial profession, work in the finance industry doing statistical analysis and financial modeling, pursue an MBA or a Masters in Mathematics or Statistics, teach math in the secondary schools, or go for a PhD in math and create the math that other people will use. In short, majoring in mathematics provides a training in rigorous thinking and problem analysis that finds broad application in many career paths. Check the careers page on this site for more information on the wide spectrum of opportunities awaiting today's math major.

What background do I need to major in math?

Math is a sequential subject, so you can't just jump in without a solid base. All math majors must complete the calculus sequence, Math 20100, 20200 and 20300. These courses cover the fundamentals of single variable and multivariable calculus. Although some advanced math courses do not use any calculus, the calculus courses cover many methods in algebra and geometry that are essential in much of higher mathematics.

What does a math major take after calculus?

We offer three options for majoring in math. What you take after calculus depends to some extent on which of these options you pursue. The options are Pure Mathematics (B.A. or B.S.), Secondary School Education (B.A. or B.S.) and Applied Mathematics (B.S.). These are described in detail in the major description.

Although there is overlapping in the requirements for these options, it is best for a student to consult with the Assistant Chair in planning a course of study and completing the documentation needed to register as a major.

Students who are pursuing Childhood Education studies in the School of Education can take a special interdisciplinary concentration in mathematics to fulfill their major requirement. The courses in this program are described in the Math/Childhood Education fact sheet.

I'm interested in majoring in math. Which option should I choose?

If you have a specific interest in public secondary education - and the demand for high school mathematics teachers is insatiable - you should pursue the option in Secondary School Education. In order to obtain certification in New York State you will need to complete some courses in education.

If you are thinking about pursuing an advanced degree in mathematics or simply wish to engage in a rigorous course of study that will prepare you for challenging professional opportunities, you should consider the Pure Mathematics option.

The Applied Mathematics program aims to give students the mathematical tools that are most needed in entry-level careers in the business and research communities. The program emphasizes probability and statistics, mathematical modeling and the use of computers in mathematical and statistical analysis.

What opportunities does the Department provide for majors?

We offer several scholarship programs that provide full tuition for one to two years for outstanding majors. The Department has additional awards, both monetary and academic, that it gives in recognition of student achievements. We also provide a number of paid summer internships for students in business and government.

Majors are also encouraged to apply directly to other institutions that sponsor residential summer research programs funded by the National Science Foundation. The Assistant Chair will assist students in selecting a suitable program.

Math majors who wish to work on campus in a job that will also enhance their academic skills are encouraged to apply as tutors for the Math Help Desk. Applicants should have completed the calculus sequence with a 3.0 average, and at least one of the following courses (Math 34600, Math 39200, or Math 39100) with a grade of B or higher and have a Math GPA (calculus and higher) of at least 2.75.

Can I take graduate math courses as an undergraduate?

Undergraduates who have the appropriate prerequisites may register for A-level graduate courses. These will be credited towards their undergraduate degree. Please check the schedule of graduate courses for the proposed offerings. Undergraduates wishing to take a graduate level course need approval from the Assistant Chair.

If you are on the Pure Mathematics track and will be taking graduate courses, you should consider applying to the Accelerated Master's Degree Program which can enable you to more quickly get a Master's degree after receiving your Bachelor's degree.

Are there opportunities for undergraduates to do research?

The Department awards a degree with honors for students who complete the appropriate course of study. Professor Thea Pignataro is the department Honors coordinator and students interested in the honors program should consult with her by the beginning of their junior year.

Can I minor in math?

Yes, students can obtain a Minor in Mathematics if they complete the following requirements. The calculus sequence (Math 20100, 20200 and 20300) plus a total of twelve credits in math at the City College in 30000-level courses or higher, which includes one of the following: Math 34600: Elements of Linear Algebra or Math 39200: Linear Algebra and Vector Analysis for Engineers. Engineering students can usually complete the Minor by taking two mathematics courses in addition to those required for their degree. Upon completion of the requirements, students must file a Minor declaration form via the Department's Major Advisor.


Mathematics: Applications and Interpretation SL WORKED SOLUTIONS

This book gives you fully worked solutions for every question in Exercises, Review Sets, Activities, and Investigations (which do not involve student experimentation) in each chapter of our textbook Mathematics: Applications and Interpretation SL.

Where applicable, each worked solution is modelled on the relevant worked example in the textbook. Correct answers can sometimes be obtained by different methods.

This product has been developed independently from and is not endorsed by the International Baccalaureate Organization. International Baccalaureate, Baccalaureát International, Bachillerato Internacional and IB are registered trademarks owned by the International Baccalaureate Organization.

Year Published: 2020
Page Count: 701
Online ISBN: 978-1-925489-84-2 (9781925489842)

Mathematics: Applications and Interpretation SL WORKED SOLUTIONS

1 APPROXIMATIONS AND ERROR 5
2 LOANS AND ANNUITIES 25
3 FUNCTIONS 51
4 MODELLING 105
5 BIVARIATE STATISTICS 133
6 QUADRATIC FUNCTIONS 180
7 DIRECT AND INVERSE VARIATION 252
8 EXPONENTIALS AND LOGARITHMS 278
9 TRIGONOMETRIC FUNCTIONS 334
10 DIFFERENTIATION 374
11 PROPERTIES OF CURVES 420
12 APPLICATIONS OF DIFFERENTIATION 468
13 INTEGRATION 505
14 DISCRETE RANDOM VARIABLES 551
15 THE NORMAL DISTRIBUTION 582
16 HYPOTHESIS TESTING 623
17 VORONOI DIAGRAMS 660

Authors

Bradley Steventon

Joseph Small

Ngoc Vo

Ngoc Vo completed a Bachelor of Mathematical Sciences at the University of Adelaide, majoring in Statistics and Applied Mathematics. Her Mathematical interests include regression analysis, Bayesian statistics, and statistical computing. Ngoc has been working at Haese Mathematics as a proof reader and writer since 2016.

What drew you to the field of mathematics?

Originally, I planned to study engineering at university, but after a few weeks I quickly realised that it wasn't for me. So I switched to a mathematics degree at the first available opportunity. I didn't really have a plan to major in statistics, but as I continued my studies I found myself growing more fond of the discipline. The mathematical rigor in proving distributional results and how they link to real-world data -- it all just seemed to click.

What are some interesting things that you get to do at work?

As the resident statistician here at Haese Mathematics, I get the pleasure of writing new statistics chapters and related material. Statistics has always been a challenging subject to both teach and learn, however it doesn't always have to be that way. To bridge that gap, I like to try and include as many historical notes, activities, and investigations as I can to make it as engaging as possible. The reasons why we do things, and the people behind them are often important things we forget to talk about. Statistics, and of course mathematics, doesn't just exist within the pages of your textbook or even the syllabus. There's so much breadth and depth to these disciplines, most of the time we just barely scratch the surface.

What interests you outside mathematics?

In my free time I like studying good typography and brushing up on my TeX skills to become the next TeXpert. On the less technical side of things, I also enjoy scrapbooking, painting, and making the occasional card.


§ 22.1-253.13:3. Standard 3. Accreditation, other standards, assessments, and releases from state regulations.

A. The Board shall promulgate regulations establishing standards for accreditation pursuant to the Administrative Process Act (§ 2.2-4000 et seq.), which shall include (i) student outcome and growth measures, (ii) requirements and guidelines for instructional programs and for the integration of educational technology into such instructional programs, (iii) administrative and instructional staffing levels and positions, including staff positions for supporting educational technology, (iv) student services, (v) auxiliary education programs such as library and media services, (vi) requirements for graduation from high school, (vii) community relations, and (viii) the philosophy, goals, and objectives of public education in the Commonwealth.

The Board shall promulgate regulations establishing standards for accreditation of public virtual schools under the authority of the local school board that enroll students full time.

The Board's regulations establishing standards for accreditation shall ensure that the accreditation process is transparent and based on objective measurements and that any appeal of the accreditation status of a school is heard and decided by the Board.

The Board shall review annually the accreditation status of all schools in the Commonwealth. The Board shall review the accreditation status of a school once every three years if the school has been fully accredited for three consecutive years. Upon such triennial review, the Board shall review the accreditation status of the school for each individual year within that triennial review period. If the Board finds that the school would have been accredited every year of that triennial review period the Board shall accredit the school for another three years. The Board may review the accreditation status of any other school once every two years or once every three years, provided that any school that receives a multiyear accreditation status other than full accreditation shall be covered by a Board-approved multiyear corrective action plan for the duration of the period of accreditation. Such multiyear corrective action plan shall include annual written progress updates to the Board. A multiyear accreditation status shall not relieve any school or division of annual reporting requirements.

Each local school board shall maintain schools that are fully accredited pursuant to the standards for accreditation as prescribed by the Board. Each local school board shall report the accreditation status of all schools in the local school division annually in public session.

The Board shall establish a review process to assist any school that does not meet the standards established by the Board. The relevant school board shall report the results of such review and any annual progress reports in public session and shall implement any actions identified through such review and utilize them for improvement planning.

The Board shall establish a corrective action plan process for any school that does not meet the standards established by the Board. Such process shall require (a) each school board to submit a corrective action plan for any school in the local school division that does not meet the standards established by the Board and (b) any school board that fails to demonstrate progress in developing or implementing any such corrective action plan to enter into a memorandum of understanding with the Board.

When the Board determines through its review process that the failure of schools within a division to meet the standards established by the Board is related to division-level failure to implement the Standards of Quality or other division-level action or inaction, the Board may require a division-level academic review. After the conduct of such review and within the time specified by the Board, each school board shall enter into a memorandum of understanding with the Board and shall subsequently submit to the Board for approval a corrective action plan, consistent with criteria established by the Board setting forth specific actions and a schedule designed to ensure that schools within its school division meet the standards established by the Board. If the Board determines that the proposed corrective action plan is not sufficient to enable all schools within the division to meet the standards established by the Board, the Board may return the plan to the local school board with directions to submit an amended plan pursuant to Board guidance. Such corrective action plans shall be part of the relevant school division's comprehensive plan pursuant to § 22.1-253.13:6.

B. The Superintendent of Public Instruction shall develop, subject to revision by the Board, criteria for determining and recognizing educational performance in the Commonwealth's local school divisions and public schools. The portion of such criteria that measures individual student growth shall become an integral part of the accreditation process for schools in which any grade level in the grade three through eight range is taught. The Superintendent of Public Instruction shall annually report to the Board on the accreditation status of all school divisions and schools. Such report shall include an analysis of the strengths and weaknesses of public education programs in the various school divisions in Virginia and recommendations to the General Assembly for further enhancing student learning uniformly across the Commonwealth. In recognizing educational performance and individual student growth in the school divisions, the Board shall include consideration of special school division accomplishments, such as numbers of dual enrollments and students in Advanced Placement and International Baccalaureate courses, and participation in academic year Governor's Schools.

The Superintendent of Public Instruction shall assist local school boards in the implementation of action plans for increasing educational performance and individual student growth in those school divisions and schools that are identified as not meeting the approved criteria. The Superintendent of Public Instruction shall monitor the implementation of and report to the Board on the effectiveness of the corrective actions taken to improve the educational performance in such school divisions and schools.

C. With such funds as are available for this purpose, the Board shall prescribe assessment methods to determine the level of achievement of the Standards of Learning objectives by all students. Such assessments shall evaluate knowledge, application of knowledge, critical thinking, and skills related to the Standards of Learning being assessed. The Board shall, with the assistance of independent testing experts, conduct a regular analysis and validation process for these assessments. In lieu of a one-time end-of-year assessment, the Board shall establish, for the purpose of providing measures of individual student growth over the course of the school year, a through-year growth assessment system, aligned with the Standards of Learning, for the administration of reading and mathematics assessments in grades three through eight. Such through-year growth assessment system shall include at least one beginning-of-year, one mid-year, and one end-of-year assessment in order to provide individual student growth scores over the course of the school year, but the total time scheduled for taking all such assessments shall not exceed 150 percent of the time scheduled for taking a single end-of-year proficiency assessment. The Department shall ensure adequate training for teachers and principals on how to interpret and use student growth data from such assessments to improve reading and mathematics instruction in grades three through eight throughout the school year. With such funds and content as are available for such purpose, such through-year growth assessment system shall provide accurate measurement of a student's performance, through computer adaptive technology, using test items at, below, and above the student's grade level as necessary.

The Board shall also provide the option of industry certification and state licensure examinations as a student-selected credit.

The Department shall make available to school divisions Standards of Learning assessments typically administered by high schools by December 1 of the school year in which such assessments are to be administered or when newly developed assessments are available, whichever is later.

The Board shall make publicly available such assessments in a timely manner and as soon as practicable following the administration of such tests, so long as the release of such assessments does not compromise test security or deplete the bank of assessment questions necessary to construct subsequent tests, or limit the ability to test students on demand and provide immediate results in the web-based assessment system.

The Board shall prescribe alternative methods of Standards of Learning assessment administration for children with disabilities, as that term is defined in § 22.1-213, who meet criteria established by the Board to demonstrate achievement of the Standards of Learning. An eligible student's Individual Education Program team shall make the final determination as to whether an alternative method of administration is appropriate for the student.

The Board shall include in the student outcome and growth measures that are required by the standards of accreditation the required assessments for various grade levels and classes, including the completion of the alternative assessments implemented by each local school board, in accordance with the Standards of Learning. These assessments shall include end-of-course or end-of-grade tests for English, mathematics, science, and history and social science and may be integrated to include multiple subject areas.

The Standards of Learning assessments administered to students in grades three through eight shall not exceed (i) reading and mathematics in grades three and four (ii) reading, mathematics, and science in grade five (iii) reading and mathematics in grades six and seven (iv) reading, writing, and mathematics in grade eight (v) science after the student receives instruction in the grade six science, life science, and physical science Standards of Learning and before the student completes grade eight and (vi) Virginia Studies and Civics and Economics once each at the grade levels deemed appropriate by each local school board. The reading and mathematics assessments administered to students in grades three through eight shall be through-year growth assessments.

Each school board shall annually certify that it has provided instruction and administered an alternative assessment, consistent with Board guidelines, to students in grades three through eight in each Standards of Learning subject area in which a Standards of Learning assessment was not administered during the school year. Such guidelines shall (a) incorporate options for age-appropriate, authentic performance assessments and portfolios with rubrics and other methodologies designed to ensure that students are making adequate academic progress in the subject area and that the Standards of Learning content is being taught (b) permit and encourage integrated assessments that include multiple subject areas and (c) emphasize collaboration between teachers to administer and substantiate the assessments and the professional development of teachers to enable them to make the best use of alternative assessments.

Local school divisions shall provide targeted mathematics remediation and intervention to students in grades six through eight who show computational deficiencies as demonstrated by their individual performance on any diagnostic test or grade-level Standards of Learning mathematics test that measures non-calculator computational skills.

The Department shall award recovery credit to any student in grades three through eight who performs below grade level on a Standards of Learning assessment in English reading or mathematics, receives remediation, and subsequently retakes and performs at or above grade level on such an assessment, including any such student who subsequently retakes such an assessment on an expedited basis.

In addition, to assess the educational progress of students, the Board shall (1) develop appropriate assessments, which may include criterion-referenced tests and other assessment instruments that may be used by classroom teachers (2) select appropriate industry certification and state licensure examinations and (3) prescribe and provide measures, which may include nationally normed tests to be used to identify students who score in the bottom quartile at selected grade levels. An annual justification that includes evidence that the student meets the participation criteria defined by the Department shall be provided for each student considered for the Virginia Grade Level Alternative. Each Individual Education Program team shall review such justification and make the final determination as to whether or not the Virginia Grade Level Alternative is appropriate for the student. The superintendent and the school board chairman shall certify to the Board, as a part of certifying compliance with the Standards of Quality, that there is a justification in the Individual Education Program for every student who takes the Virginia Grade Level Alternative. Compliance with this requirement shall be monitored as a part of the special education monitoring process conducted by the Department. The Board shall report to the Governor and General Assembly in its annual reports pursuant to § 22.1-18 any school division that is not in compliance with this requirement.

The Standards of Learning requirements, including all related assessments, shall be waived for any student awarded a scholarship under the Brown v. Board of Education Scholarship Program, pursuant to § 30-231.2, who is enrolled in a preparation program for a high school equivalency examination approved by the Board or in an adult basic education program or an adult secondary education program to obtain the high school diploma or a high school equivalency certificate.

The Department shall develop processes for informing school divisions of changes in the Standards of Learning.

The Board may adopt special provisions related to the administration and use of any Standards of Learning test or tests in a content area as applied to accreditation ratings for any period during which the Standards of Learning content or assessments in that area are being revised and phased in. Prior to statewide administration of such tests, the Board shall provide notice to local school boards regarding such special provisions.

The Board shall not include in its calculation of the passage rate for a Standards of Learning assessment or the level of achievement of the Standards of Learning objectives for an individual student growth assessment for the purposes of state accountability any student whose parent has decided to not have his child take such Standards of Learning assessment, unless such exclusions would result in the school's not meeting any required state or federal participation rate.

D. The Board may pursue all available civil remedies pursuant to § 22.1-19.1 or administrative action pursuant to § 22.1-292.1 for breaches in test security and unauthorized alteration of test materials or test results.

The Board may initiate or cause to be initiated a review or investigation of any alleged breach in security, unauthorized alteration, or improper administration of tests, including the exclusion of students from testing who are required to be assessed, by local school board employees responsible for the distribution or administration of the tests.

Records and other information furnished to or prepared by the Board during the conduct of a review or investigation may be withheld pursuant to subdivision 10 of § 2.2-3705.3. However, this section shall not prohibit the disclosure of records to (i) a local school board or division superintendent for the purpose of permitting such board or superintendent to consider or to take personnel action with regard to an employee or (ii) any requester, after the conclusion of a review or investigation, in a form that (a) does not reveal the identity of any person making a complaint or supplying information to the Board on a confidential basis and (b) does not compromise the security of any test mandated by the Board. Any local school board or division superintendent receiving such records or other information shall, upon taking personnel action against a relevant employee, place copies of such records or information relating to the specific employee in such person's personnel file.

Notwithstanding any other provision of state law, no test or examination authorized by this section, including the Standards of Learning assessments, shall be released or required to be released as minimum competency tests, if, in the judgment of the Board, such release would breach the security of such test or examination or deplete the bank of questions necessary to construct future secure tests.

E. With such funds as may be appropriated, the Board may provide, through an agreement with vendors having the technical capacity and expertise to provide computerized tests and assessments, and test construction, analysis, and security, for (i) web-based computerized tests and assessments, including computer-adaptive Standards of Learning assessments, for the evaluation of student progress during and after remediation and (ii) the development of a remediation item bank directly related to the Standards of Learning.

F. To assess the educational progress of students as individuals and as groups, each local school board shall require the use of Standards of Learning assessments, alternative assessments, and other relevant data, such as industry certification and state licensure examinations, to evaluate student progress and to determine educational performance. Each local school shall require the administration of appropriate assessments to students, which may include criterion-referenced tests and teacher-made tests and shall include the Standards of Learning assessments, the local school board's alternative assessments, and the National Assessment of Educational Progress state-by-state assessment. Each school board shall analyze and report annually, in compliance with any criteria that may be established by the Board, the results from the Stanford Achievement Test Series, Ninth Edition (Stanford Nine) assessment, if administered, industry certification examinations, and the Standards of Learning Assessments to the public.

The Board shall not require administration of the Stanford Achievement Test Series, Ninth Edition (Stanford Nine) assessment, except as may be selected to facilitate compliance with the requirements for home instruction pursuant to § 22.1-254.1.

The Board shall include requirements for the reporting of the Standards of Learning assessment data, regardless of accreditation frequency, as part of the Board's requirements relating to the School Performance Report Card. Such scores shall be disaggregated for each school by student subgroups on the Virginia assessment program as appropriate and shall be reported to the public within three months of their receipt. These reports (i) shall be posted on the portion of the Department's website relating to the School Performance Report Card, in a format and in a manner that allows year-to-year comparisons, and (ii) may include the National Assessment of Educational Progress state-by-state assessment.

G. Each local school division superintendent shall regularly review the division's submission of data and reports required by state and federal law and regulations to ensure that all information is accurate and submitted in a timely fashion. The Superintendent of Public Instruction shall provide a list of the required reports and data to division superintendents annually. The status of compliance with this requirement shall be included in the Board's annual report to the Governor and the General Assembly as required by § 22.1-18.

H. Any school board may request the Board for release from state regulations or, on behalf of one or more of its schools, for approval of an Individual School Accreditation Plan for the evaluation of the performance of one or more of its schools as authorized for certain other schools by the Standards for Accreditation pursuant to 8VAC20-131-280 C of the Virginia Administrative Code. Waivers of regulatory requirements may be granted by the Board based on submission of a request from the division superintendent and chairman of the local school board. The Board may grant, for a period up to five years, a waiver of regulatory requirements that are not (i) mandated by state or federal law or (ii) designed to promote health or safety. The school board shall provide in its waiver request a description of how the releases from state regulations are designed to increase the quality of instruction and improve the achievement of students in the affected school or schools. The Department shall provide (a) guidance to any local school division that requests releases from state regulations and (b) information about opportunities to form partnerships with other agencies or entities to any local school division in which the school or schools granted releases from state regulations have demonstrated improvement in the quality of instruction and the achievement of students.

The Board may also grant local school boards waivers of specific requirements in § 22.1-253.13:2, based on submission of a request from the division superintendent and chairman of the local school board, permitting the local school board to assign instructional personnel to the schools with the greatest needs, so long as the school division employs a sufficient number of personnel divisionwide to meet the total number required by § 22.1-253.13:2 and all pupil/teacher ratios and class size maximums set forth in subsection C of § 22.1-253.13:2 are met. The school board shall provide in its request a description of how the waivers from specific Standards of Quality staffing standards are designed to increase the quality of instruction and improve the achievement of students in the affected school or schools. The waivers may be renewed in up to five-year increments, or revoked, based on student achievement results in the affected school or schools.

1988, cc. 645, 682 1990, cc. 820, 839 1992, c. 591 1998, cc. 456, 567, 602, 627, 843, 902 1999, cc. 670, 731, 1015 2000, cc. 504, 735, 742, 750, 752, 867, 1061 2001, cc. 651, 731 2002, cc. 101, 167, 656, 732 2003, cc. 691, 1004 2004, cc. 472, 939, 955, 965 2005, cc. 331, 450, 753, 834 2006, cc. 25, 38, 95, 117, 131 2007, c. 234 2009, c. 825 2010, c. 76 2011, cc. 248, 666 2012, c. 183 2013, cc. 539, 571, 584, 728 2014, cc. 84, 585, 622 2015, cc. 145, 149, 322, 323, 558, 566 2016, cc. 386, 387, 502, 522, 720, 750 2017, cc. 328, 778 2019, c. 585 2021, Sp. Sess. I, cc. 443, 444.

The chapters of the acts of assembly referenced in the historical citation at the end of this section may not constitute a comprehensive list of such chapters and may exclude chapters whose provisions have expired.


13.3 Shifting Equilibria: Le Châtelier’s Principle

A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Châtelier’s principle : if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.

Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.

Effect of a Change in Concentration

If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction

The rate laws for the forward and reverse reactions are

When this system is at equilibrium, the forward and reverse reaction rates are equal.

If the system is stressed by adding reactant, either H2 or I2, the resulting increase in concentration causes the rate of the forward reaction to increase, exceeding that of the reverse reaction:

The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.

The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction

and a temporary net reaction in the reverse direction to re-establish equilibrium.

As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,

If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the new equilibrium mixture will have greater concentrations of H2 and I2 and a lesser concentration of HI. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left.

For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M,

and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.

At equilibrium, the reaction H 2 ( g ) + I 2 ( g ) ⇌ 2 HI ( g ) H 2 ( g ) + I 2 ( g ) ⇌ 2 HI ( g ) is described by the reaction quotient

If the volume occupied by an equilibrium mixture of these species is decreased by a factor of 3, the partial pressures of all three species will be increased by a factor of 3:

And so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.

A similar treatment of a different system, 2 NO 2 ( g ) ⇌ 2 NO ( g ) + O 2 ( g ) , 2 NO 2 ( g ) ⇌ 2 NO ( g ) + O 2 ( g ) , however, yields a different result:

In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.

These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.

Link to Learning

Check out this link to see a dramatic visual demonstration of how equilibrium changes with pressure changes.

Chemistry in Everyday Life

Equilibrium and Soft Drinks

The connection between chemistry and carbonated soft drinks goes back to 1767, when Joseph Priestley (1733–1804) developed a method of infusing water with carbon dioxide to make carbonated water. Priestly’s approach involved production of carbon dioxidey reacting oil of vitriol (sulfuric acid) with chalk (calcium carbonate).

The carbon dioxide was then dissolved in water, reacting to produce hydrogen carbonate, a weak acid that subsequently ionized to yield bicarbonate and hydrogen ions:

These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.

When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria. Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”

Effect of a Change in Temperature

Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to re-establish equilibrium without any change in the value of the equilibrium constant, K. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition that exhibits a different value for the equilibrium constant.

To understand this phenomenon, consider the elementary reaction

Since this is an elementary reaction, the rates laws for the forward and reverse may be derived directly from the balanced equation’s stoichiometry:

When the system is at equilibrium,

Substituting the rate laws into this equality and rearranging gives

The equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.

Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process:

For purposes of applying Le Chatelier’s principle, heat (q) may be viewed as a reactant:

Raising the temperature of the system is akin to increasing the amount of a reactant, and so the equilibrium will shift to the right. Lowering the system temperature will likewise cause the equilibrium to shift left. For exothermic processes, heat is viewed as a product of the reaction and so the opposite temperature dependence is observed.

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    13.3: New Page - Mathematics

    You’ve figured out the solution to the problem—fantastic! But you’re not finished. Whether you are writing solutions for a competition, a journal, a message board, or just to show off for your friends, you must master the art of communicating your solution clearly. Brilliant ideas and innovative solutions to problems are pretty worthless if you can’t communicate them. In this article, we explore many aspects of how to write a clear solution. Below is an index each page of the article includes a sample ‘How Not To’ solution and ‘How To’ solution. One common theme you’ll find throughout each point is that every time you make an experienced reader have to think to follow your solution, you lose.

    As you read the ‘How To’ solutions, you may think some of them are overwritten. Indeed, some of them could be condensed. Some steps we chose to prove could probably be cited without proof. However, it is far better to prove too much too clearly than to prove too little. Rarely will a reader complain that a solution is too easy to understand or too easy on the eye.

    One note of warning: many of the problems we use for examples are extremely challenging problems. Beginners, and even intermediate students, should not be upset if they have difficulty solving the problems on their own.

    Table of Contents:

    Have a Plan

    Your goal in writing a clear solution is to prevent the reader from having to think. You must express your ideas clearly and concisely. The experienced reader should never have to wonder where you are headed, or why any claim you make is true. The first step in writing a clear solution is having a plan. Make a simple outline of your solution. Include the items you’ll need to define, and the order in which you will write up the important parts of your solution. The outline will help ensure that you don’t skip anything and that you put your steps in an order that’s easy to follow.

    A sphere of radius (r) is inscribed in a tetrahedron. Planes tangent to this sphere and parallel to the faces of the tetrahedron cut off four small tetrahedra from the tetrahedron these small tetrahedra have inscribed spheres with radii (a), (b), (c), (d). Show that:

    Here’s a solution that looks short but is pretty tough to read:

    How Not to Write the Solution:

    Let our tetrahedron be (ABCD). The small tetrahedron which includes vertex A is similar to the big tetrahedron. Since the face of this tetrahedron parallel to face (BCD) is tangent to the sphere inscribed in (ABCD), the distance between (BCD) and this parallel face of the small tetrahedron is (2r). Let’s call that small tetrahedron (AXYZ). Hence, the altitude from (A) in (AXYZ) is (h_a – 2r), where (h_a) is the length of the altitude from (A) to side (BCD). Therefore the ratio of the altitudes from (A) in (AXYZ) and (ABCD) is ((h_a- 2r)/h_a). Since these two tetrahedrons are similar with ratio (a/r) (since that’s the ratio of the corresponding lengths, namely the radii of the inscribed spheres) we have (a/r =) ((h_a- 2r)/h_a).

    The volume of the tetrahedron is ([A]h_a/3), where ([A]) is the area of triangle (BCD). The volume of the tetrahedron can also be written (rS/3), where (S) is the surface area of (ABCD). We can prove that by letting (I) be the center of the inscribed sphere. Then the volume of the tetrahedron is the sum of the volumes of the tetrahedra (IABC), (IABD), (IBCD), and (IACD). The volume of (IABC) is (r[D]/3), where ([D]) is defined like we defined ([A]) above. We can similarly find the volumes of the other 4 pieces. When we add them all up, we get
    [ ext ABCD = ([A] + [B] + [C] + [D]) r/3 = rS/3.]

    We set that equal to our other volume expression and get (h_a= rS/[A]). If we rearrange our equation from above, we have (a = r – 2r^2/ h_a). We can then put in the (h_a) expression we just found to get:

    If we define ([B]), ([C]), and ([D]) just like we defined ([A]), we can use the same argument to get:

    Adding these and our expression for (A), we get:

    $a + b + c + d = 4r – 2rcdot([A]+[B]+[C]+[D])/S = 2r,$

    The main problem with the above solution is one of organization. We defined variables after they popped up. Midway through the solution we sidetracked to prove the volume of ABCD is (rS/3). Sometimes we wrote important equations right in our paragraphs instead of highlighting them by giving them their own lines.

    If we outline before writing the solution, we won’t have these problems. We can list what we need to define, decide what items we need to prove before our main proof (we call these lemmas), and list the important steps so we know what to highlight.

    Our scratch sheet with the outline might have the following:

    Stuff to define: (ABCD, h_a, S, [A], AXYZ).

    1. Volume (ABCD = rS/3) (lemma)
    2. Show altitude (AXYZ = h_a – 2r)
    3. Use similarity to get (a = r – 2r^2/h_a)
    4. Equate volumes to get (1/h_a= A/(rS))
    5. sub (4) into (3) and add

    This list looks obvious once you have it written up, but if you just plow ahead with the solution without planning, you may end up skipping items and having to wedge them in as we did in our ‘How Not to Write the Solution’.

    How to Write the Solution:

    Let our original tetrahedron be (ABCD). We define:

    ([A]) = the area of the face of (ABCD) opposite (A)
    (h_a) = the length of the altitude from (A) to (BCD)
    (S) = the surface area of (ABCD)
    (AXYZ) = one of the small tetrahedrons formed as described

    Define ([B]), ([C]), ([D]) and (h_b), (h_c), (h_d) similarly.

    Since face (XYZ) of small tetrahedron (AXYZ) is parallel to face (BCD), tetrahedron (AXYZ) is similar to (ABCD). The ratio of corresponding lengths in these tetrahedra equals the ratio of the radii of their inscribed spheres, or (a/r).

    Since (XYZ) is tangent to the sphere inscribed in (ABCD), the distance between (BCD) and (XYZ) is (2r). Hence, the altitude from (A) to (XYZ) is (h_a – 2r). Therefore the ratio of the altitudes from (A) in the two tetrahedra is ((h_a – 2r)/h_a). Hence,

    (a/r = (h_a – 2r)/h_a), or (a = r – 2r^2/h_a).

    The volume of the tetrahedron is (h_a[A]/3). Setting this equal to the expression from Lemma 1 yields:

    and substituting this into equation (1), we get:

    By the same argument, we have:

    Adding these and our expression for (A), we get:

    $a + b + c + d = 4r – 2rcdot ([A]+[B]+[C]+[D])/S = 2r,$

    Readers Are Not Interpreters

    The first thing a reader sees on your paper isn’t the structure of your solution. It isn’t the answer, it isn’t the words you choose. It’s how the solution sits on the paper. If the reader has to decipher scrawl, you’re going to lose him. Ideally, you’ll typeset your solution with a program like LaTeX. However, in most contests you don’t have the luxury of turning to a computer and you’ll have to write it out by hand. There are few very important rules of thumb when writing a solution by hand. Many are obvious, some are less so. You should follow them all.

    1. Use blank paper. Don’t use graph paper or lined paper – the lines often make solutions harder to read. Never use paper that is torn out of a spiral notebook.
    2. Respect margins. If you are starting with a completely blank piece of paper, draw the margins on all four sides (top, bottom, right, left). Make your margins at least 0.5 inches, and preferably a full inch.
    3. Write horizontally never turn your writing when you reach the end of a line in order to jam in a little more information. You can always start a new line or a new page.
    4. Leave space at the top for a ‘Page _ of _’ so the reader knows how many pages there are, and what page she’s on. You probably won’t know how many pages you’ll write when you start, but you can fill these out when you’re finished. If you get to the bottom of a page and your solution must continue on another page, write ‘Continued’ at the bottom of that page so the reader knows we’re not finished. (This also helps readers know if they’re missing pages.)
    5. Don’t write in cursive. Print. And print clearly.
    6. Use pen. If you must use pencil, do not erase – the smudges from erasers make a mess.
    7. When you make a mistake you’d like to omit, draw a single line through it and move on. If it’s a large block to omit, draw an ‘X’ through it and move on. Don’t scribble out large blocks of text.
    8. If you left something out and want to add it at the end, put a simple symbol, like a (*), at the point where you would like the new text to be considered added, and leave a brief note, such as ‘Proof below.’ Below, you can write ‘(*) Addendum:’ and proceed with the proof. Don’t use a bunch of arrows to direct the reader all over the page.

    Problem: Let (S(n)) be the sum of the digits of (n). Find

    Below are two solutions. Neither solution is picture-perfect when you’re under time pressure, it’s hard to write perfect-looking proofs. You should find the second one much more enjoyable to read. When you’re writing solutions, keep the above tips in mind, and just remember, ‘If they can’t read it, it’s not right.’

    How Not to Write the Solution:

    That solution above is a mess. The one below took me just as long to write, and is much easier to read.

    How to Write the Solution:

    U s e S p a c e

    When you write your solution you should:

    1. Give each important definition or equation its own line.
    2. Don’t bury too much algebra in a paragraph. You can write line after line of algebra, but put each step on its own line. Don’t cram the algebra in a paragraph.
    3. Label equations or formulas or lemmas or cases you will use later very clearly.
    4. Remember that there’s always more paper.

    Problem: Let (p(x)) be a polynomial with degree (98) such that (p(n) = 1/n) for (n =) (1, 2, 3,) (4, ldots , 99). Determine (p(100)).

    Have fun reading this solution:

    How Not to Write the Solution:

    Let (r(x) =) (x (p(x) – 1/x)) (= x p(x) – 1.) Since (p(x)) is a polynomial with degree (98), (r(x)) is a polynomial with degree (99). Since (r(x) =) (x (p(x) – 1/x),) and we are given that ((p(x) – 1/x) = 0) for (x = 1,) (2, 3, ldots , 99), (r(x)) has roots (1, 2, ldots, 99.) Since (r(x)) has degree (99), these are the only roots of (r(x)), which must thus have the form (r(x) =) (c(x – 1)(x – 2)) ((x – 3) cdots (x – 99)) for some constant (c). To find (c,) we first let (x = 0) in equation (r(x) =) (x p(x) – 1,) yielding (r(0) = -1.) Letting (x = 0) in (r(x) =) (c(x – 1)(x – 2)) ((x – 3) cdots (x – 99)) yields (r(0) = -c(99!)) hence, (c = 1/99!.) Thus, we have (r(x) =) ((x – 1)(x – 2)) ((x – 3) cdots (x – 99)/99!.) We can combine the equations (r(x) =) (x p(x) – 1) and (r(x) =) ((x – 1)(x – 2)) ((x – 3) cdots (x – 99)/99!) and let (x = 100) to find (100p(100) – 1 =) ( (100 – 1)(100 – 2)) ((100 – 3) ldots(100 – 99)/99!,) so (100p(100) – 1 =) ( 99!/99! = 1,) so (p(100) = 1/50.)

    Here’s the same solution, with nearly the same wording.

    How to Write the Solution:

    Since (p(x)) is a polynomial with degree (98), (r(x)) is a polynomial with degree (99). Since (r(x) =) (x (p(x) – 1/x)), and we are given that ((p(x) – 1/x) = 0) for (x = ) (1, 2,) (3, ldots, 99,)

    (r(x)) has roots (1, 2, ldots , 99.)

    Since (r(x)) has degree (99), these are the only roots of (r(x)), which must thus have:

    for some constant (c). To find (c), we first let (x = 0) in equation ((1)), yielding (r(0) = -1.) Letting (x = 0) in ((2)) yields (r(0) = -c(99!)) hence, (c = 1/99!). Thus, we have:

    We can combine equations ((1)) and ((3)) and let (x = 100) to find

    [100p(100) – 1 = (100 – 1)(100 – 2)(100 – 3) cdots (100 – 99)/99!][100p(100) – 1 = 99!/99! = 1][p(100) = 1/50.]

    Which would you rather read?

    SdrawkcaB knihT, Write Forwards

    The following is an excerpt from a cookbook that was never written:

    “Figuring out how to make an omelette is easy. Anybody who has eaten an omelette knows that an omelette is typically made with several eggs filled with various foods such as ham, peppers, onions, and bacon and is often cooked with cheese. The fact that all these ingredients end up inside the egg means that we should begin cooking the eggs flatly on a pan and then add the ingredients. We can then roll part of the egg over the ingredients so as to trap them on the inside. If we needed some of the ingredients precooked we could do that before adding them to the eggs…”

    It is one thing to figure out how to make an omelette. It is another to explain to somebody else how to make one. Starting our explanation from the beginning is much clearer than starting with the finished omelette.

    “Prepare vegetables and other desired omelette fillings. Beat eggs. Start cooking the eggs. Add your fillings in the middle so that part of the egg can be pulled over the ingredients. When the omelette is closed, continue to cook and flip the omelette until the eggs look well-cooked.”

    The reader doesn’t care how the process of cooking an omelette was unraveled by the author. The reader just wants to know how to make an omelette.

    Think of solutions as recipes. Start at the beginning and move forward. List the ingredients and explain how and when to add them to the pot.

    Problem: Let (a), (b), and (c) denote the lengths of the sides of a triangle. Show that

    $a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c)leq 3abc.$

    This solution might be a good way to see how we might come up with a solution from scratch, but it’s not a particularly well-written proof:

    How Not to Write the Solution:

    We note that the inequality contains the factors ((-a + b + c)), ((a – b + c)), and ((a + b – c)). These factors point to using the triangle inequality so it seems natural to leave the factors alone and invoke the fact that each is nonnegative.

    Since each of these three factors is multiplied by the square of the length of a side it might be possible to manipulate the inequality into something involving these nonnegative triangle inequality factors multiplied by perfect squares. We could then argue that this sum must also be nonnegative. We begin by moving (3abc) to the left hand side:

    $a^2(-a + b + c) = b^2(a – b + c) + c^2 (a + b – c) – 3abc le 0.$

    If we were to view (3abc) as the sum of (3) terms that are each the product of (ab), (bc), or (ca) and one of the triangle inequality factors, we begin to get an idea as to how the inequality can be reorganized. Since the inequality is cyclic, it seems natural to take these products in a way that preserves the cyclic nature. For instance, we multiply (ab) with ((a + b – c)) because (a) and (b) have the same sign in ((a + b – c)):

    egin
    ab( a + b – c) &= a^2b +ab^2 -abc ge 0,
    ab (-a + b +c) &= -abc + b^2c + bc^2 ge 0,
    ca (a – b + c) &=a^2c -abc + ac^2 ge 0.
    end

    We see the (-3abc) in the sum of these products. Examining the other terms in

    we notice that (a^2b + ab^2) are factors that would pop out of ((a – b)^2(a + b – c) ). Expanding the squared parts of expressions like ((a-b)^2(a+b-c)) we get

    egin
    (a – b)^2(a + b – c) &= a^2(a + b – c) – 2ab(a + b – c) + b^2(a + b – c) ge 0,
    (b – c)^2(-a + b + c) &= b^2(-a + b + c) – 2bc(-a + b + c) + c^2(-a + b + c) ge 0,
    (c – a)^2(a – b + c) &= c^2(a – b + c) – 2ca(a – b + c) + a^2(a – b + c) ge 0.end

    Adding these inequalities together we begin to see the inequality take shape:

    egin
    &a^2(a + b – c + a – b + c) + b^2(a + b – c – a + b + c) + c^2(-a + b + c + a – b + c)
    &- 2a^2b – 2ab^2 + 2abc + 2abc – 2b^2c – 2bc^2 – 2a^2c + 2abc – 2ac^2
    &= a^2(2a – 2b – 2c) + b^2(-2a + 2b – 2c) + c^2(-2a – 2b + 2c) + 6abc ge 0.
    end

    Multiplying this inequality by (-1/2) reverses the inequality sign and gives us

    $a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c) – 3abc ge 0.$

    Adding (3abc) to both sides gives us

    $a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c) ge 3abc$

    The cookbook style is easier to read and far more convincing:

    How to Write the Solution:

    According to the triangle inequality, the sum of any two sides of a triangle is at least as great as the length of the third side. Thus, we have the inequalities

    Multiplying these by perfect squares leave each left-hand side nonpositive, so

    Adding these inequalities we get

    egin
    &a^2(a + b – c + a – b + c) + b^2(a + b – c – a + b + c) + c^2(-a + b + c + a – b + c)
    &- 2a^2b – 2ab^2 + 2abc + 2abc – 2b^2c – 2bc^2 – 2a^2c + 2abc – 2ac^2
    &= a^2(2a – 2b – 2c) + b^2(-2a + 2b – 2c) + c^2(-2a – 2b + 2c) + 6abc ge 0.
    end

    Adding (6abc) to both sides and dividing by (2) we have the desired

    $a^2(-a + b + c) + b^2(a – b + c) + c^2(a + b – c)leq3abc.$

    Name Your Characters

    A large thin-shelled vehicle for a young fowl that was created by a huge female bird sat on a wall. The large thin-shelled vehicle for a young fowl that was created by a huge female bird had a great fall. All the horses of the great man who lived in a large castle that ruled over the people in the land and all the men of the great man who lived in a large castle that ruled over the people in the land couldn’t put the large thin-shelled vehicle for a young fowl that was created by a huge female bird back together again.

    Proofs are a lot like stories. When writing a solution your job is tell a math story in a way your audience will understand and enjoy. Instead of writing about ‘A large thin-shelled vehicle for a young fowl that was created by a huge female bird,’ we call that big egg ‘Humpty-Dumpty’ and tell the story. Likewise, a well-written proof often involves naming the important quantities or ideas that play a part in the story of your solution. Naming your characters can also help you find solutions to problems, so it’s not something you should wait until proof-writing time to do.

    When you do name your characters, you name them simply, clearly, and write up front, so the reader knows exactly where to go to find out exactly who this (n) person is and what that (f(x)) function stands for.

    The solution below is hard to read because the integers and the sums that are the key to the solution remain unnamed.

    How Not to Write the Solution:

    Suppose we put the numbers in our set in some fixed order. If we start from the beginning, there are (100) sums we can make by just adding up starting from the beginning number. We could add up the first (2) numbers, or the first (4), the first (57), or whatever. If one of these sums is a multiple of (100), then we are done. If none of the sums is a multiple of (100), then we need to consider the remainders when each of these terms is divided by (100). There are (100) total remainders since there are (100) sums. Since none of them is (0), there are at most (99) different remainders among these (100) remainders. Therefore, two of these remainders must be the same since if there weren’t at least two the same there could only be (99) total remainders (since we know none is zero). Now, take the difference between the two sums which have the same remainder when divided by (100). This difference must have a remainder of (0) when divided by (100). Suppose we have subtracted the sum with fewer numbers from the one with more numbers. When we take this difference, all the numbers in the second sum cancel with numbers in the first sum, because each sum is just adding up numbers in our set starting with the first one but the second sum is shorter. Due to this cancellation, the difference of these two sums which have the same remainder when divided by (100) results in a sum of numbers in the original set. We have shown that this difference has a remainder of (0) when divided by (100), so this is our desired sum of numbers in the set that are divisible by (100).

    The solution below is easy to read because the main characters have names. Specifically, we name the integers in the set and the sums of the elements in subsets that we examine. These names allow us to follow the characters throughout the story. They also allow the writer to describe the characters more completely and succinctly.

    How to Write the Solution:

    Call the 100 integers (n_1,) (n_2,) (ldots,) (n_<100>).

    Let (S_k = n_1 + n_2 + dots + n_k) for (k = 1,) (2, dots, 100.)

    Case 1: If (S_1, S_2,dots, S_<100>) are all distinct (mod 100), then exactly one of them must be a multiple of 100.

    Case 2: Otherwise, the 100 sums, (S_k), have at most 99 distinct residues (mod 100) and by the Pigeonhole Principle two of the sums, (S_k), have the same residue (mod 100).

    Thus means there exist some integers (j) and (k), (0 < j < k < 101), such that

    Now, consider the subset with elements (n_,) (n_,) (ldots,) (n_k). The sum of the elements of this subset is

    egin
    & n_ + n_ + dots + n_k
    & = (n_1 + n_2 + dots + n_k) – (n_1 + n_2 + dots + n_j)
    S_k& equiv 0 pmod<100>.
    end

    Thus this sum is a multiple of 100 and we are done.

    A Picture is Worth a Thousand Words

    When you’re writing a solution to a geometry problem, or any problem involving a picture, you should include the diagram. If you don’t include the diagram, you often make the grader have to draw it for you. Even if the diagram is given in the problem, you should include it in your solution. If you make your reader go looking somewhere else for a diagram, you are very likely to lose their attention.

    Draw your diagram precisely. Use a geometry rendering program if you are typesetting your solution, or use a ruler and compass if you are writing your solution by hand.

    Here’s a solution without a diagram:

    How to Write the Solution:

    Let our circles be (O) and (O’). Let (M) and (M’) be on (O) and (O’) such that (MM’) is the common tangent through (K). Let (L) and (L’) be the points where (AD) meets circles (O) and (O’), respectively. Let (N) and (N’) be the points where (BC) meets circles (O) and (O’), respectively.

    We will show that (AK = ) (<(AB + AC – BC)>/2), and thus show that the length of (AK) is independent of (D).

    Since tangents from a point to a circle are equal, we have both (DN = DL) and (DN’ = DL’). Thus,

    Since (MM’ = NN’) by symmetry, we conclude that (KL = DL). Hence,

    We can compute (NN’), and hence (KD), in terms of (AD) and the sides of the triangle:

    as desired. Since (A), (B), and (C) are independent of (D), we conclude that the length (AK) is independent of (D).

    Here’s a solution that includes the diagram:

    How to Write the Solution:

    Let our circles be (O) and (O’). Let (M) and (M’) be on (O) and (O’) such that (MM’) is the common tangent through (K). Let (L) and (L’) be the points where (AD) meets circles (O) and (O’), respectively. Let (N) and (N’) be the points where (BC) meets circles (O) and (O’), respectively.

    We will show that (AK = ) (<(AB + AC – BC)>/2), and thus show that the length of (AK) is independent of (D).

    Since tangents from a point to a circle are equal, we have both (DN = DL) and (DN’ = DL’). Thus,

    Since (MM’ = NN’) by symmetry, we conclude that (KL = DL). Hence,

    We can compute (NN’), and hence (KD), in terms of (AD) and the sides of the triangle:

    as desired. Since (A), (B), and (C) are independent of (D), we conclude that the length (AK) is independent of (D).

    (Solution method found by community member 3cnfsat in the Olympiad Geometry class)

    In our solution above, we used the fact that the length of the segments from a vertex of a triangle (like vertex (D) of triangle (ABD) above) to the points of tangency of the inscribed circle with the sides of the triangle from that vertex (segments (DN) and (DL) above) equals half the perimeter of the triangle minus the opposite side of the triangle. Applying this principle to find length (DN) in triangle (ABD) gives us:

    If you aren’t familiar with this fact, try to prove it yourself (and write a nice solution). Every good geometer reaches for this fact as easily as they reach for the Pythagorean Theorem.

    Solution Readers, not Mindreaders

    A full solution does not just mean a correct answer. You should justify every notable step of your solution. An experienced reader should never wonder ‘Why is that true?’ while reading your solution. She should also never be left in doubt as to whether or not you know why it is true.

    It’s not always clear what steps you can assume the reader understands and what steps you have to explain. Here area few guidelines:

    1. If you can cite a theorem that has a name, then you don’t have to prove the theorem. You can cite the theorem and move on, as in, ‘By the Pythagorean Theorem, (AC = 3).’
    2. If you are very confident the step is well known but you don’t know a name, you can say ‘By a well-known theorem, the area of (ABC) equals (rs), where (r) is the inradius and (s) is the semiperimeter.’ You can also leave out the ‘By a well-known theorem’ bit, particularly for extremely common results such as the one just stated. (If you don’t know that result, try proving it on your own.)
    3. If you still aren’t sure whether to prove a certain step or assume it’s well-known, you have a decision to make. If you can prove it in one or two lines, go ahead and do so. If it’s going to take a lot of work to prove but you know how to do it, then at least outline the proof (and give a more thorough one if you have time). If you’re taking a contest and have no idea how to prove it, cite it and move on. Maybe you’ll get lucky and it will be a ‘well-known theorem’. If you’re writing an educational paper and you don’t know how to prove it, then your paper isn’t finished until you figure it out.
    4. When writing a string of algebraic steps, each step should follow obviously from the one before it. Don’t write something like, ‘Thus, we have [x^2(x – 4) + <(x +1)>^2 + 5x – 4(x +2) = -2,] so (x=1) is the only solution.’ You should include clear simple steps that make it clear that the above is equivalent to ((x – 1)^3 = 0).
    5. You can invoke symmetry or analogy when the cases are precisely the same. For example, suppose you want to prove that the area of any triangle (ABC) is given by [ [ABC] = (ab sin C + bc sin A + ac sin B)/6,] where (a = BC), (b = AC), and (c = AB), and ([ABC]) is the area of (ABC). If you prove that [[ABC] = (ab/2)(sin C),] then you can just write, ‘Similarly, we have [[ABC] = (bc/2)(sin A) = (ac/2)(sin B).’]
    6. When in doubt, explain it. Many of the solutions presented in this article have a little overkill in them. It’s better to prove too much than too little.

    This is totally unacceptable:

    How Not to Write the Solution 1:

    The above is an answer, not a solution. This ‘solution’ lacks any evidence that these solutions actually work, and doesn’t show that there are no other solutions. Moreover, it brings the reader no closer to understanding the solution.

    How Not to Write the Solution 2:

    The given equation rearranges to $(m + 1)(n – 1) = pm 2,$ so the solutions are [(1, 2), (-3, 0), (0, 3), (-2, -1), (1, 0), (-3, 2), (0, -1), (-2, 3).]

    The above solution is better than the first one a motivated reader at least has a glimmer of a path to the solution, but it’s not at all clear how the original equation rearranges to the given equation, nor how the show solutions follow.

    How to Write the Solution:

    We expand the first term and the right-hand side then regroup terms: $(m^2+ 1)(n^2 + 1) + 2(m – n)(1 – mn) = 4(mn + 1)$ $m^2n^2+ m^2 + n^2+ 1 + 2(m – n)(1 – mn) = 4mn + 4$ $m^2n^2- 2mn + 1 + m^2 – 2mn + n^2 + 2(m – n)(1 – mn) = 4$ $(mn – 1)^2 + (m – n)^2 – 2(m – n)(mn – 1) = 4$ This left side is the square of ([(mn – 1) – (m – n)]), so we have: $[(mn – 1)(m – n)]^2 = 4$ $[mn – m + n – 1]^2 = 4$ $[(m + 1)(n – 1)]^2 = 4$ $(m + 1)(n – 1) = pm2,$ Case 1: For ((m + 1)(n – 1) = 2), we have the systems of equations: egin m + 1 &= 1 n – 1 &= 2 &(0, 3) end egin m + 1 &= 2 n – 1 &= 1 &(1, 2) end egin m + 1 &= -1 n – 1 &= -2 &(-2, -1) end egin m + 1 &= -2 n – 1 &= -1 &(-3, 0) end Case 2: For ((m + 1)(n – 1) = -2), we have the systems of equations: egin m + 1 &= 1 n – 1 &= -2 &(0, -1) end egin m + 1 &= -2 n – 1 &= 1 &(-3, 2) end egin m + 1 &= -1 n – 1 &= 2 &(-2, 3) end egin m + 1 &= 2 n – 1 &= -1 &(1, 0) end Thus, the solutions are ((1, 2),) ((-3, 0),) ((0, 3),) ((-2, -1),) ((1, 0),) ((-3, 2),) ((0, -1),) ((-2, 3).)

    Follow the lemmas

    Often you will have to prove multiple preliminary items before tackling the main problem. In writing a proof, we often choose to separate these parts from the main proof by labeling each as a ‘Lemma’ and clearly delimiting the lemma and its proof from the rest of the solution.

    Here’s a sample problem with two different solutions that employ lemmas. We’ve used a little overkill in writing the solutions with lemmas to highlight how well we can clarify solutions with lemmas. Both of these solutions are made significantly easier to read by clearly breaking the solution into pieces.

    Problem: From vertex (A) of triangle (ABC), perpendiculars (AM) and (AN) are drawn to the bisectors of the exterior angles of the triangle at (B) and (C). Prove that (MN) is equal to half the perimeter of (ABC).

    How Not to Write the Solution 1:

    Let the line through (A) parallel to (BC) meet line (BM) at (J). Let the line through (J) parallel to (AB) meet line (BC) at (K). Let (MN) hit (AB) at (X) and (AC) at (Y).

    Since (JK parallel AB) and (AJ parallel BK), (JKBA) is a parallelogram. Since

    From right triangle (BAM), we find

    $angle BAM = 90 – angle ABM = angle B/2.$

    Since (AJ parallel BC), we have (angle JAB = angle B), so (MA) bisects (angle BAJ). Thus, (angle JAM =) (angle BAM) and triangles (BAM) and (JAM) are congruent by ASA. Thus, (AJ = AB) and parallelogram (JKBA) is a rhombus.

    The diagonals of a rhombus bisect each other, so triangles (AMJ) and (KMB) are congruent. Thus, the altitudes from (M) to (AJ) and (BK) are equal and (M) is equidistant from lines (AJ) and line (BK). By symmetry, this is also true for (N). Thus, (MN parallel BC) and (MN) is equidistant from (AJ) and (BC).

    Since (MX parallel BK), ( riangle AMX sim riangle AKB). Since (AM =) (AK/2), we have (MX =) (KB/2). Since (JKBA) is a rhombus, (KB = AB), so (MX =) (AB/2), as desired.

    By symmetry, we also have (NY =) (AC/2). Since (XY) is the midline of (ABC) parallel to side (BC), we have (XY =) (BC/2). Thus,

    $MN = MX + XY + YN = AB/2 + AC/2 + BC/2,$

    Clear Casework

    Sometimes the solution to a problem comes down to investigating a few different cases. In your solution, you should identify the cases clearly and show that these cases cover all possibilities.

    Problem: How many positive 3-digit integers are such that one digit equals the product of the other 2 digits?

    How Not to Write the Solution:

    The solution above is short, and the answer is correct, but it’s not at all clear that all possibilities have been discovered. Also, it’s pretty tough to see that we have found exactly 52 solutions – the reader is forced to go through and count themselves.

    The solution below clearly covers all possible cases and leaves no doubt that the total is 52.

    How to Write the Solution:

    We divide our investigation into cases based on the smallest digit of each number.

    Case 1: The smallest digit is (0).

    If the smallest digit is (0), then the number must contain a second (0). Thus, this case consists of numbers of the form (n00), where (1 leq n leq 9) is any digit from (1) to (9). There are thus 9 numbers with smallest digit (0) that satisfy the problem.

    Case 2: The smallest digit is (1).

    If the smallest digit is (1), the number must be of the form (nn1), or permutations of this form (i.e. (n1n) or (1nn)). However, these 3 permutations are the same when (n = 1). Hence, we have 3 permutations each for (2 leq n leq 9) and only (1) for (n = 1), for a total of (1 +) (3(8) =) (mathbf<25>) numbers with smallest digit (1) that satisfy the problem.

    Case 3: The smallest digit is (2).

    If the smallest digit is (2), then the number is of the form (2mn), where (n = 2m), and permutations of this form. Our only options here are ((m,n) = (2,4)), which gives us 3 numbers ((224, 242, 422)), ((m,n) = (3,6)), which gives us 6 numbers (permutations of (236)), and ((m,n) = (4,8)), which also gives us 6 numbers. Hence, there are (3 +) (6 +) (6 =) (mathbf<15>) numbers with smallest digit (2).

    Case 4: The smallest digit is (3).

    There are 3 solutions in this case: (339), (393), (933).

    Case 5: The smallest digit is larger than (3).

    If the smallest digit is larger than (3), the smallest product we can form with two of the digits is (4(4) = 16), which is not a single digit number. Hence, there are no numbers that satisfy the problem with smallest digit larger than (3).

    Since every possible 3-digit number falls in exactly one of these cases, we conclude that there are

    numbers that satisfy the problem.

    Proofreed

    Comunicacating complex idas is not ease and can b even harder wen don’t edit the presentaion of those ideas for our adience. It pays to oganize are work in ways taht are easy to read to be sur that the audiense gets the point, and to bee sure that your saying what you meen.

    If I always wrote that way, nobody would ever read anything I wrote.

    Proof-read and edit your work. God may do crosswords in pen, but you’re going to make mistakes. Making sure that you wrote in a way that expresses your ideas clearly and correctly is second in importance only to having the right answer.

    Make sure your equations and inequalities use your variables the way you intend. You don’t want to write “ abc + bcd” when you mean “ abd + acd .” This not only makes deciphering the rest of your proof difficult but might also throw off your own calculations.

    Practice writing proofs. We all make occasional spelling or grammar errors, but the effects of errors multiply and too many of them make otherwise good ideas unreadable. Remember that “repetition is the mother of all skill.”

    Problem: (x), (y), and (z) are real numbers such that

    [x + y + z = 5 quad extquad xy + yz + zx = 3.]

    Determine with proof the largest value that any one of the three numbers can be.

    If all proofs were written this poorly I would cry:

    How Not to Write the Solution:

    We will manipulate the given equations to make use of the fact that the square of any real number is negative:

    We can substitute for both (x + y) and (xy) giving us an inequality involving only the variable (z):

    $0 le (x + y)^2 – 4xy = 25 – 10z + z^2 – 12 + 20z – 4z^2 = 3z^2 + 10z + 13.$

    Since this inequality holds for (z) we can determine all possible values of (z):

    $0 ge -3z^2 + 10z + 13 = -(z + 1)(3z – 13).$

    The iequality holds when (-1 ge z ge 13/3).

    Graders will be happier when reading this solution:

    How to Write the Solution:

    We will manipulate the given equations to make use of the fact that the square of any real number is nonnegative:

    We can substitute for both (x + y) and (xy) giving us an inequality involving only the variable (z):

    $0 le (x + y)^2 – 4xy = 25 – 10z + z^2 – 12 + 20z – 4z^2= -3z^2 + 10z + 13.$

    Since this inequality holds for z we can determine all possible values of z:

    $0 ≤ -3z^2 + 10z + 13 = -(z + 1)(3z – 13).$

    The inequality holds when (-1 ≤ z ≤ 13/3).

    Bookends

    We have several shelves full of math books in our offices. When we don’t have bookends on either end, eventually the books at the ends fall over. Then more fall over, then more, and it’s a hassle to find and retrieve books without spilling others all over the place.

    Similarly, when you have a complicated solution, you should place bookends on your solution so the reader doesn’t get lost in the middle. Start off saying what you’re going to do, then do it, then say what you did. Explaining your general method before doing it is particularly important with standard techniques such as contradiction or induction. For example, you might start with, ‘We will show by contradiction that there are infinitely many primes. Assume the opposite, that there are exactly (n) primes (dots ).’

    When you finish your solution, make it clear you are finished. State the final result, which should be saying that you did exactly what the problem asked you to do, e.g. ‘Thus, we have shown by contradiction that there are infinitely many prime numbers.’ You can also decorate the end of proofs with such items as (QED) or (AYD) or (WWWWW) or ( lacksquare ) or //.

    Problem: Let (I) be the incenter of triangle (ABC). Prove

    where (R) is the circumradius of (ABC) and (r) is the inradius of (ABC).

    As this is our last problem, we’ll include many of our no-nos in the ‘How Not’ solution. Good luck piecing it together.

    How Not to Write the Solution:

    From ( riangle AIC) we have (angle AIC =) ( 180^ – angle ACI – angle CAI =) ( 180^ – alpha/2 – gamma/2 =) ( 180^ – (180^ – eta)/2 =) ( 90^ + eta/2) and from ( riangle EBC) we have (angle EBC =) ( angle ABC + angle ABQ =) ( eta + (180^ – eta)/2 =) ( 90^ + eta/2), so (angle AIC = angle EBC). Thus, ( riangle AIC sim riangle EBC) by Angle-Angle Similarity. By symmetry, we conclude ( riangle BIC sim riangle EAC).

    ([AIE] = (AI)(AE)/2 =) ((x)(by/z)/2 = bxy/2z), ([AIC] = br/2), and ([EBC] = [AIC](BC/IC)^2 =) ((br/2)(a/z)^2 =) ( a^2br/2z^2), so ([EACB] =) (bxy/2z + br/2+) (a^2br/2z^2 =) ( axy/2z + ar/2 +) (ab^2r/2z^2). Thus, ((b – a)(xy/2z + r + abr/2z^2) = 0). If (b = a), then (ab – z^2 = xyz/r) follows from the Pythagorean Theorem and the Angle Bisector Theorem. Otherwise, (ab – z^2 = xyz/r) follows immediately.

    Draw altitude (IE) of (AIC). ([EIC] =) ((z^2/2) sin gamma =) ( r(s – c)/2), and ([ABC] =) ((ab/2) sin gamma =) ( rs), so ([(ab – z^2)/2] sin gamma = rc). Then the Law of Sines and the earlier equation give our result.

    Short, ugly, and completely incomprehensible.

    How to Write the Solution:

    [a = BC, b = AC, c = AB][s = (a + b + c)/2][alpha = angle BAC, eta =angle ABC, gamma= angle ABC,]
    [[ABC] = extABC][x = IA, y = IB, z = IC]

    Let the external bisectors of angles (A) and (B) of ( riangle ABC) meet at (E) as shown. Point (E) is equidistant from lines (AB), (AC), and (BC), so it is on angle bisector (CI) as well. We will show

    [[EACB] = bxy/2z + br/2 + a^2br/2z^2 = axy/2z + ar/2 + ab^2r/2z^2², ag <1>]

    From (1) we will show that (ab – z^2 =) (xyz / r), which we will combine with (2) and known triangle relationships to show the desired result.

    Lemma 1: ( riangle AIC sim riangle EBC) and ( riangle BIC sim riangle EAC).
    Proof: By symmetry, the two results are equivalent. We will show the first. Since (CI) bisects ( angle ACB), we have ( angle ACI = angle BCE).

    From ( riangle AIC) we have

    and from ( riangle EBC) we have

    [angle EBC = angle ABC + angle ABQ = eta + (180^ – eta)/2 = 90^ + eta/2,]

    so ( riangle AIC sim riangle EBC) by Angle-Angle Similarity. By symmetry, we conclude ( riangle AIC sim riangle EBC.) (lacksquare)

    Lemma 2: [ [EACB] = bxy/2z + br/2 + a^2br/2z^2 = axy/2z + ar/2 + ab^2r/2z^2] Proof: We find the area of (EACB) by splitting it into pieces:

    First we tackle ([AIE]) by showing it is a right triangle with legs (x) and (by/z). From Lemma 1, we have ( riangle BIC sim riangle EAC). Hence, (AE/IB = AC/IC), or

    [[AIE] = (AI)(AE)/2 = (x)(by/z)/2 = bxy/2z. ag <3>]

    For triangle (AIC) we note that the altitude from (I) to (AC) is the inradius of (ABC), so

    Finally, since ( riangle AIC sim riangle EBC), we have

    [[EBC] = [AIC](BC/IC)^2 = (br/2)(a/z)^2 = a^2br/2z^2. ag <5>]

    [[EACB] = bxy/2z + br/2 + a^2br/2z^2 ag <6>]

    By symmetry, we note that ([EACB]) also equals our expression in (6) with (a) and (b) interchanged and (x) and (y) interchanged. Hence, we have the desired

    $[EACB] = bxy/2z + br/2 + a^2br/2z^2 = axy/2z + ar/2 + ab^2r/2z^2$

    Lemma 3 : (ab – z^2 = xyz / r).
    Proof: Rearranging our result from Lemma 1 yields

    [(bxy/2z + br/2 + a^2br/2z^2) – (axy/2z + ar/2 + ab^2r/2z^2) = 0][(bxy/2z – axy/2z) + (br/2 – ar/2) + (a^2br/2z^2 – ab^2r/2z ^2) = 0][(b-a)(xy/2z) + (b – a)(r/2) – (b – a) (abr/2z^2) = 0][(b – a)(xy/2z + r/2 – abr/2z^2) = 0]

    Thus, one of the terms in this product equals 0.

    Case 1: (b – a = 0).

    If (b = a) then (ABC) is isosceles and (alpha = eta). Hence, the extension of angle bisector (CI) is perpendicular to (AB) at point (D) as shown. Since (I) is the incenter of (ABC) and (ID perp AB), (ID = r) since (ID) is an inradius of (ABC). Also,

    [angle IAB = alpha/2 = eta/2 = angle IBA,]

    so (IB = IA) (i.e. (x = y)). Thus, the equation we wish to prove, (ab – z^2 =) (xyz / r), is in this case equivalent to

    From right triangles (CAD) and (IAD), we have

    [(c/2)^2 + r^2 = x^2, ag <8>][(c/2)^2 + (z + r)^2 = a^2. ag <9>]

    The Angle Bisector Theorem gives us (a/z =) (AC/CI =) (AD/DI =) ((c/2)/r), or

    Substituting (10) into (8) yields

    [a^2r^2/z^2 + r^2 = x^2,][a^2r^2 + r^2z^2 = x^2z^2,][(r/z)(a^2 + z^2) = x^2z/r. ag <11>]

    Substituting (10) into (9) gives

    [a^2r^2/z^2 + (z + r)^2 = a^2,][a^2r^2 + z^2(z + r)^2 = a^2z^2,][z^2(z + r)^2 = a^2z^2 – a^2r^2,][z^2(z + r)^2 = a^2(z^2 – r^2),][z^(z + r) = a^2(z – r),][a^2r + z^2r = a^2z – z^3,][(r/z)(a^2 + z^2) = a^2 – z^2. ag <12>]

    Combining (11) and (12) gives us the desired (a^2 – z^2 = x^2z/r).

    Case 2: (xy / 2z + r / 2 – abr / 2z^2 = 0.)

    Multiplying this equation by (2z^2 / r) yields

    from which the desired (ab – z^2 = xyz / r) immediately follows.

    Thus, the lemma is proved. (lacksquare)

    Lemma 4: ((sin gamma)(ab – z^2) / 2 = rc).

    We draw altitude (IF) perpendicular to (BC) as shown. We employ the following known triangle relationships:

    Just as ([ABC] = (ab/2) sin gamma), we have

    egin
    CF & = [(CF)(IC)/2] sin (gamma/2)
    CF & = [z cos (gamma /2)][z/2 sin (gamma/2)]
    CF & = (z^2 / 2) cos(gamma / 2) sin(gamma / 2)
    CF & = (z^2 /4) sin gamma
    end

    where we have used (sin 2gamma =) (2 sin gamma cos gamma) in the last step.

    Since (CFI) is right, we have ([CFI] = r(s – c)/2). Hence, we have two expressions for ([ABC] – 2[CFI]):

    egin
    (ab / 2) sin gamma – 2(z^2 / 4) sin gamma & = rs – 2[r(s – c / 2]
    [(ab – z^2 / 2] sin gamma & = rs
    end

    We now complete our proof. Dividing the result of Lemma 4 by ((sin gamma) / 2) gives

    Since Lemma 3 gives us (ab – z^2 = xyz / r) and the Extended Law of Sines gives us (sin gamma = c / 2R), the equation above becomes

    Thus, we have shown that if (I) is the incenter of triangle (ABC), we have

    where (R) is the circumradius of (ABC) and (r) is the inradius of (ABC).

    Note, we proved some intermediate results we probably didn’t have to (such as the fact that (E) is on ray (CI)) when the results were quick and easy to prove. Others we stated by fiat, such as ([ABC]= rs), since the proofs are more involved, and we feel pretty safe that these results can be cited as known results without proof.

    The above is a pretty daunting proof. What our solution doesn’t give is any indication of how we might have come up with this solution. If you didn’t find the above solution on your own, see if you can figure out how you might have come up with it now that you have seen it.


    Watch the video: ΜΑΘΗΜΑΤΙΚΑ Α ΔΗΜΟΤΙΚΟΥ (October 2021).