Writing Math Solutions

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How to Write a Solution
U s e S p a c e
by Richard Rusczyk

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When you write your solution, you should:

Give each important definition or equation its own line.
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.
Label equations or formulas or lemmas or cases you will use later very clearly.
Remember that there's always more paper.

Here's a sample problem:

Problem: Let p(x) be a polynomial with degree 98 such that p(n) = 1/n for n= 1, 2, 3, 4, . . . , 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, . . . , 99, r(x) has roots 1, 2, . . . , 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) . . . (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) . . . (x - 99) yields r(0) = -c(99!); hence, c = 1/99!. Thus, we have r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99!. We can combine the equations r(x) = x p(x) - 1 and r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99! and let x = 100 to find 100p(100) - 1 = (100 - 1)(100 - 2)(100 - 3) . . . (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: Let

r(x) = x (p(x) - 1/x) = x p(x) - 1.

(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, . . . , 99,

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

(2)

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

r(x) = (x - 1)(x - 2)(x - 3) . . . (x - 99)/99!

(3)

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

100p(100) - 1 = (100 - 1)(100 - 2)(100 - 3) . . . (100 - 99)/99!
100p(100) - 1 = 99!/99! = 1
p(100) = 1/50.

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Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!