Mathematical Food for Thought

Problem: Factor .

Solution: Consider the polynomial . We have

,

so we want to factor the second term when . Call it so that . Consider the relation

.

Since is a root of the LHS, we factor it out of the RHS as well to get

.

Dividing through by and rearranging, we obtain the nice expression

.

Letting , this becomes

which is conveniently enough a difference of two squares. And we’ll leave it as this because the factors are not particularly nice or anything. QED.

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Comment: This “natural” factorization was basically the one crucial step to solving the USAMO #5 this year and most people did not see it, unsurprisingly. Taking the difference was the trickiest/cleverest part, and there were definitely a limited number of approaches to this factorization. Oh well, at least it seems sort of cool after you know about it.

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Practice Problem: (2007 USAMO – #5) Prove that for every nonnegative integer n, the number is the product of at least (not necessarily distinct) primes.

Good luck to everyone on the USAMO tomorrow and Wednesday!

Problem: A coin is repeatedly flipped. What is the expected number of flips to get two heads in a row? heads in a row?

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I will be out of town for the next four days, so have fun with the problem!

Problem: (1993 Canada National Olympiad – #2) Show that is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence .

Solution: We will prove the “if” statement first, i.e. is rational implies we can find the geometric progression. Let . The terms of the sequence are then

, , , .

We simply choose the terms

, , which are the same as , , ,

clearly a geometric progression.

Now for the other direction, assume are a geometric progression with positive integers. Then

.

Solving for yields which is clearly rational, as desired. QED.

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Comment: An interesting “definition” of a rational number, pretty neat in fact. Though it may not seem so at first, the “if” direction was actually considerably harder than the “only if” direction because the approach was not as straightforward. The “only if” proof required simple algebra, while the “if” proof needed a little creative picking and choosing. Another approach for the “if” direction is to set and try to find corresponding but that makes it a little more difficult than necessary.

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Practice Problem: (1993 Canada National Olympiad – #3) In triangle , medians to the sides and are perpendicular. Show that .

Problem: (2001 Poland Finals – #1) Prove the following inequality:

where are positive reals.

Solution: Like the title says, that triangular number looks really fishy… let’s write it as and pair up the terms on the RHS.

.

We claim that for . We’ll use our good friend AM-GM to show this; in fact, it is quite simple.

so

as desired. Sum them up to get our inequality. QED.

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Comment: Just a clever little application of AM-GM; apparantly not a strong inequality at all, and the only equality case is for all . Nevertheless, slightly on the tricky side which makes it sufficiently satisfying to solve.

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Practice Problem: (2006 Poland Finals – #1) Solve in reals:

.