CONTESTS, CONTRADICTION, AND PIGEONHOLES
–A MATH CIRCLE BY MICHAEL GRIFFIN
Math contests often require proofs rather than simple calculations. A proof requires a
careful, logical argument. every claim must be supported. If you ever have to ask yourself
if something is obvious, then it probably isn’t. Explain yourself. Proof by contradiction
and the pigeon hole principal are two important tools for a lot of competition mathematics.
1. Contradiction
Proof by contradiciton simply means assuming a statement is false, press forward until you arrive at a conclusion that either contradict one of the starting hypotheses, or
contradicts some well known fact.
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Example. 2 is irrational.
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Proof. Suppose 2 is rational. Then 2 = ab for two integers a and b. We can take a and
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b to be in lowest terms, which means they have no common factors. Then 2 = ab2 which
means 2b2 = a2 , which means a2 is even (divisible by 2), which means that a is as well.
Let’s say a = 2 · c. Then 2b2 = 4c2 or b2 = 2c2 . But then b is even, which means a and
b share a factor (2). This contradicts our earlier assumption that a and b were written in
lowest terms.
Example. Suppose f is a function from N to N with the property that if x and y are in
N, then
x · f (y) + y · f (x) = (x + y)f (x2 + y 2 ).
Show that f must be a constant function.
2. The pigeon hole principal
The pigeon hole principal is simply the idea that if we have more apples than baskets,
and all the apples have to go in a basket, then at least one basket contains more than one
apple.
More generally, if we have more than k · n apples and n baskets, then at least one basket
will contain at least k apples.
Example. Suppose I give you a collection of 10 distinct positive two-digit integers. Show
that we can find two disjoint subsets which both have the same sum when you add up all
the elements. Note, not all the numbers need to belong to one of these two sets.
Proof. There are 210 possible distinct subsets (why?). Throw out the empty set, and the
whole set. This leaves 210 − 2. These will be the “pigeons”. For each of these subsets,
if we add all the elements, we find the sum is between 1 and 855 (why?). These are the
baskets. By the pigeon hole principal, there are two subsets with the same sum. These
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2 CONTESTS, CONTRADICTION, AND PIGEONHOLES
–A MATH CIRCLE BY MICHAEL GRIFFIN
subsets may not be disjoint, so we throw out all common elements. The result will be two
non-empty subsets, which still have the same sum. (why?)
Sometimes these problems require some geometry.
Example. Nine darts have been thrown at a square dartboard, which has area of one
square foot (all nine darts hit the board). Prove that some three of the darts form a
triangle with area no larger than 1/8 square foot.
Proof. Split the square into four quarters. The darts are the ‘apples’, and the quarters
are the baskets. The Pigeon hole principle tells us that three darts will be in the same
quadrant. The area of the quadrant is 1/4, so the area of any triangle inside is less than
1/8 (why?).