1998 USAMO Problems/Problem 6
Problem
Let be an integer. Find the largest integer (as a function of ) such that there exists a convex -gon for which exactly of the quadrilaterals have an inscribed circle. (Here .)
Solution
Lemma: If quadrilaterals and in an equiangular -gon are tangential, and is the longest side quadrilateral for all , then quadrilateral is not tangential.
Proof:
If quadrilaterals and are tangential, then must have side length of , and must have side length of (One can see this from what is known as walk-around). Suppose quadrilateral is tangential. Then, again, we see that must have side length . We assumed by lemma that for all , so we have , , and . If we add up the side lengths , we get:
However, by the lemma, we assumed that , , and . Adding these up, we get:
which is a contradiction. Thus, quadrilateral is not tangential, proving the lemma.
By lemma, the maximum number of quadrilaterals in a -gon occurs when the tangential quadrilaterals alternate, giving us .
Note that one can find the ratio of side of an equiangular -gon in order for alternating quadrilaterals to be tangential.
Since exterior angles of a equiangular -gon have degree measure , one can write the equation:
Thus, we can find the ratio of sides of an equiangular -gon which fits the maximum to be . Note that if is even, we can easily alternate them, but if is odd, we must have two adjacent sides be the same length, and that length must be the larger side in the ratio of adjacent sides. The proof is left as an exercise for the reader.
See Also
1998 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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