# 1998 USAMO Problems/Problem 6

## Problem

Let $n \geq 5$ be an integer. Find the largest integer $k$ (as a function of $n$) such that there exists a convex $n$-gon $A_{1}A_{2}\dots A_{n}$ for which exactly $k$ of the quadrilaterals $A_{i}A_{i+1}A_{i+2}A_{i+3}$ have an inscribed circle. (Here $A_{n+j} = A_{j}$.)

## Solution

Lemma: If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ in an equiangular $n$-gon are tangential, and $A_iA_{i+3}$ is the longest side quadrilateral $A_iA_{i+1}A_{i+2}A_{i+3}$ for all $i$, then quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential.

Proof:

$[asy] import geometry; size(10cm); pair A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U; A = (-1,0); B = (1,0); draw(Circle(A,1)^^Circle(B,1)); C = (sqrt(2)/2-1,sqrt(2)/2); D = (-sqrt(3)/2 - 1, .5); E = (-sqrt(3)/2 - 1, -.5); F = (-1,-1); G = (1,-1); H = (sqrt(3)/2 + 1, -.5); I = (sqrt(3)/2 + 1, .5); J = (1-sqrt(2)/2, sqrt(2)/2); K = (-1-2/sqrt(3), 0); L = extension(K,E,F,G); M = (1+2/sqrt(3), 0); N = extension(M,H,F,G); O = extension(K,D,C,N); P = extension(M,I,L,J); Q = midpoint(F--G); R = midpoint(K--O); S = midpoint(P--M); T = midpoint(O--C); U = midpoint(J--P); draw(O--K--L--N--M--P--L^^K--M^^O--N); label("A_i", O, NW); label("A_{i+1}", K, W); label("A_{i+2}", L, SW); label("A_{i+3}", N, SE); label("A_{i+4}", M, dir(0)); label("A_{i+5}", P, NE); label("j", R, W); label("u", E, SW); label("y", Q, S); label("n", H, SE); label("h", S, NE); label("j + y - u", T, NE); label("h + y - n", U, SW); [/asy]$

If quadrilaterals $A_iA_{i+1}A_{i+2}A_{i+3}$ and $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ are tangential, then $A_iA_{i+3}$ must have side length of $j+y-u$, and $A_{i+2}A_{i+5}$ must have side length of $h + y - n$ (One can see this from what is known as walk-around). Suppose quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is tangential. Then, again, we see that $A_{i+1}A_{i+4}$ must have side length $u + n - y$. We assumed by lemma that $A_iA_{i+3} > A_{i}A_{i+1}$ for all $i$, so we have $A_iA_{i+3} > j$, $A_{i+1}A_{i+4} > y$, and $A_{i+2}A_{i+5} > h$. If we add up the side lengths $A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5}$, we get: $$A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} = j + y - u + h + y - n + u + n - y$$ $$A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} = j + h + y$$

However, by the lemma, we assumed that $A_iA_{i+3} > j$, $A_{i+1}A_{i+4} > y$, and $A_{i+2}A_{i+5} > h$. Adding these up, we get: $$A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} > j + h + y,$$

which is a contradiction. Thus, quadrilateral $A_{i+1}A_{i+2}A_{i+3}A_{i+4}$ is not tangential, proving the lemma.

By lemma, the maximum number of quadrilaterals in a $n$-gon occurs when the tangential quadrilaterals alternate, giving us $k = \lfloor \frac{n}{2} \rfloor$.

Note that one can find the ratio of side of an equiangular $n$-gon in order for alternating quadrilaterals to be tangential.

$[asy] import geometry; size(10cm); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (1+(1-cos(2pi/9))*cos(2pi/9), (1-cos(2pi/9))*sin(2pi/9)); D = (-(1-cos(2pi/9))*cos(2pi/9), (1-cos(2pi/9))*sin(2pi/9)); E = midpoint(D--A); F = midpoint(A--B); draw(A--B--C--D--A); label("A_i", D, NW); label("A_{i+1}", A, SW); label("A_{i+2}", B, SE); label("A_{i+3}", C, NE); label("x", E, W); label("y", F, S); [/asy]$

Since exterior angles of a equiangular $n$-gon have degree measure $\frac{2pi}{n}$, one can write the equation:

$$2x = y + y + 2x \cos \frac{2pi}{n}$$ $$y = x \left( 1- \cos \frac{2pi}{n} \right)$$ $$\frac{y}{x} = \frac{1}{1- \cos \frac{2pi}{n}}$$

Thus, we can find the ratio of sides of an equiangular $n$-gon which fits the maximum to be $1 : 1- \cos \frac{2pi}{n}$. Note that if $n$ is even, we can easily alternate them, but if $n$ 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.