Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.6 National olympiad, round 2 Find the answer

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 $A_{n+j} = A_{j}$ .)

Your answer

A number or a short expression. Spacing and $ signs are ignored.

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:

Figure (Asymptote source)

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);

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.

Figure (Asymptote source)

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);

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{2\pi}{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.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.