Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1998 USAMO Problems/Problem 6"
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
Lemma: If quadrilaterals <math>A_iA_{i+1}A_{i+2}A_{i+3}</math> and <math>A_{i+2}A_{i+3}A_{i+4}A_{i+5}</math> are tangential, and <math>A_iA_{i+3}</math> is the longest side quadrilateral <math>A_iA_{i+1}A_{i+2}A_{i+3}</math> for all <math>i</math>, then quadrilateral <math>A_{i+1}A_{i+2}A_{i+3}A_{i+4}</math> is not tangential.
+
Lemma: If quadrilaterals <math>A_iA_{i+1}A_{i+2}A_{i+3}</math> and <math>A_{i+2}A_{i+3}A_{i+4}A_{i+5}</math> in an equiangular <math>n</math>-gon are tangential, and <math>A_iA_{i+3}</math> is the longest side quadrilateral <math>A_iA_{i+1}A_{i+2}A_{i+3}</math> for all <math>i</math>, then quadrilateral <math>A_{i+1}A_{i+2}A_{i+3}A_{i+4}</math> is not tangential.
  
 
Proof:
 
Proof:
Line 59: Line 59:
  
 
By lemma, the maximum number of quadrilaterals in a <math>n</math>-gon occurs when the tangential quadrilaterals alternate, giving us <math>k = \lfloor \frac{n}{2} \rfloor</math>.
 
By lemma, the maximum number of quadrilaterals in a <math>n</math>-gon occurs when the tangential quadrilaterals alternate, giving us <math>k = \lfloor \frac{n}{2} \rfloor</math>.
 +
 +
 +
*Note that one can find the ratio of side of an equiangular <math>n</math>-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 <math>n</math>-gon have degree measure <math>\frac{2pi}{n}</math>, one can write the equation:
 +
 +
<cmath>2x = y + y + 2x \cos \frac{2pi}{n}</cmath>
 +
<cmath>y = x \left( 1- \cos \frac{2pi}{n} \right)</cmath>
 +
<cmath>\frac{y}{x} = \frac{1}{1- \cos \frac{2pi}{n}}</cmath>
 +
 +
Thus, we can find the ratio of sides of an equiangular <math>n</math>-gon which fits the maximum to be <math>1 : 1- \cos \frac{2pi}{n}</math>. Note that if <math>n</math> is even, we can easily alternate them, but if <math>n</math> 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==
 
==See Also==
 
{{USAMO newbox|year=1998|num-b=5|after=Last Question}}
 
{{USAMO newbox|year=1998|num-b=5|after=Last Question}}
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:38, 23 March 2015

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.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_USAMO_Problems/Problem_6&oldid=69441"