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Difference between revisions of "1998 USAMO Problems/Problem 6"
Line 3: Line 3:
  
 
== Solution ==
 
== 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.
  
 +
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);
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G = (1,-1);
 +
H = (sqrt(3)/2 + 1, -.5);
 +
I = (sqrt(3)/2 + 1, .5);
 +
J = (1-sqrt(2)/2, sqrt(2)/2);
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K = (-1-2/sqrt(3), 0);
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L = extension(K,E,F,G);
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M = (1+2/sqrt(3), 0);
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N = extension(M,H,F,G);
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O = extension(K,D,C,N);
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P = extension(M,I,L,J);
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Q = midpoint(F--G);
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R = midpoint(K--O);
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S = midpoint(P--M);
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T = midpoint(O--C);
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U = midpoint(J--P);
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draw(O--K--L--N--M--P--L^^K--M^^O--N);
 +
label("$A_i$", O, NW);
 +
label("$A_{i+1}$", K, W);
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label("$A_{i+2}$", L, SW);
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label("$A_{i+3}$", N, SE);
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label("$A_{i+4}$", M, dir(0));
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label("$A_{i+5}$", P, NE);
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label("$j$", R, W);
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label("$u$", E, SW);
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label("$y$", Q, S);
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label("$n$", H, SE);
 +
label("$h$", S, NE);
 +
label("$j + y - u$", T, NE);
 +
label("$h + y - n$", U, SW);
 +
</asy>
 +
 +
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, then <math>A_iA_{i+3}</math> must have side length of <math>j+y-u</math>, and <math>A_{i+2}A_{i+5}</math> must have side length of <math>h + y - n</math> (One can see this from what is known as walk-around). Suppose quadrilateral <math>A_{i+1}A_{i+2}A_{i+3}A_{i+4}</math> is tangential. Then, again, we see that <math>A_{i+1}A_{i+4}</math> must have side length <math>u + n - y</math>. We assumed by lemma that <math>A_iA_{i+3} > A_{i}A_{i+1}</math> for all <math>i</math>, so we have <math>A_iA_{i+3} > j</math>, <math>A_{i+1}A_{i+4} > y</math>, and <math>A_{i+2}A_{i+5} > h</math>. If we add up the side lengths <math>A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5}</math>, we get:
 +
<cmath>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</cmath>
 +
<cmath>A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} = j + h + y</cmath>
 +
 +
However, by the lemma, we assumed that <math>A_iA_{i+3} > j</math>, <math>A_{i+1}A_{i+4} > y</math>, and <math>A_{i+2}A_{i+5} > h</math>. Adding these up, we get:
 +
<cmath>A_iA_{i+3} + A_{i+1}A_{i+4} + A_{i+2}A_{i+5} > j + h + y,</cmath>
 +
 +
which is a contradiction. Thus, quadrilateral <math>A_{i+1}A_{i+2}A_{i+3}A_{i+4}</math> is not tangential, proving the lemma.
 +
 +
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>.
 
==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 20:13, 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 $A_{i+2}A_{i+3}A_{i+4}A_{i+5}$ 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 $k = \lfloor \frac{n}{2} \rfloor$.

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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