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Difference between revisions of "1999 AIME Problems/Problem 4"

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label("$H$", H, dir(O--H));</asy>
 
label("$H$", H, dir(O--H));</asy>
 
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== Solution ==
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== Solution 1 ==
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=== Simple Solution ===
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Triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length <math>1</math> in the circumcircle of the squares pass through <math>B</math>, etc.), and each area is <math>\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}</math>. Since the area of a triangle is <math>bh/2</math>, the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.
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Triangles <math>AOB</math>, <math>BOC</math>, <math>COD</math>, etc. are congruent, and each area is <math>\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}</math>. Since the area of a triangle is <math>bh/2</math>, the area of all <math>8</math> of them is <math>\frac{86}{99}</math> and the answer is <math>\boxed{185}</math>.
 
  
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== Solution 2 ==
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=== Other Solution ===
 
 
Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>.  
 
Define the two possible [[distance]]s from one of the labeled points and the [[vertex|corners]] of the square upon which the point lies as <math>x</math> and <math>y</math>. The area of the [[octagon]] (by [[subtraction]] of areas) is <math>1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy</math>.  
  

Latest revision as of 12:27, 12 December 2020

Problem

The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers. Find $m+n.$

[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z), C=intersectionpoint(Y--X, y--x), E=intersectionpoint(W--X, w--x), G=intersectionpoint(W--Z, w--z), B=intersectionpoint(Y--Z, y--x), D=intersectionpoint(Y--X, w--x), F=intersectionpoint(W--X, w--z), H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy]

Solution 1

Triangles $AOB$, $BOC$, $COD$, etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$, etc.), and each area is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$. Since the area of a triangle is $bh/2$, the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$.

Solution 2

Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$. The area of the octagon (by subtraction of areas) is $1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy$.

By the Pythagorean theorem, \[x^2 + y^2 = \left(\frac{43}{99}\right)^2\]

Also, \begin{align*}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}

Substituting, \begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}

Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$, so $m + n = \boxed{185}$.

See also

1999 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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