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Difference between revisions of "2001 AIME II Problems/Problem 6"

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== Problem ==
 
== Problem ==
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Square <math>ABCD</math> is inscribed in a circle. Square <math>EFGH</math> has vertices <math>E</math> and <math>F</math> on <math>\overline{CD}</math> and vertices <math>G</math> and <math>H</math> on the circle. The ratio of the area of square <math>EFGH</math> to the area of square <math>ABCD</math> can be expressed as <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers and <math>m < n</math>. Find <math>10n + m</math>.
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[[Square]] <math>ABCD</math> is inscribed in a [[circle]]. Square <math>EFGH</math> has vertices <math>E</math> and <math>F</math> on <math>\overline{CD}</math> and vertices <math>G</math> and <math>H</math> on the circle. The [[ratio]] of the area of square <math>EFGH</math> to the area of square <math>ABCD</math> can be expressed as <math>\frac {m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers and <math>m < n</math>. Find <math>10n + m</math>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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Let <math>O</math> be the center of the circle, and <math>2a</math> be the side length of <math>ABCD</math>, <math>2b</math> be the side length of <math>EFGH</math>. By the [[Pythagorean Theorem]], the radius of <math>\odot O = OC = a\sqrt{2}</math>.
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<center><asy>
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size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7);
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pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2);
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D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A));
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D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d);
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</asy></center>
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Now consider [[right triangle]] <math>OGI</math>, where <math>I</math> is the midpoint of <math>\overline{GH}</math>. Then, by the Pythagorean Theorem,
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<cmath>\begin{align*}
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OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\
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0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b)
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\end{align*}
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</cmath>
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Thus <math>a = 5b</math> (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so <math>\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}</math>, and the answer is <math>10n + m = \boxed{251}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2001|n=II|num-b=5|num-a=7}}
 
{{AIME box|year=2001|n=II|num-b=5|num-a=7}}
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[[Category:Intermediate Geometry Problems]]

Revision as of 13:47, 26 July 2008

Problem

Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. The ratio of the area of square $EFGH$ to the area of square $ABCD$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n + m$.

Solution

Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$, $2b$ be the side length of $EFGH$. By the Pythagorean Theorem, the radius of $\odot O = OC = a\sqrt{2}$.

[asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1), B=(-1,1), A=(-1,-1), E= (1, -0.2), F=(1, 0.2), G=(1.4, 0.2), H=(1.4, -0.2); D(MP("A",A)--MP("B",B,N)--MP("C",C,N)--MP("D",D)--cycle); D(MP("E",E,SW)--MP("F",F,NW)--MP("G",G,NE)--MP("H",H,SE)--cycle); D(CP(D(MP("O",(0,0))), A)); D((0,0) -- (2^.5, 0), d); D((0,0) -- G -- (G.x,0), d); [/asy]

Now consider right triangle $OGI$, where $I$ is the midpoint of $\overline{GH}$. Then, by the Pythagorean Theorem,

\begin{align*} OG^2 = 2a^2 &= OI^2 + GI^2 = (a+2b)^2 + b^2 \\ 0 &= a^2 - 4ab - 5b^2 = (a - 5b)(a + b) \end{align*}

Thus $a = 5b$ (since lengths are positive, we discard the other root). The ratio of the areas of two similar figures is the square of the ratio of their corresponding side lengths, so $\frac{[EFGH]}{[ABCD]} = \left(\frac 15\right)^2 = \frac{1}{25}$, and the answer is $10n + m = \boxed{251}$.

See also

2001 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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