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

(→‎Solution 2 (Coordinates))
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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. If the area of square <math>ABCD</math> is <math>1</math>, then the area of square <math>EFGH</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>.
 
[[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. If the area of square <math>ABCD</math> is <math>1</math>, then the area of square <math>EFGH</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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== Solution ==
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== Solution 1(Pythagorean Theorem)==
 
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>.
 
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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<cmath>0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2</cmath>
 
<cmath>0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2</cmath>
 
This is a quadratic in <math>\frac{b}{a}</math>, and solving it gives <math>\frac{b}{a} = \frac{1}{5},-1</math>. The negative solution is extraneous, and so the ratio of the areas is <math>\left(\frac{1}{5}\right)^2 = \frac{1}{25}</math> and the answer is <math>10\cdot 25 + 1 = \boxed{251}</math>.
 
This is a quadratic in <math>\frac{b}{a}</math>, and solving it gives <math>\frac{b}{a} = \frac{1}{5},-1</math>. The negative solution is extraneous, and so the ratio of the areas is <math>\left(\frac{1}{5}\right)^2 = \frac{1}{25}</math> and the answer is <math>10\cdot 25 + 1 = \boxed{251}</math>.
 +
 
== Solution 2 (Coordinates) ==
 
== Solution 2 (Coordinates) ==
 
Let point <math>A</math> be the top-left corner of square <math>ABCD</math> and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let <math>D</math> have coordinates <math>(0,0)</math> and the side length of square <math>ABCD</math> be <math>a</math>. Let <math>DF</math> = <math>b</math> and diameter <math>HI</math> go through <math>J</math> the midpoint of <math>EF</math>. Since a diameter always bisects a chord perpendicular to it, <math>DJ</math> = <math>JC</math> and since <math>F</math> and <math>E</math> must be symmetric around the diameter, <math>FJ = JE</math> and it follows that <math>DF = EC = b.</math> Hence <math>FE</math> the side of square <math>EFGH</math> has length <math>a - 2b</math>. <math>F</math> has coordinates <math>(b,0)</math> and <math>G</math> has coordinates <math>(b, 2b - a).</math> We know that point <math>G</math> must be on the circle <math>O</math> - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square <math>(a/2, a/2)</math> and has radius <math>a * </math><math>\sqrt{2} / 2</math>, half the diagonal of the square,
 
Let point <math>A</math> be the top-left corner of square <math>ABCD</math> and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let <math>D</math> have coordinates <math>(0,0)</math> and the side length of square <math>ABCD</math> be <math>a</math>. Let <math>DF</math> = <math>b</math> and diameter <math>HI</math> go through <math>J</math> the midpoint of <math>EF</math>. Since a diameter always bisects a chord perpendicular to it, <math>DJ</math> = <math>JC</math> and since <math>F</math> and <math>E</math> must be symmetric around the diameter, <math>FJ = JE</math> and it follows that <math>DF = EC = b.</math> Hence <math>FE</math> the side of square <math>EFGH</math> has length <math>a - 2b</math>. <math>F</math> has coordinates <math>(b,0)</math> and <math>G</math> has coordinates <math>(b, 2b - a).</math> We know that point <math>G</math> must be on the circle <math>O</math> - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square <math>(a/2, a/2)</math> and has radius <math>a * </math><math>\sqrt{2} / 2</math>, half the diagonal of the square,

Revision as of 10:49, 13 June 2022

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. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n + m$.

Solution 1(Pythagorean Theorem)

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 $10n + m = \boxed{251}$.

Another way to proceed from $0 = a^2 - 4ab - 5b^2$ is to note that $\frac{b}{a}$ is the quantity we need; thus, we divide by $a^2$ to get

\[0 = 1 - 4\left(\frac{b}{a}\right) - 5\left(\frac{b}{a}\right)^2\] This is a quadratic in $\frac{b}{a}$, and solving it gives $\frac{b}{a} = \frac{1}{5},-1$. The negative solution is extraneous, and so the ratio of the areas is $\left(\frac{1}{5}\right)^2 = \frac{1}{25}$ and the answer is $10\cdot 25 + 1 = \boxed{251}$.

Solution 2 (Coordinates)

Let point $A$ be the top-left corner of square $ABCD$ and the rest of the vertices be arranged, in alphabetical order, in a clockwise arrangement from there. Let $D$ have coordinates $(0,0)$ and the side length of square $ABCD$ be $a$. Let $DF$ = $b$ and diameter $HI$ go through $J$ the midpoint of $EF$. Since a diameter always bisects a chord perpendicular to it, $DJ$ = $JC$ and since $F$ and $E$ must be symmetric around the diameter, $FJ = JE$ and it follows that $DF = EC = b.$ Hence $FE$ the side of square $EFGH$ has length $a - 2b$. $F$ has coordinates $(b,0)$ and $G$ has coordinates $(b, 2b - a).$ We know that point $G$ must be on the circle $O$ - hence it must satisfy the circle equation. Since the center of the circle is at the center of the square $(a/2, a/2)$ and has radius $a *$$\sqrt{2} / 2$, half the diagonal of the square, $(x - a/2)^2 + (y - a/2)^2 = 1/2a^2$ follows as the circle equation. Then substituting coordinates of $G$ into the equation, $(b - a/2)^2 + (2b - a - a/2)^2 = a^2/2$. Simplifying and factoring, we get $2a^2-7ab+5b^2 = (2a-5b)(a-b) = 0.$ Since $a = b$ would imply $m = n$, and $m < n$ in the problem, we must use the other factor. We get $b = 2/5a$, meaning the ratio of areas $((a-2b)/a)^2$ = $(1/5)^2$ = $1/25$ = $m/n.$ Then $10n + m = 25 * 10 + 1 = \boxed{251}$.

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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