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Difference between revisions of "1983 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
 
Diameter <math>AB</math> of a circle has length a <math>2</math>-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord <math>CD</math>. The distance from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</math>.
 
Diameter <math>AB</math> of a circle has length a <math>2</math>-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord <math>CD</math>. The distance from their intersection point <math>H</math> to the center <math>O</math> is a positive rational number. Determine the length of <math>AB</math>.
  
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[[File:pdfresizer.com-pdf-convert-aimeq12.png]]
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<asy>
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draw(circle((0,0),4));
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draw((-4,0)--(4,0));
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draw((-2,-2*sqrt(3))--(-2,2*sqrt(3)));
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draw((-2.6,0)--(-2.6,0.6));
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draw((-2,0.6)--(-2.6,0.6));
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dot((0,0));
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dot((-2,0));
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dot((4,0));
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dot((-4,0));
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dot((-2,2*sqrt(3)));
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dot((-2,-2*sqrt(3)));
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label("A",(-4,0),W);
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label("B",(4,0),E);
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label("C",(-2,2*sqrt(3)),NW);
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label("D",(-2,-2*sqrt(3)),SW);
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label("H",(-2,0),SE);
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label("O",(0,0),NE);</asy>
  
 
== Solution ==
 
== Solution ==
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Let <math>AB=10x+y</math> and <math>CD=10y+x</math>. It follows that <math>CO=\frac{AB}{2}=\frac{10x+y}{2}</math> and <math>CH=\frac{CD}{2}=\frac{10y+x}{2}</math>. Applying the [[Pythagorean Theorem]] on <math>CO</math> and <math>CH</math>, we deduce
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Let <math>AB=10x+y</math> and <math>CD=10y+x</math>. It follows that <math>CO=\frac{AB}{2}=\frac{10x+y}{2}</math> and <math>CH=\frac{CD}{2}=\frac{10y+x}{2}</math>. Scale up this triangle by 2 to ease the arithmetic. Applying the [[Pythagorean Theorem]] on <math>2CO</math>, <math>2OH</math> and <math>2CH</math>, we deduce
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<cmath>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}=\frac{3}{2}\sqrt{11(x+y)(x-y)}</cmath>  
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<cmath>(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)</cmath>
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Because <math>OH</math> is a positive rational number and <math>x</math> and <math>y</math> are integral, the quantity <math>99(x+y)(x-y)</math> must be a perfect square. Hence either <math>x-y</math> or <math>x+y</math> must be a multiple of <math>11</math>, but as <math>x</math> and <math>y</math> are different digits, <math>1+0=1 \leq x+y \leq 9+9=18</math>, so the only possible multiple of <math>11</math> is <math>11</math> itself. However, <math>x-y</math> cannot be 11, because both must be digits. Therefore, <math>x+y</math> must equal <math>11</math> and <math>x-y</math> must be a perfect square. The only pair <math>(x,y)</math> that satisfies this condition is <math>(6,5)</math>, so our answer is <math>\boxed{065}</math>. (Therefore <math>CD = 56</math> and <math>OH = \frac{33}{2}</math>.)
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== Alternate start to solution 1 ==
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Since H is the midpoint of <math>CD</math>, by [[Power of a Point]], <math>CH^2=(AH)(BH)</math>. Because <math>AH=r-OH</math> and <math>BH=r+OH</math>, where <math>r</math> is the radius of the circle, we deduce the relation <math>CH^2=r^2-OH^2</math>. Thus <math>OH^2=\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2</math>. Continue as above.
  
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Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> must be rational, so <math>11(x+y)(x-y)</math> must be a perfect square. Hence either <math>x-y</math> or <math>x+y</math> must be a multiple of <math>11</math>, but as <math>x</math> and <math>y</math> are digits, <math>1+0=1 \leq x+y \leq 9+9=18</math>, so the only possible multiple of <math>11</math> is <math>11</math> itself. However, <math>x-y</math> cannot be 11, because both must be digits. Therefore, <math>x+y</math> must equal <math>11</math> and <math>x-y</math> must be a perfect square. The only pair <math>(x,y)</math> that satisfies this condition is <math>(6,5)</math>, so our answer is <math>\boxed{065}</math>.
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~anduran
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:23, 1 January 2024

Problem

Diameter $AB$ of a circle has length a $2$-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord $CD$. The distance from their intersection point $H$ to the center $O$ is a positive rational number. Determine the length of $AB$.

[asy] draw(circle((0,0),4)); draw((-4,0)--(4,0)); draw((-2,-2*sqrt(3))--(-2,2*sqrt(3))); draw((-2.6,0)--(-2.6,0.6)); draw((-2,0.6)--(-2.6,0.6)); dot((0,0)); dot((-2,0)); dot((4,0)); dot((-4,0)); dot((-2,2*sqrt(3))); dot((-2,-2*sqrt(3))); label("A",(-4,0),W); label("B",(4,0),E); label("C",(-2,2*sqrt(3)),NW); label("D",(-2,-2*sqrt(3)),SW); label("H",(-2,0),SE); label("O",(0,0),NE);[/asy]

Solution

Let $AB=10x+y$ and $CD=10y+x$. It follows that $CO=\frac{AB}{2}=\frac{10x+y}{2}$ and $CH=\frac{CD}{2}=\frac{10y+x}{2}$. Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on $2CO$, $2OH$ and $2CH$, we deduce \[(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)\]

Because $OH$ is a positive rational number and $x$ and $y$ are integral, the quantity $99(x+y)(x-y)$ must be a perfect square. Hence either $x-y$ or $x+y$ must be a multiple of $11$, but as $x$ and $y$ are different digits, $1+0=1 \leq x+y \leq 9+9=18$, so the only possible multiple of $11$ is $11$ itself. However, $x-y$ cannot be 11, because both must be digits. Therefore, $x+y$ must equal $11$ and $x-y$ must be a perfect square. The only pair $(x,y)$ that satisfies this condition is $(6,5)$, so our answer is $\boxed{065}$. (Therefore $CD = 56$ and $OH = \frac{33}{2}$.)

Alternate start to solution 1

Since H is the midpoint of $CD$, by Power of a Point, $CH^2=(AH)(BH)$. Because $AH=r-OH$ and $BH=r+OH$, where $r$ is the radius of the circle, we deduce the relation $CH^2=r^2-OH^2$. Thus $OH^2=\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2$. Continue as above.

~anduran

See Also

1983 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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