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

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== Alternate start to solution 1 ==
 
== 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.
+
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.
  
 
~anduran
 
~anduran

Latest revision as of 17:23, 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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