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

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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>  
 
<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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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\leqx+y\leq9+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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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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 19:55, 15 February 2019

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 $AB$.

Pdfresizer.com-pdf-convert-aimeq12.png

Solution

Let $AB=10x+y$ and $CD=10y+x$. It follows that $CO=\frac{AB}{2}=\frac{10x+y}{2}$ and $CH=\frac{CD}{2}=\frac{10y+x}{2}$. Applying the Pythagorean Theorem on $CO$ and $CH$, we deduce \[OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}=\frac{3}{2}\sqrt{11(x+y)(x-y)}\]

Because $OH$ is a positive rational number, the quantity $\sqrt{11(x+y)(x-y)}$ must be rational, so $11(x+y)(x-y)$ must be a perfect square. Hence either $x-y$ or $x+y$ must be a multiple of $11$, but as $x$ and $y$ are 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 $\boxed{065}$.

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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