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1983 AIME Problems/Problem 12: Difference between revisions

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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>, <math>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}</math> <math>=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}</math> <math>=\frac{3}{2}\sqrt{11(x+y)(x-y)}</math>.  
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>, <math>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}</math> <math>=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}</math> <math>=\frac{3}{2}\sqrt{11(x+y)(x-y)}</math>.  


Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> cannot contain any square roots. Therefore, <math>x+y</math> must equal eleven and <math>x-y</math> must be a perfect square (since <math>x+y>x-y</math>). The only pair <math>(x,y)</math> that satisfies this condition is <math>(6,5)</math>, so our answer is <math>\boxed{065}</math>.
Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> cannot contain any square roots. Either <math>x-y</math> or <math>x+y</math> must be 11. However, <math>x-y</math> cannot be 11, because both must be digits. Therefore, <math>x+y</math> must equal eleven and <math>x-y</math> must be a perfect square (since <math>x+y>x-y</math>). 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 22:42, 11 March 2009

Problem

The length of diameter $AB$ is a two digit integer. Reversing the digits gives the length of a 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] pointpen=black; pathpen=black+linewidth(0.65); pair O=(0,0),A=(-65/2,0),B=(65/2,0); pair H=(-((65/2)^2-28^2)^.5,0),C=(H.x,28),D=(H.x,-28); D(CP(O,A));D(MP("A",A,W)--MP("B",B,E));D(MP("C",C,N)--MP("D",D)); dot(MP("H",H,SE));dot(MP("O",O,SE)); [/asy]

Solution

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

Because $OH$ is a positive rational number, the quantity $\sqrt{11(x+y)(x-y)}$ cannot contain any square roots. Either $x-y$ or $x+y$ must be 11. However, $x-y$ cannot be 11, because both must be digits. Therefore, $x+y$ must equal eleven and $x-y$ must be a perfect square (since $x+y>x-y$). 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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