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

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== Problem ==
 
== Problem ==
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The length of diameter <math>AB</math> is a two digit integer. Reversing the digits gives the length of a 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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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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<asy>
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[[File:pdfresizer.com-pdf-convert-aimeq12.png]]
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pointpen=black; pathpen=black+linewidth(0.65);
 
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pair O=(0,0),A=(-65/2,0),B=(65/2,0);
 
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pair H=(-((65/2)^2-28^2)^.5,0),C=(H.x,28),D=(H.x,-28);
 
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D(CP(O,A));D(MP("A",A,W)--MP("B",B,E));D(MP("C",C,N)--MP("D",D));
 
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dot(MP("H",H,SE));dot(MP("O",O,SE));
 
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</asy><!-- Asymptote replacement for Image:1983number12.JPG by bpms -->
 
  
 
== Solution ==
 
== Solution ==

Revision as of 18:52, 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 $CH$, $OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}$ $=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}$ $=\frac{3}{2}\sqrt{11(x+y)(x-y)}$.

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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