Difference between revisions of "1983 AIME Problems/Problem 12"
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== Solution == | == Solution == | ||
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 | 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 | ||
| − | <cmath>(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2 | + | <cmath>(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)</cmath> |
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 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>.) | 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 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>.) | ||
Revision as of 23:38, 17 July 2020
Problem
Diameter 
of a circle has length a 
-digit integer (base ten). Reversing the digits gives the length of the perpendicular chord 
. The distance from their intersection point 
to the center 
is a positive rational number. Determine the length of .
Solution
Let 
and 
. It follows that 
and 
. Scale up this triangle by 2 to ease the arithmetic. Applying the Pythagorean Theorem on 
, 
and 
, we deduce
![\[(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)\]](http://latex.artofproblemsolving.com/d/f/2/df2e6f18a0704de86aef307253101f4de9b08556.png)
Because 
is a positive rational number and 
and 
are integral, the quantity 
must be a perfect square. Hence either 
or 
must be a multiple of 
, but as and are digits, 
, so the only possible multiple of is itself. However, cannot be 11, because both must be digits. Therefore, must equal and must be a perfect square. The only pair 
that satisfies this condition is 
, so our answer is 
. (Therefore 
and 
.)
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 | ||
