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>. Applying the [[Pythagorean Theorem]] on <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>. 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> | + | <cmath>(2OH)^2=\left(10x+y\right)^2-\left(10y+x\right)^2=99(x+y)(x-y)=99(x+y)(x-y)</cmath> |
− | Because <math>OH</math> is a positive rational number | + | 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>.) |
== See Also == | == See Also == |
Revision as of 18:35, 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
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 |