1983 AIME Problems/Problem 12
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 different 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 .)
Alternate start to solution 1
Since H is the midpoint of , by Power of a Point, . Because and , where is the radius of the circle, we deduce the relation . Thus . Continue as above.
~anduran
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 |