Difference between revisions of "1983 AIME Problems/Problem 12"
Hashtagmath (talk | contribs) m |
m (→Problem) |
||
Line 4: | Line 4: | ||
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>. | 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>. | ||
− | + | <asy> | |
+ | draw(circle((0,0),2)); | ||
+ | draw((-2,0)--(2,0)); | ||
+ | draw((-1,-sqrt(3))--(-1,sqrt(3))); | ||
+ | draw((-1.3,0)--(-1.3,0.3)); | ||
+ | draw((-1,0.3)--(-1.3,0.3)); | ||
+ | dot((0,0)); | ||
+ | dot((-1,0)); | ||
+ | dot((2,0)); | ||
+ | dot((-2,0)); | ||
+ | dot((-1,sqrt(3))); | ||
+ | dot((-1,-sqrt(3))); | ||
+ | label("A",(-2,0),W); | ||
+ | label("B",(2,0),E); | ||
+ | label("C",(-1,sqrt(3)),NW); | ||
+ | label("D",(-1,-sqrt(3)),SW); | ||
+ | label("H",(-1,0),SE); | ||
+ | label("O",(0,0),NE); | ||
+ | </asy> | ||
== Solution == | == Solution == |
Revision as of 19:59, 14 January 2023
Contents
[hide]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 .)
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 |