Difference between revisions of "1983 AIME Problems/Problem 12"
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<asy> | <asy> | ||
− | draw(circle((0,0), | + | draw(circle((0,0),4)); |
− | draw((- | + | draw((-4,0)--(4,0)); |
− | draw((- | + | draw((-2,-2*sqrt(3))--(-2,2*sqrt(3))); |
− | draw((- | + | draw((-2.6,0)--(-2.6,0.6)); |
− | draw((- | + | draw((-2,0.6)--(-2.6,0.6)); |
dot((0,0)); | dot((0,0)); | ||
− | |||
− | |||
dot((-2,0)); | dot((-2,0)); | ||
− | dot((- | + | dot((4,0)); |
− | dot((- | + | dot((-4,0)); |
− | label("A",(- | + | dot((-2,2*sqrt(3))); |
− | label("B",( | + | dot((-2,-2*sqrt(3))); |
− | label("C",(- | + | label("A",(-4,0),W); |
− | label("D",(- | + | label("B",(4,0),E); |
− | label("H",(- | + | label("C",(-2,2*sqrt(3)),NW); |
− | label("O",(0,0),NE); | + | label("D",(-2,-2*sqrt(3)),SW); |
− | </asy> | + | label("H",(-4,0),SE); |
+ | label("O",(0,0),NE);</asy> | ||
== Solution == | == Solution == |
Revision as of 21:03, 14 January 2023
Contents
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 |