Difference between revisions of "1983 AIME Problems/Problem 12"
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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>CO</math> and <math>CH</math>, <math>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}</math> <math>=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}</math> <math>=\frac{3}{2}\sqrt{11(x+y)(x-y)}</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>. Applying the [[Pythagorean Theorem]] on <math>CO</math> and <math>CH</math>, <math>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}</math> <math>=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}</math> <math>=\frac{3}{2}\sqrt{11(x+y)(x-y)}</math>. | ||
− | Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> cannot contain any square roots. Therefore, <math>x+y</math> must equal eleven and <math>x-y</math> must be a perfect square (since <math>x+y>x-y</math>). The only pair <math>(x,y)</math> that satisfies this condition is <math>(6,5)</math>, so our answer is <math>\boxed{065}</math>. | + | Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> cannot contain any square roots. Either <math>x-y</math> or <math>x+y</math> must be 11. However, <math>x-y</math> cannot be 11, because both must be digits. Therefore, <math>x+y</math> must equal eleven and <math>x-y</math> must be a perfect square (since <math>x+y>x-y</math>). The only pair <math>(x,y)</math> that satisfies this condition is <math>(6,5)</math>, so our answer is <math>\boxed{065}</math>. |
== See also == | == See also == |
Revision as of 23:42, 11 March 2009
Problem
The length of diameter is a two digit integer. Reversing the digits gives the length of a 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
. Applying the Pythagorean Theorem on
and
,
.
Because is a positive rational number, the quantity
cannot contain any square roots. Either
or
must be 11. However,
cannot be 11, because both must be digits. Therefore,
must equal eleven and
must be a perfect square (since
). The only pair
that satisfies this condition is
, so our answer is
.
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 |