Difference between revisions of "1983 AIME Problems/Problem 12"
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<cmath>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}=\frac{3}{2}\sqrt{11(x+y)(x-y)}</cmath> | <cmath>OH=\sqrt{\left(\frac{10x+y}{2}\right)^2-\left(\frac{10y+x}{2}\right)^2}=\sqrt{\frac{9}{4}\cdot 11(x+y)(x-y)}=\frac{3}{2}\sqrt{11(x+y)(x-y)}</cmath> | ||
− | Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> must be rational, so <math>11(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\ | + | Because <math>OH</math> is a positive rational number, the quantity <math>\sqrt{11(x+y)(x-y)}</math> must be rational, so <math>11(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>. |
== See Also == | == See Also == |
Revision as of 18:55, 15 February 2019
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
. Applying the Pythagorean Theorem on
and
, we deduce
Because is a positive rational number, the quantity
must be rational, so
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
.
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 |