Difference between revisions of "1982 USAMO Problems/Problem 5"
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== Solution == | == Solution == | ||
− | {{ | + | Let the two tangent spheres be <math>S_1</math> and <math>S_2</math>, and let <math>O, O_1, O_2</math> and <math>R, R_1, R_2</math> be the origins and radii of <math>S, S_1, S_2</math> respectively. Then <math>AO</math> stands normal to the plane <math>P</math> through <math>\Delta ABC</math>. Because both spheres go through <math>A</math>, <math>B</math>, and <math>C</math>, the line <math>O_1 O_2</math> also stands normal to <math>P</math>, meaning <math>AO</math> and <math>O_1 O_2</math> are both coplanar and parallel. Therefore the problem can be flattened to the plane <math>P'</math> through <math>A</math>, <math>O</math>, <math>O_1</math> and <math>O_2</math>. |
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+ | Let <math>X, Y, Z, M, N</math> be points on <math>P'</math> such that <math>X = S \cap S_1, \quad Y = S_1 \cap S_2 \neq A, \quad Z = S \cap S_2, \quad M = O_1 Y \cap AO, \quad N = O_2 Y \cap AO</math> | ||
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+ | Let <math>J</math> be the three circles radical center, meaning <math>JX</math> and <math>JZ</math> are tangent segments to <math>S</math> and <math>J \in AY</math>. | ||
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+ | Because <math>Y \in P \iff \angle NAY = \angle YAM = 90 ^{\circ},</math> we have that <math>\overline{YN}</math> and <math>\overline{YM}</math> are diameters. | ||
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+ | This means that <math>\angle OZN = \angle ZNY = \angle JZY</math> and <math>\angle MXO = \angle YMX= \angle YXJ</math>. | ||
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+ | And because <math>\angle ZNO = 180^{\circ} - \angle AYZ = \angle ZYJ</math> and <math>\angle OMX = 180^{\circ} -\angle XYA = \angle JYX,</math> we have that <math>\Delta JZY \sim \Delta OZN</math> and <math>\Delta JYX \sim \Delta OMX</math>. | ||
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+ | We then conclude that <math>\overline{OM} = \overline{OX} \enspace \frac {\overline{JY} }{ \overline{JX}} = \overline{OZ} \enspace \frac {\overline{JY} }{ \overline{JX}} = \overline{ON}</math> | ||
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+ | Let <math>a \top b</math> denote line <math>a</math> bisecting line segment <math>b</math>. | ||
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+ | Since <math>O_1 O_2 || MN</math> it follows that <math>OY \, \top \, \overline{MN} \Rightarrow OY \, \top \, \overline{O_1 O_2}</math>. | ||
+ | Similarly we have that <math>O_1 O_2 \, \top \, \overline{YM} \Rightarrow O_1 O_2 \, \top \, \overline{OY}</math>. | ||
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+ | And so <math>O_1 O O_2 Y</math> is a parallelogram because <math>\overline{OY}</math> and <math>\overline{O_1 O_2}</math> bisect each other, meaning <math>R = \overline{OZ} = \overline{O O_2} + \overline{O_2 Z} = \overline{O_1 Y}+ R_2 = R_1 + R_2</math> | ||
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+ | <math>Quod \enspace Erat \enspace Demonstrandum</math> | ||
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== See Also == | == See Also == |
Revision as of 15:01, 11 July 2018
Problem
, and
are three interior points of a sphere
such that
and
are perpendicular to the diameter of
through
, and so that two spheres can be constructed through
,
, and
which are both tangent to
. Prove that the sum of their radii is equal to the radius of
.
Solution
Let the two tangent spheres be and
, and let
and
be the origins and radii of
respectively. Then
stands normal to the plane
through
. Because both spheres go through
,
, and
, the line
also stands normal to
, meaning
and
are both coplanar and parallel. Therefore the problem can be flattened to the plane
through
,
,
and
.
Let be points on
such that
Let be the three circles radical center, meaning
and
are tangent segments to
and
.
Because we have that
and
are diameters.
This means that and
.
And because and
we have that
and
.
We then conclude that
Let denote line
bisecting line segment
.
Since it follows that
.
Similarly we have that
.
And so is a parallelogram because
and
bisect each other, meaning
See Also
1982 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.