Difference between revisions of "1982 USAMO Problems/Problem 5"
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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>. | 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>. | ||
− | We then conclude that <math>\overline{OM} = \overline{OX} \enspace \frac {\overline{JY} }{ \overline{JX}} = \overline{OZ} \enspace \frac {\overline{JY} }{ \overline{ | + | We then conclude that <math>\overline{OM} = \overline{OX} \enspace \frac {\overline{JY} }{ \overline{JX}} = \overline{OZ} \enspace \frac {\overline{JY} }{ \overline{JZ}} = \overline{ON}</math> |
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<math>Quod \enspace Erat \enspace Demonstrandum</math> | <math>Quod \enspace Erat \enspace Demonstrandum</math> | ||
− | |||
== See Also == | == See Also == |
Revision as of 15:02, 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.