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1982 USAMO Problems/Problem 5: Difference between revisions

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== Solution ==
== 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>.
 
 
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> 
 
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>.
 
 
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.
 
This means that <math>\angle OZN = \angle ZNY = \angle JZY</math> and <math>\angle MXO = \angle YMX= \angle YXJ</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{JX}} = \overline{ON}</math>
 
 
Let <math>a \top b</math> denote line <math>a</math> bisecting line segment <math>b</math>.
 
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>.
 
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>
 
 
<math>Quod \enspace Erat \enspace Demonstrandum</math>
 


== See Also ==
== See Also ==

Revision as of 15:01, 11 July 2018

Problem

$A,B$, and $C$ are three interior points of a sphere $S$ such that $AB$ and $AC$ are perpendicular to the diameter of $S$ through $A$, and so that two spheres can be constructed through $A$, $B$, and $C$ which are both tangent to $S$. Prove that the sum of their radii is equal to the radius of $S$.

Solution

Let the two tangent spheres be $S_1$ and $S_2$, and let $O, O_1, O_2$ and $R, R_1, R_2$ be the origins and radii of $S, S_1, S_2$ respectively. Then $AO$ stands normal to the plane $P$ through $\Delta ABC$. Because both spheres go through $A$, $B$, and $C$, the line $O_1 O_2$ also stands normal to $P$, meaning $AO$ and $O_1 O_2$ are both coplanar and parallel. Therefore the problem can be flattened to the plane $P'$ through $A$, $O$, $O_1$ and $O_2$.


Let $X, Y, Z, M, N$ be points on $P'$ such that $X = S \cap S_1, \quad Y = S_1 \cap S_2 \neq A, \quad Z = S \cap S_2, \quad M = O_1 Y \cap AO, \quad N = O_2 Y \cap AO$

Let $J$ be the three circles radical center, meaning $JX$ and $JZ$ are tangent segments to $S$ and $J \in AY$.


Because $Y \in P \iff \angle NAY = \angle YAM = 90 ^{\circ},$ we have that $\overline{YN}$ and $\overline{YM}$ are diameters.

This means that $\angle OZN = \angle ZNY = \angle JZY$ and $\angle MXO = \angle YMX= \angle YXJ$.

And because $\angle ZNO = 180^{\circ} - \angle AYZ = \angle ZYJ$ and $\angle OMX = 180^{\circ} -\angle XYA = \angle JYX,$ we have that $\Delta JZY \sim \Delta OZN$ and $\Delta JYX \sim \Delta OMX$.

We then conclude that $\overline{OM} = \overline{OX} \enspace \frac {\overline{JY} }{ \overline{JX}} = \overline{OZ} \enspace \frac {\overline{JY} }{ \overline{JX}} = \overline{ON}$


Let $a \top b$ denote line $a$ bisecting line segment $b$.

Since $O_1 O_2 || MN$ it follows that $OY \, \top \, \overline{MN} \Rightarrow OY \, \top \, \overline{O_1 O_2}$. Similarly we have that $O_1 O_2 \, \top \, \overline{YM} \Rightarrow O_1 O_2 \, \top \, \overline{OY}$.

And so $O_1 O O_2 Y$ is a parallelogram because $\overline{OY}$ and $\overline{O_1 O_2}$ bisect each other, meaning $R = \overline{OZ} = \overline{O O_2} + \overline{O_2 Z} = \overline{O_1 Y}+ R_2 = R_1 + R_2$


$Quod \enspace Erat \enspace Demonstrandum$


See Also

1982 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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