Difference between revisions of "1992 USAMO Problems/Problem 4"

Revision as of 18:19, 22 April 2010

Problem

Chords $AA'$ , $BB'$ , and $CC'$ of a sphere meet at an interior point $P$ but are not contained in the same plane. The sphere through $A$ , $B$ , $C$ , and $P$ is tangent to the sphere through $A'$ , $B'$ , $C'$ , and $P$ . Prove that $AA'=BB'=CC'$ .

Solution

Consider the plane through $A,A',B,B'$ . This plane, of course, also contains $P$ . We can easily find the $\triangle APB$ is isosceles because the base angles are equal. Thus, $AP=BP$ . Similarly, $A'P=B'P$ . Thus, $AA'=BB'$ . By symmetry, $BB'=CC'$ and $CC'=AA'$ , and hence $AA'=BB'=CC'$ as desired.

$\mathbb{QED.}$

Add-on

By another person ^v^

The person that came up with the solution did not prove that $\triangle APB$ is isosceles nor the base angles are congruent. I will add on to the solution.

There is a common tangent plane that pass through $P$ for the $2$ spheres that are tangent to each other.

Since any cross section of sphere is a circle. It implies that $A$ , $A'$ , $B$ , $B'$ be on the same circle ( $\omega_1$ ), $A$ , $B$ , $P$ be on the same circle ( $\omega_2$ ), and $A'$ , $B'$ , $P$ be on the same circle ( $\omega_3$ ).

$m\angle APB= m\angleA'PB'$ (Error compiling LaTeX. ! Undefined control sequence.) because they are vertical angles. By power of point, $(AP)(A'P)=(BP)(B'P)\rightarrow\frac{AP}{B'P}=\frac{BP}{A'P}$

By the SAS triangle simlarity theory, $\triangle APB \sim\triangle B'PA'$ . That implies that $\angle ABP\cong\angleB'PA'$ (Error compiling LaTeX. ! Undefined control sequence.).

Let's call the interception of the common tangent plane and the plane containing $A$ , $A'$ , $B$ , $B'$ , $P$ , line $l$ .

$l$ must be the common tangent of $\omega_2$ and $\omega_3$ .

The acute angles form by $l$ and $\overbar{AA'}$ (Error compiling LaTeX. ! Undefined control sequence.) are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord $\overbar{AP}$ (Error compiling LaTeX. ! Undefined control sequence.) and $\overbar{A'P}$ (Error compiling LaTeX. ! Undefined control sequence.) are equal.

Similarly the central angle of chord $\overbar{BP}$ (Error compiling LaTeX. ! Undefined control sequence.) and $\overbar{B'P}$ (Error compiling LaTeX. ! Undefined control sequence.) are equal.

The length of any chord with central angle $2\theta$ and radius $r$ is $2r\sin\left({\theta}\right)$ , which can easily been seen if we drop the perpendicular from the center to the chord.

Thus, $\frac{AP}{A'P}=\frac{BP}{B'P}$ .

By the SAS triangle simlarity theory, $\triangle APB \sim\triangle A'PB'$ . That implies that $\angle ABP\cong\angleB'PA'$ (Error compiling LaTeX. ! Undefined control sequence.).

That implies that $\angle ABP\cong\angleA'PB'\cong\angleB'PA'$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, $\triangle A'PB'$ is an isosceles triangle and since $\triangle APB \sim\triangle A'PB'$ , $\triangle APB$ is an isosceles triangle too.

Resources

1992 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

Discussion on AoPS/MathLinks

@@ Line 6: / Line 6: @@
 <math>\mathbb{QED.}</math>
+=== Add-on===
+By another person ^v^
+The person that came up with the solution did not prove that  <math>\triangle APB</math> is isosceles nor the base angles are congruent. I will add on to the solution.
+There is a common tangent plane that pass through <math>P</math> for the <math>2</math> spheres that are tangent to each other.
+<br/>
+Since any cross section of sphere is a circle. It implies that <math>A</math>, <math>A'</math>, <math>B</math>, <math>B'</math> be on the same circle (<math>\omega_1</math>), <math>A</math>, <math>B</math>, <math>P</math> be on the same circle (<math>\omega_2</math>), and <math>A'</math>, <math>B'</math>, <math>P</math> be on the same circle (<math>\omega_3</math>).
+<math>m\angle APB= m\angleA'PB'</math> because they are vertical angles. By power of point, <math>(AP)(A'P)=(BP)(B'P)\rightarrow\frac{AP}{B'P}=\frac{BP}{A'P}</math>
+By the SAS triangle simlarity theory, <math>\triangle APB \sim\triangle B'PA'</math>. That implies that <math>\angle ABP\cong\angleB'PA'</math>.
+<br/>
+Let's call the interception of the common tangent plane and the plane containing <math>A</math>, <math>A'</math>, <math>B</math>, <math>B'</math>, <math>P</math>, line <math>l</math>.
+<math>l</math> must be the common tangent of <math>\omega_2</math> and <math>\omega_3</math>.
+The acute angles form by <math>l</math> and <math>\overbar{AA'}</math> are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord <math>\overbar{AP}</math> and <math>\overbar{A'P}</math> are equal.
+Similarly the central angle of chord <math>\overbar{BP}</math> and <math>\overbar{B'P}</math> are equal.
+The length of any chord with central angle <math>2\theta</math> and radius <math>r</math> is <math>2r\sin\left({\theta}\right)</math>, which can easily been seen if we drop the perpendicular from the center to the chord.
+Thus, <math>\frac{AP}{A'P}=\frac{BP}{B'P}</math>.
+By the SAS triangle simlarity theory, <math>\triangle APB \sim\triangle A'PB'</math>. That implies that <math>\angle ABP\cong\angleB'PA'</math>.
+<br/>
+That implies that <math>\angle ABP\cong\angleA'PB'\cong\angleB'PA'</math>. Thus, <math>\triangle A'PB'</math> is an isosceles triangle and since <math>\triangle APB \sim\triangle A'PB'</math>,<math>\triangle APB</math> is an isosceles triangle too.
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