Difference between revisions of "1992 USAMO Problems/Problem 4"
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<math>\mathbb{QED.}</math> | <math>\mathbb{QED.}</math> | ||
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+ | === 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>). | ||
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+ | <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> | ||
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+ | 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>. | ||
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+ | <math>l</math> must be the common tangent of <math>\omega_2</math> and <math>\omega_3</math>. | ||
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+ | 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. | ||
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+ | Similarly the central angle of chord <math>\overbar{BP}</math> and <math>\overbar{B'P}</math> are equal. | ||
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+ | 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. | ||
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+ | 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. | ||
== Resources == | == Resources == |
Revision as of 18:19, 22 April 2010
Contents
Problem
Chords , , and of a sphere meet at an interior point but are not contained in the same plane. The sphere through , , , and is tangent to the sphere through , , , and . Prove that .
Solution
Consider the plane through . This plane, of course, also contains . We can easily find the is isosceles because the base angles are equal. Thus, . Similarly, . Thus, . By symmetry, and , and hence as desired.
Add-on
By another person ^v^
The person that came up with the solution did not prove that is isosceles nor the base angles are congruent. I will add on to the solution.
There is a common tangent plane that pass through for the spheres that are tangent to each other.
Since any cross section of sphere is a circle. It implies that , , , be on the same circle (), , , be on the same circle (), and , , be on the same circle ().
$m\angle APB= m\angleA'PB'$ (Error compiling LaTeX. ! Undefined control sequence.) because they are vertical angles. By power of point,
By the SAS triangle simlarity theory, . 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 , , , , , line .
must be the common tangent of and .
The acute angles form by 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 and radius is , which can easily been seen if we drop the perpendicular from the center to the chord.
Thus, .
By the SAS triangle simlarity theory, . 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, is an isosceles triangle and since , 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 |