Difference between revisions of "1992 USAMO Problems/Problem 4"
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That implies that <math>\angle ABP\cong\angleA'PB'\cong\angle B'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. | That implies that <math>\angle ABP\cong\angleA'PB'\cong\angle B'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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+ | ==Solution 2== | ||
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+ | Call the large sphere <math> O_1</math>, the one containing A <math> O_2</math>, and the one containing <math> A'</math> O_3. The centers are <math> c_1</math>, <math> c_2,</math> and <math> c_3</math>. | ||
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+ | Since two spheres always intersect in a circle , we know that A,B, and C must lie on a circle (<math> w_1</math>)completely contained in <math> O_1</math> and <math> O_2</math> | ||
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+ | Similarly, A', B', and C' must lie on a circle (<math>w_2</math>) completely contained in <math> O_1</math> and <math> O_3</math>. | ||
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+ | So, we know that 3 lines connecting a point on <math> w_1</math> and P hit a point on <math> w_2</math>. This implies that <math> w_1</math> projects through P to <math> w_2</math>, which in turn means that <math> w_1</math> is in a plane parallel to that of <math> w_2</math>. Then, since <math> w_1</math> and <math> w_2</math> lie on the same sphere, we know that they must have the same central axis, which also must contain P (since the center projects through P to the other center). | ||
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+ | So, all line from a point on <math> w_1</math> to P are of the same length, as are all lines from a point on <math> w_2</math> to P. Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal. | ||
== Resources == | == Resources == | ||
{{USAMO box|year=1992|num-b=3|num-a=5}} | {{USAMO box|year=1992|num-b=3|num-a=5}} |
Revision as of 18:19, 25 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 .
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 .
That implies that $\angle ABP\cong\angleA'PB'\cong\angle B'PA'$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, is an isosceles triangle and since , is an isosceles triangle too.
Solution 2
Call the large sphere , the one containing A , and the one containing O_3. The centers are , and .
Since two spheres always intersect in a circle , we know that A,B, and C must lie on a circle ()completely contained in and
Similarly, A', B', and C' must lie on a circle () completely contained in and .
So, we know that 3 lines connecting a point on and P hit a point on . This implies that projects through P to , which in turn means that is in a plane parallel to that of . Then, since and lie on the same sphere, we know that they must have the same central axis, which also must contain P (since the center projects through P to the other center).
So, all line from a point on to P are of the same length, as are all lines from a point on to P. Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal.
Resources
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |