Difference between revisions of "1992 USAMO Problems/Problem 4"
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Consider the plane through <math>A,A',B,B'</math>. This plane, of course, also contains <math>P</math>. We can easily find the <math>\triangle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By symmetry, <math>BB'=CC'</math> and <math>CC'=AA'</math>, and hence <math>AA'=BB'=CC'</math> as desired. | Consider the plane through <math>A,A',B,B'</math>. This plane, of course, also contains <math>P</math>. We can easily find the <math>\triangle APB</math> is isosceles because the base angles are equal. Thus, <math>AP=BP</math>. Similarly, <math>A'P=B'P</math>. Thus, <math>AA'=BB'</math>. By symmetry, <math>BB'=CC'</math> and <math>CC'=AA'</math>, and hence <math>AA'=BB'=CC'</math> as desired. | ||
<math>\mathbb{QED.}</math> | <math>\mathbb{QED.}</math> | ||
− | === Add-on=== | + | ==== Add-on==== |
By another person ^v^ | By another person ^v^ | ||
Revision as of 07:53, 19 July 2016
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
Solution 1
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 ().
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 are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord and are equal.
Similarly the central angle of chord and 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 . Thus, is an isosceles triangle and since , then must be 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.
See Also
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.