Difference between revisions of "1992 USAMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
+ | ===Solution 1=== | ||
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==== | ||
+ | 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\angle A'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\angle B'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>\overline{AA'}</math> are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord <math>\overline{AP}</math> and <math>\overline{A'P}</math> are equal. | ||
+ | |||
+ | Similarly the central angle of chord <math>\overline{BP}</math> and <math>\overline{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\angle B'PA'</math>. | ||
+ | |||
+ | <br/> | ||
+ | That implies that <math>\angle ABP\cong\angle A'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>, then<math>\triangle APB</math> must be an isosceles triangle too. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | |||
+ | Call the large sphere <math> O_1</math>, the one containing <math>A</math> <math> O_2</math>, and the one containing <math> A'</math> <math>O_3</math>. The centers are <math> c_1</math>, <math> c_2,</math> and <math> c_3</math>. | ||
+ | |||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | 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). | ||
+ | |||
+ | |||
+ | 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. | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{USAMO box|year=1992|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 18:37, 25 June 2021
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 , and the one containing . 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.