Difference between revisions of "1992 USAMO Problems/Problem 4"
(→Solution) |
m (→Add-on) |
||
Line 19: | Line 19: | ||
<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> | <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\ | + | 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/> | <br/> | ||
Line 35: | Line 35: | ||
Thus, <math>\frac{AP}{A'P}=\frac{BP}{B'P}</math>. | 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\ | + | 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/> | <br/> | ||
− | That implies that <math>\angle ABP\cong\angleA'PB'\cong\ | + | 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. |
== Resources == | == Resources == |
Revision as of 18:21, 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. Unknown error_msg) 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. Unknown error_msg) are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord $\overbar{AP}$ (Error compiling LaTeX. Unknown error_msg) and $\overbar{A'P}$ (Error compiling LaTeX. Unknown error_msg) are equal.
Similarly the central angle of chord $\overbar{BP}$ (Error compiling LaTeX. Unknown error_msg) and $\overbar{B'P}$ (Error compiling LaTeX. Unknown error_msg) 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. Unknown error_msg). 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 |