Let $A$ and $B$ be points on a circle $\mathcal{C}$ with center $O$ such that $\angle AOB = \dfrac {\pi}2$. Circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are internally tangent to $\mathcal{C}$ at $A$ and $B$ respectively and are also externally tangent to one another. The circle $\mathcal{C}_3$ lies in the interior of $\angle AOB$ and it is tangent externally to $\mathcal{C}_1$, $\mathcal{C}_2$ at $P$ and $R$ and internally tangent to $\mathcal{C}$ at $S$. Evaluate the value of $\angle PSR$.
Problem
Source: Romanian IMO Team Selection Test TST 1996, problem 5
Tags: geometry, rectangle, geometry proposed
Amir.S
28.09.2005 16:19
Actually I'm sure that the problem is wrong. Don't say Valentin that you solved a Wrong Problem :
Valentin Vornicu
28.09.2005 16:41
amir2 wrote: Actually I'm sure that the problem is wrong. Don't say Valentin that you solved a Wrong Problem : I forgot to mention that the third circle lies inside the $\angle AOB$. Statement is corrected now
grobber
28.09.2005 17:04
Should the answer be a fixed value? I always thought (ever since I first saw the proble, which was stated exactly as you have posted it) that $\mathcal C_1,\mathcal C_2$ should also be externally tangent to each other, in which case it's not hard to prove that the angle has $45^{\circ}$ .
Valentin Vornicu
28.09.2005 17:09
Grrr, I missed that line again. Thanks Grobber, circles $\mathcal{C}_1$ and $\mathcal{C}_2$ are externally tangent to one another!
Ashegh
28.09.2005 19:54
valentin ,it appears that the angle should be equal to 45.but in our geometry class ,we prove that the problem is wrong.by inversion i found that its wrong.i will post the method ,as soon as i can
Valentin Vornicu
28.09.2005 20:00
Please do ...
ehsan2004
29.09.2005 16:24
Solution: Suppose that $O_1$ and $O_2$ were center of circles $C_1,C_2$ respectively and $D$ be the fourth vertic of the rectangle $O_2OO_1D$.If the circles $C_1,C_2$ were tangent together in $K$ then the points $O_1,K,O_2$ are concurrent , so $O_1O_2=R_1+R_2$ s.t. $R_i , i=1,2$ are the radius of circle $C_i$ then $DO=O_1O_2$ and in the triangle $\triangle O_1OO_2$ we have $OO_1=R-R_1$ and $OO_2=R-R_2$ where $R$ is the radius of circle $C$. We know that in triangle $\triangle O_1OO_2 , O_1O_2<OO_1+OO_2$ so that $R>R_1+R_2$ so the point $D$ is in the circle $C$. We know that $S=(OD, circle C)$ and it's obviously to show that $DP=DR=DS=R-(R_1+R_2)$ so that $D$ is the center of circle $C_3$ ,then $\angle PSR=45$.