Let $ \triangle ABC$ be a triangle with incircle $ \omega(I,r)$and circumcircle $ \zeta(O,R)$.Let $ l_{a}$ be the angle bisector of $ \angle BAC$.Denote $ P=l_{a}\cap\zeta$.Let $ D$ be the point of tangency $ \omega$ with $ [BC]$.Denote $ Q=PD\cap\zeta$.Show that $ PI=QI$ if $ PD=r$.
Problem
Source: SRMC 2002
Tags: geometry, circumcircle, angle bisector, geometry proposed
22.09.2007 14:44
From angles $ \angle DQB =\angle PQB =\angle PAB =\angle PAC =\angle PBC =\angle PBD$ So circumcircle of $ \bigtriangleup QDB$ is tangent to $ PB$. From tangent teorem and $ PB = PI$ (it is well-known) we get $ PD\cdot PQ = PB^{2}= PI^{2}\Rightarrow$ circumcircle of $ \bigtriangleup DIQ$ ıs tangent to $ PI\Rightarrow\angle IQP =\angle IQD =\angle DIP =\angle IPD =\angle IPQ\Rightarrow IP = IQ$
22.09.2007 14:55
Dear,cefer,Did you take participant in this olympiad,i mean not exactly in 2002's,but in SRMC at all.Thank you for your nice solution.
22.09.2007 15:30
I participated in 2005-2006, but i dont know my teachers sent our results or not. I think in 2006 they were late for sendıng answers.
22.09.2007 15:39
Where must your teacher sent your results?I suppose that in your country,as in mine,APMO and SRMC come together,APMO results are sent to Korea,but i don't know about SRMC..
22.09.2007 16:21
Very nice problem and the Cefer's proof ! Here is another proof. Erken wrote: Let $ ABC$ be a triangle with the incircle $ \omega(I,r)$ and the circumcircle $ \Omega (O,R)$. Let $ l_{a}$ be the angle bisector of the angle $ \widehat{BAC}$. Denote $ P = l_{a}\cap\Omega$. Let $ D$ be the point of tangency of the circle $ \omega$ with the side $ [BC]$. Denote the second intersection $ Q\in PD\cap\Omega$. Show that $ PD = r\implies IQ = IP$. Proof. Denote the second intersection $ A'\in AO\cap\Omega$. Since $ \widehat{DIP}\equiv\widehat{IAA'}$ and the power $ p_{\Omega}(I)$ of the point $ I$ w.r.t. the circle $ \Omega$ is $ -p_{\Omega}(I) = AI\cdot IP = 2Rr = ID\cdot AA'$, i.e. $ \frac{ID}{IP}=\frac{AI}{AA'}$ results $ \triangle IDP\sim\triangle AIA'$. Thus $ \widehat{IPD}\equiv\widehat{AA'I}$ what means that the point $ \boxed{\ Q\in PD\cap IA'\ }$. Therefore, $ \widehat{DQI}\equiv\widehat{IAA'}\equiv\widehat{DIP}$. In conclusion, $ PD = r$ $ \Longleftrightarrow$ $ PD = DI$ $ \Longleftrightarrow$ $ \widehat{DPI}\equiv\widehat{DIP}$ $ \Longleftrightarrow$ $ \widehat{DPI}\equiv\widehat{DQI}$ $ \Longleftrightarrow$ $ IP = IQ$. Remark. $ \boxed{\ PD = r\Longleftrightarrow IP = IQ\Longleftrightarrow IO\perp IA\Longleftrightarrow b+c = 2a\ }$.
22.09.2007 16:34
Erken wrote: Where must your teacher sent your results?I suppose that in your country,as in mine,APMO and SRMC come together,APMO results are sent to Korea,but i don't know about SRMC.. I think they send SRMC results to Turkey