Given are three pairwise externally tangent circles $ K_{1}$ , $ K_{2}$ and $ K_{3}$. denote by $ P_{1}$ tangent point of $ K_{2}$ and $ K_{3}$ and by $ P_{2}$ tangent point of $ K_{1}$ and $ K_{3}$. Let $ AB$ ($ A$ and $ B$ are different from tangency points) be a diameter of circle $ K_{3}$. Line $ AP_{2}$ intersects circle $ K_{1}$ (for second time) at point $ X$ and line $ BP_{1}$ intersects circle $ K_{2}$(for second time) at $ Y$. If $ Z$ is intersection point of lines $ AP_{1}$ and $ BP_{2}$ prove that points $ X$, $ Y$ and $ Z$ are collinear.
Problem
Source: Federation of Bosnia and Herzegovina, 4th grades, 2008.
Tags: trigonometry, geometry, circumcircle, geometry proposed
TTsphn
03.05.2008 06:06
Here is my solution . $ YZ$ cut $ AB$ at M . Apply Menelaus law : $ \frac {\overline{MA}}{\overline{MB}}.\frac {\overline{YB}}{\overline{YP_1}}.\frac {\overline{XP_1}}{\bar{XA}} = 1$ We have : ${ \frac {\overline{YB}}{\bar{YP_1}} = 1 + \frac {R_2}{R_3}}$ $ \frac {\overline{XP_1}}{\overline{XA}} = \frac {S_[BXP_1]}{S_[BAX]} = - \frac {BP_1}{AP_2}.\frac {P_1P_2}{AB}$ But we have $ \frac {P_1P_2}{AB} = \sin{\frac{\angle O_1O_3O_2}{2}} = \sqrt {\frac {R_1R_2}{(R_3 + R_1)(R_3 + R_2)}}$ This formula only depend on $ R_1 , R_2,R_3,AP_2,BP_1$ so if $ ZX$ cut $ AB$ at N then $ \frac {\overline{NA}}{\overline{NB}} = \frac {\overline{MA}}{\overline{MB}}$ It mean that $ M\equiv N$ so $ Y,Z,X$ is collinear .
April
03.05.2008 08:34
Denote $ P_3$ the tangent point of $ K_1$ and $ K_2$, $ T$ the intersection of $ AP_2$, $ BP_1$. We can easily to prove that $ X$, $ Y$, $ P_3$ are collinear and quadrilateral $ P_1TP_2P_3$ is cyclic, denote $ Z'$ the intersection of its circumcircle and $ YZ$.
We have $ \angle P_1P_3T = \angle P_1P_2T = \angle P_1BA = \angle P_1YS = \angle P_1P_3S$, where $ S$ is the second intersection of $ K_2$ and $ AP_1$. It implies $ T$, $ S$, $ P_3$ are collinear, i.e., $ \angle TP_3Z' = 90^{\circ}$. In conclusion, $ \angle TP_1Z' = \angle TP_2Z' = 90^{\circ}$, hence $ Z\equiv Z'$.
Hope someone can rewrite it in more details. Thanks
Attachments:
t=202955.pdf (10kb)
tmrfea
06.05.2008 15:35
April wrote, that quadrilateral $ P_{1}P_{2}TP_{3}$ is cyclic. If we assume this is true, then the task becomes trivial. Therefore I think, that the major problem is to prove that $ P_{1}P_{2}TP_{3}$ is cyclic. Can anybody prove that?
The QuattoMaster 6000
07.05.2008 06:17
tmrfea wrote: April wrote, that quadrilateral $ P_{1}P_{2}TP_{3}$ is cyclic. If we assume this is true, then the task becomes trivial. Therefore I think, that the major problem is to prove that $ P_{1}P_{2}TP_{3}$ is cyclic. Can anybody prove that?
Let the centers of $ K_1$, $ K_2$, and $ K_3$ be $ O_1$, $ O_2$, and $ O_3$ respectively. Let $ \angle O_1O_3O_2 = 180 - 2\alpha - 2\beta$ and $ \angle O_3O_1O_2 = 2\alpha$, so $ \angle O_1O_2O_3 = 2\beta$. Now, notice that $ O_1$, $ P_3$, and $ O_2$ are collinear as are $ O_1$, $ P_2$, and $ O_3$ and as are $ O_3$, $ P_1$, and $ O_2$. Now, notice that $ \angle O_1P_3P_2 = 90 - \frac {\angle P_3O_1P_2}{2} = 90 - \alpha$. Similarly, $ \angle P_1P_3O_2 = 90 - \beta$. It follows that $ \angle P_2P_3P_1 = \alpha + \beta$. Now, notice that $ \angle P_2TP_1 = \angle AP_2B + \angle P_2AP_3$. Since $ O_3$ is the center of $ K_3$, we have that $ \angle P_2AP_1 = \frac {\angle P_2O_3P_1}{2} = 90 - \alpha - \beta$ and since $ AB$ is a diameter, we can also notice that $ \angle BP_2A = 90$. Hence, $ \angle P_2TP_1 + \angle P_2P_3P_1 = \alpha + \beta + 180 - \alpha - \beta = 180$, so $ TP_2P_3P_1$ is cyclic.
tmrfea
07.05.2008 10:38
Thank you The QuattoMaster 6000. Now I can give the complete solution. There are two possible configurations, with different possitions of points A and B. I will attach the sketch to show them. Let us denote $ AP_{2}\cap K_{1}=X_{1}, AP_{1}\cap K_{2}=X{2}, BP_{1}\cap K_{2}=Y_{1}, BP_{2}\cap K_{1}=Y_{2}, AP_{1}\cap BP_{2}=Z_{1}, AP_{2}\cap BP_{1}=Z_{2}.$ Therefore we shall prove that $ X_{1}, Y_{1}, Z_{1}$ are collinear and $ X_{2}, Y_{2}, Z_{2}$, as well. Now we denote $ K_{1}\cap K_{2}=P_{3}$. First I will prove that $ X_{1}, P_{3}, Y_{1}$ and $ X_{2}, P_{3}, Y_{2}$ are collinear. Since $ \angle Y_{2}P_{3}X_{2}=\angle Y_{2}P_{3}P_{2}+\angle P_{2}P_{3}P_{1}+\angle P_{1}P_{3}X_{2}=(180-\angle P_{3}Y_{2}P_{2}- \angle P_{3}P_{2}Y_{2})+ \angle P_{2}P_{3}P_{1}+ (180-\angle P_{3}P_{1}X_{2}- \angle P_{3}X_{2}P_{1})= 180-\angle P_{3}Y_{2}P_{2}-(180-\angle P_{3}P_{2}Z_{1})+ \angle P_{2}P_{3}P_{1} + 180- (180- \angle P_{3}P_{1}Z_{1})- \angle P_{3}X_{2}P_{1}= \angle P_{3}P_{2}Z_{1} + \angle P_{3}P_{1}Z_{1}+\angle P_{2}P_{3}P_{1} - \angle P_{3}Y_{2}P_{2} - \angle P_{3}X_{2}P_{1}=\angle P_{3}P_{2}Z_{1} + \angle P_{3}P_{1}Z_{1}+\angle P_{2}P_{3}P_{1} - (180-\angle P_{2}Z_{1}P_{3})=P_{3}P_{2}Z_{1} + \angle P_{3}P_{1}Z_{1}+\angle P_{2}P_{3}P_{1}+\angle P_{2}Z_{1}P_{3}-180= 180 \rightarrow X_{2}, P_{3}, Y_{2}$ are collinear. The second collinearity: because $ Y_{1}X_{2} (\angle Y_{1}P_{1}X_{2}=90)$ and $ X_{1}Y_{2}(\angle X_{1}P_{2}Y{2}=90$ are diameters of $ K_{2}$ and $ K_{1}$, respectively, $ \angle Y_{1}P_{3}X_{2}=\angle X_{1}P_{3}Y_{2}=90$. Now, since $ X_{2}, P_{3}, Y_{2}$ and $ \angle Y_{1}P_{3}X_{2}=\angle X_{1}P_{3}Y_{2}$, also $ X_{1}, P_{3}, Y_{1}$ are collinear. Therefore to show collinearities $ X_{1}, Y_{1}, Z_{1}$ and $ X_{2}, Y_{2}, Z_{2}$, it sufficies to prove collinearities $ P_{3}, Y_{1}, Z_{1}$ and $ X_{2}, Z_{2}, P_{3}$. Now we use the fact, that $ P_{1}P_{2}P_{3}Z_{1}Z_{2}$ is cyclic (which was proved by The QuattoMaster 6000). First we show that $ P_{3}, Y_{1}, Z_{1}$ are collinear. $ P_{3}, Y_{1}, Z_{1}$ are collinear $ \Leftrightarrow \angle Y_{1}P_{3}P_{1}= \angle Z_{1}P_{3}P_{1}$. Now $ \angle Z_{1}P_{3}P_{1} = \angle Z_{1}P_{2}P_{1} =\angle BP_{2}P_{1}=\angle BAP_{1}= \angle P_{1}X_{2}Y_{1}$ (where the last is true, since $ \Delta P_{1}BA \approx \Delta P_{1}YX_{2}$). Now $ \angle P_{1}X_{2}Y_{1}= \angle P_{1}P_{3}Y_{1}= \angle P_{1}P_{3}Z_{1} \rightarrow P_{3}, Y_{1}, Z_{1}$ are collinear $ \rightarrow X_{1}, Y_{1}, Z_{1}$ are collinear. $ X_{2}, Z_{2}, P_{3}$ are collinear, since $ \angle Z_{2}P_{3}Z_{1}=\angle Z_{2}P_{3}Y_{1}=90$ and $ \angle Y_{2}P_{3}Y_{1}= \angle Y_{1}P_{3}X_{2}=90$. Hence $ \angle Z_{2}P_{3}Y_{1}= Y_{2}P_{3}Y_{1} \rightarrow Y_{2}, Z_{2}, P_{3}$ are collinear $ \rightarrow Y_{2}, Z_{2}, X_{2}$ are collinear. I apologize but I cannot attach the sketch. In fact it is the same as April one`s, just that $ P_{2}$ and $ P_{1}$ are changed.