Two circles with centers $ S_{1}$ and $ S_{2}$ are externally tangent at point $ K$. These circles are also internally tangent to circle $ S$ at points $ A_{1}$ and $ A_{2}$, respectively. Denote by $ P$one of the intersection points of $ S$ and common tangent to $ S_{1}$ and $ S_{2}$ at $ K$.Line $ PA_{1}$ intersects $ S_{1}$ at $ B_{1}$ while $ PA_{2}$ intersects $ S_{2}$ at $ B_{2}$. Prove that $ B_{1}B_{2}$ is common tangent of circles $ S_{1}$ and $ S_{2}$.
Problem
Source: Federation of Bosnia and Heryegovina, 3rd grades, 2008.
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The QuattoMaster 6000
29.04.2008 02:31
delegat wrote: Two circles with centers $ S_{1}$ and $ S_{2}$ are externally tangent at point $ K$. These circles are also internally tangent to circle $ S$ at points $ A_{1}$ and $ A_{2}$, respectively. Denote by $ P$one of the intersection points of $ S$ and common tangent to $ S_{1}$ and $ S_{2}$ at $ K$.Line $ PA_{1}$ intersects $ S_{1}$ at $ B_{1}$ while $ PA_{2}$ intersects $ S_{2}$ at $ B_{2}$. Prove that $ B_{1}B_{2}$ is common tangent of circles $ S_{1}$ and $ S_{2}$. Just out of curiosity, by 3rd graders, do you mean that they are around 8 years old? These problems would seem extremely tough to those young students. Anyways, here is the solution:
Let the common tangent at $ A_2$ to $ S_2$ meet $ B_1B_2$ at $ Q$. Furtermore, let the tangent from $ Q$ to $ S_2$ meet $ S_2$ at $ B'$ so that $ B'\ne A_2$. It follows that $ PB_1\cdot PA_1 = PK^2 = PB_2\cdot PA_2$, so $ A_1A_2B_2B_1$ is cyclic. Thus,
\[ \angle B_2A_2Q = \angle A_2A_1P = \angle B_1B_2P = \angle B_2A_2Q
\]
so $ A_2Q = B_2Q$. Yet, $ B'Q$ and $ QA_2$ are common tangents from the same circle. We have that $ B_2Q = A_2Q = B'Q$, which means that the circle with radius $ QA_2$ and center $ Q$ meets $ S_2$ at $ A_2$, $ B'$, and $ B_2$. If all of these points were different, then two circles intersect at three points, which is impossible. Thus, $ B_2 = B'$ since $ A_2\ne B_2$ and $ B'\ne A_2$. It follows that $ QB_1$ is tangent to $ S_2$, so $ B_1B_2$ is tangent to $ S_2$. Similarly, it is tangent to $ S_1$, so it is a common tangent of the two circles.
delegat
29.04.2008 03:43
Quote: Just out of curiosity, by 3rd graders, do you mean that they are around 8 years old? But it is Bosnian competition We prefer easier problems 3rd grades is equivalent to 11th grades (we hae 8+4 grades -primary + secondary schools) so students that were asked this question are aroun 17 years old. Thanks for solution The QuattoMaster 6000 Kind regards
April
30.04.2008 01:29
delegat wrote: Two circles with centers $ S_{1}$ and $ S_{2}$ are externally tangent at point $ K$. These circles are also internally tangent to circle $ S$ at points $ A_{1}$ and $ A_{2}$, respectively. Denote by $ P$ one of the intersection points of $ S$ and common tangent to $ S_{1}$ and $ S_{2}$ at $ K$. Line $ PA_{1}$ intersects $ S_{1}$ at $ B_{1}$ while $ PA_{2}$ intersects $ S_{2}$ at $ B_{2}$. Prove that $ B_{1}B_{2}$ is common tangent of circles $ S_{1}$ and $ S_{2}$. As dear The QuattoMaster 6000 showed: $ A_1A_2B_2B_1$ is cyclic. Let's call $ C_1$, $ C_2$ be the second intersections of $ A_1A_2$ with $ S_1$, $ S_2$, respectively. It's well-known (?) that $ B_1C_1$ and $ PA_2$ are parallel. It implies $ \angle B_2A_2A_1 = \angle B_1C_1A_1 = \angle PB_1B_2$, i.e., $ B_1B_2$ is tangent to $ S_1$ at $ B_1$. Similarly, $ B_1B_2$ is tangent to $ S_2$ at $ B_2$, this completes our proof.
Dilshodbek
21.12.2016 20:40
Good solutions