midpoint wanted, equilateral triangles on sides of a triangle related

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Source: 2006 Oral Moscow Geometry Olympiad grades 8-9 p5

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#1 h Oct 17, 2020, 10:21 AM

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Equilateral triangles $ABC_1, BCA_1, CAB_1$ are built on the sides of the triangle $ABC$ to the outside. On the segment $A_1B_1$ to the outer side of the triangle $A_1B_1C_1$ , an equilateral triangle $A_1B_1C_2$ is constructed. Prove that $C$ is the midpoint of the segment $C_1C_2$ .

(A. Zaslavsky)

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#2 h May 13, 2023, 2:36 PM

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The solution will consist of two parts. In the first part, we will prove that the points $C$ , $C_1$ , $C_2$ are collinear using Torrichelli point. In the second part we will be proving that $CC_1=CC_2$ using rotations

Part 1: Proving that $C$ , $C_1$ and $C_2$ are collinear
Let $AA_1\cap BB_1=T$ . Then $T$ is the Torrichelli point of triangle $ABC$ and $T$ is the Torichelli point of triangle $A_1B_1C$ . So $C$ , $T$ , $C_1$ are collinear. Analogously, points $C$ , $C_2$ and $T$ are collinear. This means that points $C$ , $C_1$ , $C_2$ all lie on line $CT$ , hence are collinear

Part 2: Proving $CC_1=CC_2$
Let's prove that $AA_1=BB_1$ . Consider rotation centered at point $C$ taking $B_1$ to $A$ , then $B$ is taken to $AA_1$ , meaning that the segment $BB_1$ is taken to $AA_1$ . Analogously, rotating around $B$ you can get $CC_1=AA_1$ . Thus, $AA_1=BB_1=CC_1$ . Note that the same can be done with vertices of triangle $A_1B_1C_2$ proving $CC_2=AA_1=BB_1$ . From this follows, that $CC_1=CC_2$

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This post has been edited 2 times. Last edited by Bexultan, Jul 19, 2023, 11:20 AM
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