55rd Ukrainian National Mathematical Olympiad - Third Round (Second Tour)

Question

A straight line that contains the center of rectilinear triangle

A B C

intersects straight lines

A B

,

B C

and

C A

in points

C_{1}

,

A_{1}

and

B_{1}

respectively. Let

A_{2}

be the symmetric point to

A_{1}

about the middle of

B C

; points

B_{2}

and

C_{2}

are defined in a similar manner. Prove that points

A_{2}

,

B_{2}

and

C_{2}

are on a line that touches the circle inscribed in the triangle

A B C

.

(Serdiuk Nazar)

Fig. 25

Accepted Answer

Let points B_1 and C_1 be on the sides, and A_1 is on the extension of the side BC in direction of point B. By the Menelaus' theorem CB_2 AC_2 BA_2B_2A C_2B A_2C = AB_1 BC_1 CA_1B_1C C_1A A_1B = 1, so points A_2, B_2 and C_2 are collinear (fig. 25). Let M be the middle of AB, N be the middle of AC, K be the base of the perpendicular dropped from a point I, the center of the inscribed circle, to a straight line B_2C_2. Triangles C_2IM and C_1IM are equal, so C_2I = IC_1, C_2IC_1 = 180^ - 2 B_1C_1A. Similarly, B_2I = B_1I and B_2IB_1 = 180^ - 2 C_1B_1A. Then, B_2IC_2 = 2 B_1C_1A + 2 C_1B_1A - 180^ = 2(180^ - A) - 180^ = 60^. Since B_2IC_2 = B_1AC_1 and B_2IC_2I = B_1IC_1I = B_1AC_1A (I is a base of the bisectrix AI of the triangle B_1AC_1), triangles B_2IC_2 and B_1AC_1 are homothetic, henc…