Competition problem — Geometry — Maths Olympiad Prep

Library / /12 of 18

Geometry Difficulty 6.5 National olympiad Prove it Belarus

Let $AM$ be the median of the triangle $ABC$ , $B_1$ be the foot of the perpendicular $BB_1$ from $B$ onto the bisector of the angle $BMA$ , $C_1$ be the foot of the perpendicular $CC_1$ from $C$ onto the bisector of the angle $AMC$ . The ray $MA$ intersects the segment $B_1C_1$ at a point $A_1$ .
Find the value of the ratio $B_1A_1/A_1C_1$ .

Solution

Answer: 1.
Let $\angle BMA = 2x$ , then $\angle AMC = 180^\circ - 2x$ . Since $MB_1$ is the bisector of the angle $BMA$ , we have $\angle BMB_1 = \angle B_1MA = x$ .
Similarly, $\angle AMC_1 = \angle C_1MC = 90^\circ - x$ . The triangle $BMB_1$ is a right-angled triangle, so $\angle B_1BM = 90^\circ - \angle B_1MB = 90^\circ - x$ . The right-angled triangles $BMB_1$ and $MCC_1$ are equal ( $BM = MC$ , $\angle B_1BM = \angle C_1MC = 90^\circ - x$ ). So $BB_1 = MC_1$ . Since $\angle B_1BM = \angle C_1MC$ , we have $BB_1 \parallel MC_1$ . Therefore $BMC_1B_1$ is a parallelogram. Since $B_1C_1 \parallel BM$ , we have $\angle BMB_1 = \angle A_1B_1M = \angle B_1MA_1$ .
So, the triangle $B_1MA_1$ is an isosceles triangle: $MA_1 = A_1B_1$ . Similarly, the triangle $MC_1A_1$ is an isosceles triangle: $MA_1 = C_1A_1$ . Therefore, we have $B_1A_1/A_1C_1 = MA_1/MA_1 = 1$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.