Competition problem — Geometry — Maths Olympiad Prep

Library / /147 of 520

Geometry Difficulty 6.2 National olympiad Prove it

95. As shown in the figure, triangle $ABC$ is a right triangle, $\angle ACB=90^{\circ}$ . $M_{1}, M_{2}$ are any two points inside $\triangle ABC$ , $M$ is the midpoint of segment $M_{1} M_{2}$ , lines $BM_{1}$ , $BM_{2}$ , $BM$ intersect side $AC$ at points $N_{1}$ , $N_{2}$ , $N$ respectively. Prove: $\frac{M_{1} N_{1}}{B M_{1}}+\frac{M_{2} N_{2}}{B M_{2}} \geqslant 2 \frac{M N}{B M}$ . (2010 China National High School Mathematics Competition)

Solution

95. Proof Let $H_{1}, H_{2}, H$ be the projections of $M_{1}, M_{2}, M$ on the line $B C$ . Then $\frac{M_{1} N_{1}}{B M_{1}}=$ $\frac{H_{1} C}{B H_{1}}, \frac{M_{2} N_{2}}{B M_{2}}=\frac{H_{2} C}{B H_{2}}, \frac{M N}{B M}=\frac{H C}{B H}=\frac{H_{1} C+H_{2} C}{B H_{1}+B H_{2}}$ . Without loss of generality, let $B C=1, B H_{1}=x, B H_{2}=y$ , then $\frac{M_{1} N_{1}}{B M_{1}}=\frac{H_{1} C}{B H_{1}}=\frac{1-x}{x}, \frac{M_{2} N_{2}}{B M_{2}}=\frac{H_{2} C}{B H_{2}}=\frac{1-y}{y}, \frac{M N}{B M}=\frac{1-x+1-y}{x+y}$ .

Thus, the original inequality is equivalent to $\frac{1-x}{x}+\frac{1-y}{y} \geqslant 2 \frac{1-x+1-y}{x+y}$ .

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: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.