Competition problem — Geometry — Maths Olympiad Prep

Library / /364 of 520

Geometry Difficulty 5.6 AIME, harder Prove it

4. As shown in Figure 10, in the acute triangle $\triangle ABC$ , $AM$ is the median on side $BC$ , and $X$ is the intersection of the lines passing through points $B$ and $C$ and tangent to the circumcircle $\odot O$ of $\triangle ABC$ . Prove that: $\frac{AM}{AX}=\cos \angle BAC$ .

Solution

(提示: Set $A X$ intersects $\odot O$ at point $A_{1}$ , connect $O B$ , $O C$ , $O A_{1}$ , $O X$ . First prove $\triangle X M A$ $\backsim \triangle X A_{1} O$ , then $\frac{A M}{A X}=\frac{O A_{1}}{O X}=\frac{O B}{O X}$ . It is also easy to know that $\angle B O X=$ $\angle B A C$ , then
$\left.\frac{A M}{A X}=\frac{O B}{O X}=\cos \angle B A C .\right)$

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.