Library / /14 of 25

, 2008

Geometry Difficulty 5.7 AIME, harder Prove it Ukraine

Point $M$ is placed on the side $BC$ of the triangle $ABC$ so that $BM = AC$ . $H$ is the foot of the perpendicular dropped on $AM$ from point $B$ . We know that $BH = CM$ , and $\angle MAC = 30^\circ$ . Find degree measure of the $\angle ACB$ .

Answer: $15^\circ$ or $105^\circ$ .

Solution

Let $AC = BM = b$ , $BH = MC = h$ . We can easily show that point $H$ can not be on the segment $AM$ . Let's consider two cases.

1) Point $H$ is on the ray $MA$ (fig.1).
Let's draw perpendicular $LC \perp AM$ , point $L$ is on the line $AM$ . As one of the angles of the right-angled triangle $ALC$ is $30^\circ$ , $LC = \frac{1}{2}b$ . Therefore, $\angle MCL = \angle MBH \Rightarrow$

Fig.1

$\cos \angle MCL = \frac{LC}{MC} = \frac{b}{2h} = \cos \angle MBH = \frac{BH}{BM} = \frac{h}{b} \Rightarrow \frac{b}{2h} = \frac{h}{b}$ and $b^2 = 2h^2$ , which means that $\cos \angle MCL = \frac{b}{2h} = \frac{1}{\sqrt{2}}$ and $\angle MCL = 45^\circ$ . Since $\angle LCA = 60^\circ$ , $\angle ACB = 15^\circ$ .

2) The point $H$ is on the ray $AM$ (fig.2). Let's draw perpendicular $LC \perp AM$ , the point $L$ is on the line $AM$ . Further we solve the problem as in the first case, and find that $\angle MCL = 45^\circ$ . Since $\angle LCA = 60^\circ$ , $\angle ACB = 15^\circ$ .

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.