Library / /462 of 1394

, 2015

Geometry Difficulty 5.1 AIME, harder Prove it United States

Problem:

In triangle $ABC$ , $AB = 2$ , $AC = 1 + \sqrt{5}$ , and $\angle CAB = 54^{\circ}$ . Suppose $D$ lies on the extension of $AC$ through $C$ such that $CD = \sqrt{5} - 1$ . If $M$ is the midpoint of $BD$ , determine the measure of $\angle ACM$ , in degrees.

Solution

Solution:

Let $E$ be the midpoint of $\overline{AD}$ . $EC = \sqrt{5} + 1 - \sqrt{5} = 1$ , and $EM = 1$ by similar triangles ( $ABD \sim EMD$ ). $\triangle ECM$ is isosceles, with $m \angle CEM = 54^{\circ}$ . Thus $m \angle ACM = m \angle ECM = \frac{180 - 54}{2} = 63^{\circ}$ .

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 reproduced verbatim; metadata (topic, difficulty) added by this project.