2021/22 AHSMC Part 1 Problem 15

Problem 15 of 16

In triangle $ABC$ , $AB = 4$ , $AC = 2\sqrt{23}$ and the measure of $\angle BAC = 46^\circ$ . $D$ is a point on $AC$ such that $AD = \sqrt{23} + 2$ and $M$ is the midpoint of $BC$ . The measure of $\angle ADM$ is

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Solution

Let $E$ be the midpoint of $AC$ . $ME$ is a midline in $\triangle ABC$ so $ME = \dfrac{AB}{2} = 2$ and $ME \parallel AB$ , and thus $\angle MED = \angle BAC = 46^\circ$ . Meanwhile $ED = AD - AE = \sqrt{23} + 2 - \sqrt{23} = 2$ . Hence $\triangle MED$ is isosceles and the measure of $\angle ADM = \dfrac{180^\circ - 46^\circ}{2} = 67^\circ$ . The answer is (C).

University of Alberta

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