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Problem 1875

National Olympiad, first round

Geometry Difficulty 6.4 Prove it Romanian Mathematical Olympiad · Romania

a) Show that in a right triangle with an angle of $30^\circ$ , the leg opposite to the $30^\circ$ -angle has half of the length of the hypotenuse.

b) Inside the triangle $ABC$ , with $m(\angle A) = 100^\circ$ and $m(\angle B) = 20^\circ$ , we consider the point $D$ , such that $m(\angle DAB) = 30^\circ$ and $m(\angle DBA) = 10^\circ$ . Determine $m(\angle ACD)$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

a) If $M$ is on the hypotenuse of the triangle $ABC$ , with a right angle in $A$ , with $m(\angle B) = 30^\circ$ such that $m(\angle BAM) = 30^\circ$ , then the triangle $MAC$ is equilateral. Hence $MA = CM = CA = \frac{1}{2}BC$ , and $MB = \frac{1}{2}BC$ , also.

b) We construct $D' \in (AD)$ , such that $CA = CD'$ , $D' \in \text{Int}(ABC)$ , $m(\angle ACD') = 40^\circ$ , $m(\angle D'CB) = 20^\circ$ .
If $CG \perp AD'$ , $G \in (AD')$ , and $D'F \perp BC$ , $F \in (BC)$ , then $\triangle CD'G \equiv \triangle CD'F$ (H.A.), so $D'G = D'F$ .

$D'E \perp AB$ , $E \in (BC)$ , $G$ is the midpoint of $[AD']$ , $m(\angle D'AE) = 30^\circ$ , $D'F = D'G = D'E$ , hence $D'$ is on the angle bisector of the angle $B$ , $m(\angle D'BA) = 10^\circ$ .
$D' \in (AD)$ , $D' \in (BD)$ , $D' = D$ , hence $m(\angle ACD) = 40^\circ$ .

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