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Austria competition problems, 2010

Geometry Difficulty 5.7 AIME, harder Prove it Austria

Let $ABC$ be a right-angled triangle with the right angle at $C$ such that the side $BC$ is longer than the side $AC$ . The perpendicular bisector of $AB$ intersects the line $BC$ in $D$ and the line $AC$ in $E$ . We assume that $DE$ and the side $AB$ have the same length.

Determine the angles of the triangle $ABC$ .

Solution

The angle $\angle ABC$ is denoted by $\beta$ . As $BC$ is normal to $AE$ and $DE$ is normal to

Abbildung 1: Problem 4.
$AB$ , the angles $\angle ABC$ and $\angle AED$ are of equal measure. As we have $\angle ACB = \angle DCE = 90^\circ$ and, by assumption, $\overline{AB} = \overline{DE}$ , the triangles $ABC$ and $DEC$ are congruent.

This yields $\overline{BC} = \overline{CE}$ which implies that the triangle $BCE$ is an isosceles right-angled triangle with $\angle CEB = \angle CBE = 45^\circ$ .

Furthermore, we have $\beta = \angle CED = \angle DEB$ , as $E$ lies on the perpendicular bisector of $AB$ .

Thus we obtain $45^\circ = \angle CEB = 2\beta$ and therefore $\beta = 22.5^\circ$ and $\alpha = \angle CAB = 67.5^\circ$ .

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