Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.8 AIME, harder Prove it New Zealand

Problem:

Let $ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$ , $\angle ABC = 70^{\circ}$ , and $AB = 1$ . Let $M$ be the midpoint of $BC$ . Let $D$ be the point on the extension of $AM$ beyond $M$ such that $\angle CDA = 110^{\circ}$ . Find the length of $CD$ .

Solution

Solution:

Construct point $E$ so that $ABEC$ is a rectangle. The diagonals of any rectangle bisect each other, that is, they meet at each other's midpoints. Hence $AE$ and $BC$ meet at $M$ , i.e. $E$ lies on line $AM$ .

By symmetry in rectangle $ABEC$ , we have
$\angle CEA = \angle ABC = 70^{\circ}.$
By angles on a line,
$\angle CDE = 180^{\circ} - \angle CDA = 180^{\circ} - 110^{\circ} = 70^{\circ}.$
So triangle $CDE$ is isosceles with $CD = CE$ , because
$\angle CED = \angle CDE = 70^{\circ}.$
Opposite sides in a rectangle are equal so $CE = AB = 1$ , hence $CD$ has length $1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.