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Annual Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Given $\triangle ABC$ with $AB < AC$ , the altitude $AD$ , angle bisector $AE$ , and median $AF$ are drawn from $A$ , with $D, E, F$ all lying on $\overline{BC}$ . If $\measuredangle BAD = 2 \measuredangle DAE = 2 \measuredangle EAF = \measuredangle FAC$ , what are all possible values of $\measuredangle ACB$ ?

Solution

Solution:

$30^{\circ}$ or $\pi / 6$ radians

Let $H$ and $O$ be the orthocenter and circumcenter of $ABC$ , respectively: it is well-known (and not difficult to check) that $\measuredangle BAH = \measuredangle CAO$ . However, note that $\measuredangle BAH = \measuredangle BAD = \measuredangle CAF$ , so $\measuredangle CAF = \measuredangle CAO$ , that is, $O$ lies on median $AF$ , and since $AB < AC$ , it follows that $F = O$ . Therefore, $\measuredangle BAC = 90^{\circ}$ .

Now, we compute $\measuredangle ACB = \measuredangle BAD = \frac{2}{6} \measuredangle BAC = 30^{\circ}$ .

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