Maths Olympiad Prep

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HMMT February

Geometry Difficulty 5.1 AIME, harder Find the answer

Given △ABC\triangle A B C with AB<ACA B<A C, the altitude ADA D, angle bisector AEA E, and median AFA F are drawn from AA, with D,E,FD, E, F all lying on B‾\overline{B} C}.If∡BAD=2∡DAE=2∡EAF=∡FAC. If \measuredangle B A D=2 \measuredangle D A E=2 \measuredangle E A F=\measuredangle F A C, what are all possible values of ∡\measuredangle A C B$ ?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let HH and OO be the orthocenter and circumcenter of ABCA B C, respectively: it is well-known (and not difficult to check) that ∡\measuredangle B A H=∡H=\measuredangle C A O.However,notethat∡BAH=∡BAD=∡CAF. However, note that \measuredangle B A H=\measuredangle B A D=\measuredangle C A F, so ∡\measuredangle C A F=∡F=\measuredangle C A O,thatis,, that is, Oliesonmedian lies on median A F,andsince, and since A B<A C,itfollowsthat, it follows that F=O.Therefore,∡BAC=90∘. Therefore, \measuredangle B A C=90^{\circ}. Now, we compute ∡\measuredangle A C B=∡B=\measuredangle B A D=26∡D=\frac{2}{6} \measuredangle B A C=30∘$.C=30^{\circ}\$.

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