Competition problem — Geometry — Maths Olympiad Prep

Library / /176 of 520

Geometry Difficulty 5.0 AIME, harder Find the answer

6. In an acute triangle $ABC$ , the altitudes from $AC$ and $AB$ are $BE$ and $CF$ respectively, with $E$ and $F$ being the feet of the perpendiculars, and $BE$ and $CF$ intersect at $H$ . If $\angle BAC=60^{\circ}, BC=2$ , then the length of $AK$ is:

Pick one

Solution

6. (B).

In the right-angled $\triangle A B E$ , $A E$ $=A B \cdot \cos \angle B A C$ . Similarly, $A F=A C \cdot \cos \angle B A C$ .
In $\triangle A E F$ ,
$\begin{array}{l} E F^{2}= A E^{2}+A F^{2} \\ -2 A E \cdot A F \cdot \cos \angle B A C \\ = \cos ^{2} \angle B A C\left(A B^{2}+A C^{2}\right. \\ -2 A B \cdot A C \cdot \cos \angle B A C) \\ = B C^{2} \cdot \cos ^{2} \angle B A C . \end{array}$

Given $B C=2, \angle B A C=60^{\circ}$ , then $E F=2 \cos 60^{\circ}=1$ .
Since $A, F, H, E$ are concyclic, $A H$ is the diameter of the circle. Let it be $d$ , then by the Law of Sines $E F=d \cdot \sin \angle B A C$ , i.e., $d=\frac{E F}{\sin \angle B A C}=\frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.