Competition problem — Geometry — Maths Olympiad Prep

Library / /10 of 13

Geometry Difficulty 6.8 National olympiad Prove it China

As shown in the diagram, in $\triangle ABC$ , $\angle A = 60^\circ$ , $AB > AC$ , point $O$ is a circumcenter and $H$ is the intersection point of two altitudes $BE$ and $CF$ . Points $M$ and $N$ are on the line segments $BH$ and $HF$ respectively, and satisfy $BM = CN$ . Determine the value of $\frac{MH + NH}{OH}$ .

Solution

We take $BK = CH$ on $BE$ and join $OB$ , $OC$ and $OK$ .
From the property of the circumcenter of a triangle, we know that $\angle BOC = 2\angle A = 120^\circ$ . From the property of the orthocenter of a triangle, we get $\angle BHC = 180^\circ - \angle A = 120^\circ$ . So $\angle BOC = \angle BHC$ . Then four points $B$ , $C$ , $H$ and $O$ are concyclic. Hence $\angle OBH = \angle OCH$ .
In addition, $OB = OC$ and $BK = CH$ . Therefore, $\triangle BOK \cong \triangle COH$ . It follows that $\angle BOK = \angle COH$ , and $OK = OH$ .
$\text{So,} \quad \begin{aligned} \angle KOH &= \angle BOC = 120^\circ, \\ \angle OKH &= \angle OHK = 30^\circ. \end{aligned}$
In $\triangle OKH$ , by the sine rule, we get $KH = \sqrt{3}OH$ . In view of $BM = CN$ and $BK = CH$ , we get $KM = NH$ , and
$MH + NH = MH + KM = KH = \sqrt{3}OH.$
Therefore,
$\frac{MH + NH}{OH} = \sqrt{3}.$

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: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.