$\triangle ABC$ satisfies $\angle A=60^{\circ}$. Call its circumcenter and orthocenter $O, H$, respectively. Let $M$ be a point on the segment $BH$, then choose a point $N$ on the line $CH$ such that $H$ lies between $C, N$, and $\overline{BM}=\overline{CN}$. Find all possible value of \[\frac{\overline{MH}+\overline{NH}}{\overline{OH}}\]
Problem
Source: 2017 Taiwan TST Round 3
Tags: geometry, circumcircle
anantmudgal09
14.04.2018 09:19
Note that $O$ is the midpoint of arc $BHC$ hence $O$ is the rotation center for $BM \mapsto CN$; hence $O$ lies on $\odot(HMN)$. Now $\overline{HO}$ is the internal bisector of $\angle MHN$. Suppose $x=|MH|, y=|NH|$ then $HO=\frac{xy}{\frac{2xy}{x+y}\cos \tfrac{1}{2}\angle MHN}=\frac{x+y}{2\cos 30^{\circ}}$ thus the desired value is $\sqrt{3}$.
liekkas
19.04.2018 23:40
It was China 2nd round, 2006(2007?), Problem 1.
Kaskade
19.04.2018 23:46
https://artofproblemsolving.com/community/u331394h1593378p9884619
snakeaid
18.10.2020 16:56
Let $BC=a$, $CA=b$, $AB=c$, $\angle ABC=\beta$, $\angle ACB=\gamma$, $R$ be the circumradius of $\triangle ABC$.
Notice that
$$MH=BH-BM, NH=CN-CH, CN=BM \implies MH+NH=BH-CH$$Then it would suffice to prove that
$$\frac{(BH-CH)^2}{OH^2}=3$$With complex numbers we easily prove the formula
$$OH^2=9R^2-(a^2+b^2+c^2)$$By sine law in $\triangle ABC$ we get
$$R^2=\frac{a^2}{3}, b^2=\frac{4a^2sin^2\beta}{3}, c^2=\frac{4a^2sin^2\gamma}{3}$$Then
$$OH^2=\frac{2a^2(3-2sin^2\beta-2sin^2\gamma)}{3}$$Now by cosine law in $\triangle BHC$ we get
$$2BH \cdot CH=2(a^2-BH^2-CH^2)$$whence
$$(BH-CH)^2=3(BH^2+CH^2)-2a^2$$By sine law in $\triangle BHC$ we get
$$BH^2=\frac{4a^2sin^2(\angle HCB)}{3}=\frac{4a^2cos^2\beta}{3}=\frac{4a^2(1-sin^2\beta)}{3}$$and similarly
$$CH^2=\frac{4a^2(1-\sin^2\gamma)}{3}$$Then
$$(BH-CH)^2=2a^2(3-2sin^2\beta-2sin^2\gamma)$$Then
$$\bigg(\frac{BH-CH}{OH}\bigg)^2=3$$As desired.