Minimizing the distance between the circumcenter and orthocenter of a triangle

Question

Let $\triangle ABC$ be inscribed in a circle with radius $1$ and center $O$. Let $\angle AOM=150^\circ$, where $M$ is the middle of $BC$. Let $H$ be the orthocenter of the triangle. If $A$, $B$, $C$ are selected such that $OH$ has the minimum length, then the length of $BC$ is

A: $\sqrt{15}\quad$ B: $\sqrt{13}/2\quad$ C: $\sqrt{3}/2\quad$ D: $\sqrt{13}/4$

I made a sketch and tried to apply the Sylvester's theorem and to solve the problem with vectors but did not succeed.

Could you please help me?

Answer 1

Here is one approach -

Say $\angle A = 2x$. As $\angle AOM = 150$ and $\angle BOC = 2 \angle A$,
$ \ \angle AOC = 150 - \angle A = 150 - 2x \implies \angle B = 75 - x$ and $\angle C = 105-x$

Using the identity $OH^2 = 9R^2 - (a^2+b^2+c^2) = 9R^2 - 4R^2 (\sin^2A + \sin^2B+\sin^2C)$

$OH^2 = 9 - 4 (\sin^2 2x + \sin^2 (75-x) + \sin^2 (105-x)) \ $ (as $R = 1$)

$ = 9 - 2 (3 + \sqrt3 \cos 2x - \cos 4x)$

$OH$ is minimum when $\sqrt3 \cos 2x - \cos 4x$ is maximum.

I will leave for you to work through the details but that occurs when $\cos 2x = \frac{\sqrt3}{4}$.

Can you take it from here?

Answer 2

Draw a diagram. Since $AH \parallel OM$, $\angle HOA=180^\circ-\angle AOM=30^\circ$. $OH$ will be minimum when $OH \perp AH$.

Given is $OA=1$. So $AH=\sqrt{3}/2$.

It is well known that $OM=AH/2$. Hence

$$BC=2\sqrt{OB^2-OM^2}=2\sqrt{1-\left( \frac{\sqrt{3}}{4} \right)^2}=\frac{\sqrt{13}}{2}$$