# $\sin\angle HIO$ given $\triangle ABC$ with $(\overline{AB},\overline{BC},\overline{CA})=(8,13,15)$?

In triangle $$ABC, AB = 8, BC = 13$$ and $$CA = 15$$.

Let $$H, I, O$$ be the orthocenter, incenter and circumcenter of triangle $$ABC$$ respectively. Find $$\sin$$ of angle $$HIO$$ .

My attempt: By using law of cosine I can see angle $$A = 60^{\circ}$$. I also observed that angle $$BHC$$, $$BIC$$ and $$BOC$$ equals $$120^{\circ}$$. How do I proceed after this?

Recall that $$B,I,C$$ lie on a circle centered at $$D$$, midpoint of arc $$BC$$ not containing $$A$$ ($$AD$$ is the angle bisector of $$\angle A$$). We have $$O,H$$ lying on this circle too.

Since $$O,H$$ are isogonal conjugates, $$\angle OCI=\angle ICH$$. It follows, $$IH=IO$$ and $$\angle OCI=\angle HCI=\angle OHI=\angle HOI$$.

Hence $$\angle HIO=180-2\angle OCI$$.

But $$\angle OCI=\angle ACI-\angle ACO=C/2-(90-B)$$. Therefore, $$2\angle OCI=C-180+2B=C-(A+B+C)+2B=B-A$$

Finally, $$\sin \angle HIO =\sin 2\angle OCI=\sin (B-A)=\ldots$$

• Hello, I've lately seen a few problems of geometry solved using isogonal conjugates, you might even be the solver. Unfortunately, I don't know much about their properties apart from the Wikipedia definition. I also don't know anything about projective geometry. Do you have a book or handout you can recommend so I can learn more? Thank you! Apr 4 at 18:40
• @Oussema Evan Chen's Euclidean Geometry in Mathematical Olympiads is good. Apr 5 at 5:15

$$\displaystyle \small \angle IAO = \angle IAH = \frac{\angle B - \angle C}{2}$$

Now use the identity, $$\small AH = 2R \cos \angle A \implies AH = R = AO \$$ (as $$\angle A = 60^0$$)

That means $$\small \triangle HAO$$ is an isosceles triangle. As $$\small AI$$ is angle bisector of $$\small \angle HAO, IH = OI$$.

Now remember that, $$\small OI^2 = R(R-2r), OH^2 = 9R^2 - (a^2+b^2+c^2)$$

$$\small R = \displaystyle \frac{a}{2 \sin A} = \frac{13}{\sqrt3}$$

$$\small \displaystyle r = \frac{\Delta}{s} = \frac{(bc \sin A) / 2}{(a+b+c)/2} = \frac{5}{\sqrt3}$$

So $$\small OI = \sqrt {13} = IH, \ OH = 7$$

$$\small \displaystyle \cos \angle HIO = - \frac{23}{26} \implies \fbox {\sin \angle HIO = \frac{7 \sqrt3}{26}}$$

This is just an addendum showing how $$\displaystyle \small \angle IAO = \angle IAH = \frac{B - C}{2}$$

$$\small \displaystyle \angle BAH = 90^0 - B, \angle IAH = \angle IAB - \angle BAH = \frac{A}{2} - 90^0 + B$$

As $$\small \displaystyle \frac{A+B+C}{2} = 90^0, \angle IAH = \frac{B-C}{2}$$

Now $$\small \angle AOB = 2C, \angle OAB = \angle OBA = 90^0 - C$$

$$\small \displaystyle \angle IAO = \angle OAB - \angle IAB = \frac{A+B+C}{2} - C - \frac{A}{2} = \frac{B-C}{2}$$

Given $$a=13$$, $$b=15$$, $$c=8$$, we can find semiperimeter, area, inradius and circumradius ot the triangle:

\begin{align} \rho&=\tfrac12(a+b+c)=18 ,\\ S_{ABC}&= \tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} = 30\sqrt3 ,\\ r&=\frac{S_{ABC}}\rho=\tfrac53\sqrt3 ,\\ R&=\frac{abc}{4S_{ABC}}= \tfrac{1}3\sqrt3 \end{align}

And use known expressions for $$|OI|$$, $$|OH|$$, and $$|IH|$$: \begin{align} |OI|&=\sqrt{R(R-2r)}=\sqrt{13} ,\\ |OH|&=\sqrt{R^2+2((r+2R)^2-\rho^2)} =7 ,\\ |IH|&= \sqrt{(r+2R)^2+2r^2-\rho^2} =\sqrt{13} ,\\ S_{OIH}&= \tfrac14\sqrt{4|OI|^2|OH|^2-(|OI|^2+|OH|^2-|IH|^2)^2} =\tfrac74\sqrt3 ,\\ S_{OIH}&=\tfrac12|OI|\cdot|IH|\sin OIH =\tfrac{13}2\sin OIH ,\\ \sin OIH&= \frac{\tfrac74\sqrt3}{\tfrac{13}2} = \tfrac7{26}\sqrt3 . \end{align}