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.

enter image description here

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

  • $\begingroup$ 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! $\endgroup$
    – Oussema
    Apr 4 at 18:40
  • $\begingroup$ @Oussema Evan Chen's Euclidean Geometry in Mathematical Olympiads is good. $\endgroup$
    – cosmo5
    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 $\small \angle AOB = 2C, \angle OAB = \angle OBA = 90^0 - C$

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


enter image description here

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.