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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?

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

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

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

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

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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}

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