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Problem from kyiv festival

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Source: 5-th Kyiv math festival, 2006

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rogue

#1 h May 12, 2006, 5:18 PM • 1 Y

Y by Adventure10

See all the problems from 5-th Kyiv math festival
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Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.

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rogue

#2 h May 17, 2006, 2:34 PM • 2 Y

Y by Adventure10, Mango247

Here is a bit simplier twin problem:

Let $I$ be the incenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CI$ and $BI$ at points $D$ and $E$ respectively. Prove that if triangles $IDH$ and $IEH$ are isosceles then triangle $ABC$ is isosceles too.

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SCP

#3 h Oct 30, 2011, 5:19 PM • 2 Y

Y by Adventure10, Mango247

Fopr the original question:
We remark that $D,E$ can't be at the same side of $OH$ and hence the problem is equivalent with $EHD=EOD$ which is in the two cases equal to $\angle A=60^\circ$ and so the problem will become stupid, because the equilateral case gives $O=H=E=D.$

Or may $OH=OE$ happen, which makes we have three pairs of possible easul angles per triangle?

Someone can say how the question exactly goes and give a good example of such a configuration or/and the solution?

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#4 h Oct 31, 2011, 8:10 AM • 2 Y

Y by SCP, Adventure10

Well, if you assume $ABC$ is an isosceles triangle then you can easy angle chase that $O$ is the circumcenter of $DHE$ so that $OE = OH = OD$ . So this should be the case that works. I show a proof of this case.

Easy angle chase yields $\angle OBD = \angle C - \angle A$ and $OCE = \angle B - \angle A$ .

(Take into consideration that $H$ must be an interior point of triangle $BOC$ so the lines $BH$ and $CH$ can intersect segments $BO$ and $CO$ )

From this we get $\angle OEC = 180 - \angle 2A - (\angle B - \angle A) = \angle C$ and $\angle ODB = \angle B$ .

So we have, by law of sines, $OE = \frac{OC \cdot \sin \angle B - \angle A}{\sin \angle C}$ and $OD = \frac{OB \cdot \sin \angle C - \angle A}{\sin B}$

But since we assumed $OEH$ and $OHD$ isosceles with $OE = OH = OD$ we obtain $\frac{ \sin \angle B - \angle A}{\sin \angle C} = \frac{ \sin \angle C - \angle A}{\sin B} (*)$

Clearly, since $\angle A < \angle C$ and $\angle A < \angle B$ we have that $0 < \angle B+ \angle C- \angle A< 180$ so in $(*)$ we must have $\angle B - \angle A = \angle C - \angle A$ and $\angle C = \angle B$ and therefore $ABC$ is isosceles.

In here I used the well known lemma that if we have $\frac{\sin A}{\sin B}=\frac{\sin C}{\sin D}$ and $A+B = C+D = K$ with $0<K<180$ then we must have $A=C$ and $B=D$ .

The other cases in which triangles $ODH$ and $OEH$ are isosceles can be dealt with easy angle chases to get contradictions.

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