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Can you please give me a proof for this one:

Let $ABC$ be a triangle with orthocenter $H$. $P$ is a point on $BC$. Circumcircle of triangle $HBP$ cuts $AB$ again at $M$, and circumcircle of triangle $HCP$ cuts $AC$ again at $N$. Then $H$ is incenter of the triangle $PMN$.

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Are you assuming the triangle $ABC$ to be acute? At least if the angle at $B$ is obtuse you seem to get the center of an exterior tangent circle. – t.b. Aug 2 '11 at 14:06
@Theo, consider all the sides to extend indefinitely, this does not pose a problem. i.e. instead of considering AB as the line segment, consider it as the line AB. – picakhu Aug 2 '11 at 14:38
@picakhu: I don't have a problem with that and I agree with your argument below, however you don't get the center of the incircle but rather the center of an excircle, as I said. – t.b. Aug 2 '11 at 14:48
@Theo, you are correct. Thanks. My proof holds only for the acute case as the claim is incorrect for the obtuse claim. – picakhu Aug 2 '11 at 15:27
up vote 6 down vote accepted

Note: This is essentially the same as picakhu's solution, but slightly more fleshed-out.

Before addressing the problem itself, I'd like to make a simple auxiliary observation:

In an acute triangle draw the three altitudes and note that they split the angles into neighboring angles $\color{green}{\alpha'}, \color{red}{\alpha''}, \color{violet}{\beta'}, \color{green}{\beta''}, \color{red}{\gamma'}, \color{violet}{\gamma''}$ as indicated in the picture:

angles in an acute triangle

and $\color{green}{\alpha'} = \color{green}{\beta''}$, $\color{violet}{\beta'} = \color{violet}{\gamma''}$ and $\color{red}{\gamma'} = \color{red}{\alpha''}$.

Following your construction ($H$ orthocenter of $\triangle ABC$; $P$ on $BC$ arbitrary; $M$: second point of intersection of $AB$ with the circle through $BHP$; $N$: second point of intersection of $AC$ with the circle through $CHP$, using the above simple fact and the inscribed angle theorem, we see at a glance that all the angles of the same color in the following picture have the same size (ignore the dashed turquoise circle for the moment):

triangle with identified angles

Assuming that both $M$ and $N$ lie inside the segment $AB$ and $AC$, respectively (otherwise the argument is a bit simpler) $$ \angle AMH + \angle ANH = (\pi-\angle HMB) + (\pi-\angle HNC) = \angle HPB + \angle HPC = \pi $$ so that $AMHN$ is a circular quadrilateral. But from this we see that $\angle HAN = \angle HMN$ and $\angle HAM = \angle HNM$ and we're done since $HM$, $HN$, $HP$ are the angular bisectors of the orange triangle $MNP$ so that $H$ is the center of the incircle.

Here's a picture showing that the claim doesn't hold if the triangle $ABC$ is not acute.


I leave it to you to verify that you get an excircle as soon as the triangle is obtuse and treat the two right-angled cases.

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This solution holds when we consider $H$ to be INSIDE triangle $ABC$, i.e. when $ABC$ is acute.

First note that $AMHN$ lies on a circle (opposite sides add to 180). So, it suffices to check that $\angle MPH = \angle HPN$, to do this, we let the perpendicular dropped from $B$ onto $AC$ be $N'$ and the perpendicular from $C$ to $AB$ be M'. Then, $\angle MPH = \angle MBH$ and $\angle NPH = \angle NCH$. However since $N'M'BC$ lies on a circle, the conclusion follows.

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picakhu: If I'm not completely wrong (and fool myself a third time) I think you can adapt your argument to see that actually $H$ is the center of an excircle. Your observations on points lying on a circle still hold. – t.b. Aug 2 '11 at 16:03
@Theo, I believe you are right that it can be modified slightly, however, it is unnecessary as that was not the intention of the OP. – picakhu Aug 2 '11 at 16:23
As Amir was somewhat terse (as usual), I refrained from making conclusions about the intentions. However, your answer certainly is sufficient for Amir to complete the argument. – t.b. Aug 2 '11 at 16:26

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