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This question is from a list of training for math olympiad. It comes from a Brazilian olympic training called POTI.

Let ABC be a triangle of a circumcircle $w_1$, $O$ be the circumcenter of ABC and $w_2$ be the excircle relative to the BC side. If M, N and L are points are the points of tangency of $w_2$ with the lines BC, AC, AB and the radii of $w_1$ and $w_2$ are equal, show that O is the orthocenter of the triangle MNL.

I don't know if help, but the list is about congruence of triangles.

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Let $I$ be the center of excircle $w_2$ and let $E$ be the midpoint of edge $AC$. Then since $O$ is the circumcenter of triangle $ABC$, the line $OE$ is the orthogonal bisector of $AC$. Let $OE$ intersect circle $w_1$ at points $P$ and $Q$ as shown on the picture. Then quad $CBPQ$ is inscribed in $w_1$ so $$\angle \, PBC = \pi - \angle \, CQP$$ However, $\angle \, AOC = 2 \, \angle \, ABC = 2\beta$ and since $OE$ is the interior angle bisector of $\angle \, AOC$ $$\angle \, EOC = \angle \, QOC = \beta$$ Triangle $QOC$ is isosceles, $OQ=OC$, so $$\angle CQO = \angle \, CQP = \frac{\pi - \beta}{2}$$

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Consequently, $$\angle \, PBC = \pi - \angle \, CQP = \pi - \frac{\pi - \beta}{2} = \frac{\pi + \beta}{2}$$ On the other hand $BI$ is the angle bisector of angle $\angle \, LBC = \pi - \beta$ so $$\angle \,IBC = \frac{1}{2} \, \angle \, LBC = \frac{\pi - \beta}{2}$$ As a result, we obtain $$\angle \, PBI = \angle \, PBC + \angle \, IBC = \frac{\pi + \beta}{2} + \frac{\pi - \beta}{2} = \pi$$ In other words, point $P$ lies on the line $IB$.

Now, observe that by cosntruciton line $PO$ which is the same line as $OE$ is orthogonal to $AC$. Also $IN$ is orthogonal to $AC$. Therefore $PO$ is parallel to $IN$. Moreover, by assumption $PO = IN$. Thus quad $OPIN$ is a parallelogram which means that segments $IP$ and $NO$ are parallel and equal. However, line $IB$, which is the same as line $IP$, is orhtogonal to $LM$ and so $NO$ is also orthogonal to $LM$.

Absolutely analogously one shows that $LO$ is orthogonal to $NM$. This immediately implies tht $O$ is the orthocenter of triangle $LMN$.

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  • $\begingroup$ Great answer! But, there are little mistakes in your answer that don't disturb the understanding, however is better fix them to let clearer it. $\endgroup$ Commented Nov 13, 2016 at 12:59
  • $\begingroup$ @RafaelDeiga These were not mistakes. When I write "line $IB \equiv PI$" I do not mean that segments $IB = PI$. I mean that the two lines $IB$ and $PI$ are the same line. "Line $IB$" means the line passing through the two points $I$ and $B$. And the notation $IB \equiv PI$ means that the line through points $I$ and $B$ is the same line as the line through points $P$ and $I$. $\endgroup$ Commented Nov 13, 2016 at 14:58
  • $\begingroup$ Ah ok, sorry. It's because here in Brazil some books use the symbol $\equiv$ as congruence of figures, including congruence of line segments. $\endgroup$ Commented Nov 13, 2016 at 22:22

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