Cesko-Slovacko-Poljsko 2006

In any obtuse triangle $XYZ$ with obtuse angle in $Z$ and ortocentre $W$, angles $XYZ$ and $XWZ$ are equal, as complementing the angle $YXW$ to $90$ degrees (fig. 1). Moreover, points $Y$ and $W$ lie in different half-planes determined by $XZ$.

Fig. 1

Let $P$, $Q$, $R$ be ortocentres of given triangles in that order. We will show $\nleqq PQR=90^{\circ }$. Obviously all three triangles are obtuse in $C$. So $P$, $Q$, $R$ lie on extensions of altitudes going through $C$ to corresponding sides. Because of position of these sides it is also obvious the ray $CQ$ lies "between" rays $CP$ and $CR$, i.e. in angle $PCR$. So $\nleqq PQR=\nleqq RQC+\nleqq PQC$ (fig. 2). By the fact in the first paragraph $Q$, $R$ lie in

Fig. 2

the same half-plane determined by the line $BC$ and $$\nleqq BEC=\nleqq BRC\text{and}\nleqq BDC=\nleqq BQC.$$ Angles $BEC$, $BDC$ are equal being inscribed angles with the same chord $BC$. Thus also $\nleqq BRC=\nleqq BQC=\omega$ and $BCRQ$ is cyclic. So $\nleqq RQC=\nleqq RBC=\phi$. As $EC=r$ for inscribed angle we have $\nleqq EBC=30^{\circ }$. Let $U$ be the foot on $BE$ in triangle $BEC$. Counting the angles in right triangle $BUR$ we get $$\omega +\phi +30^{\circ }+90^{\circ }=180^{\circ },\text{i.e.}\nleqq RQC=\phi =60^{\circ }-\omega =60^{\circ }-\nleqq BDC.$$ In the same way we conclude $\nleqq PQC=60^{\circ }-\nleqq DBC$. So we have (using the sum of angles in triangle $BCD$ is $180^{\circ }$) $$\nleqq PQR=\nleqq RQC+\nleqq PQC=120^{\circ }-(\nleqq BDC+\nleqq DBC)=\nleqq BCD-60^{\circ }.(1)$$ But also $BD=r$, thus $\nleqq BCD=150^{\circ }$. Finally by (1) $\nleqq PQR=90^{\circ }$.