circleseuclidean-geometrygeometrytriangles
The altitudes of an acute triangle $ABC$ which is not isosceles concur at the point $H$. Let $S$ be the midpoint of the circular arc $BHC$ of the circumcenter of the triangle $BCH$. If $AS$ and $AH$ are of the same length, find the angle $\angle BAC$
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I found out two interesting equalities:
$\angle BAC =\angle BOS$
$\angle HAS = \angle HOS$
But apart from this I can't move further with the solution.
Triangles $ABH$ and $ABK$ are right triangles whose hypothenuse is $AB$. So they have the same circumcircle, whose diameter is $AB$. As a consequence, considering that $M$ is the midpoint of the diameter $AB$, we get that $MH=MK$ because they are radii of the same circumference, and then $MHK$ is isosceles.
Without using circles: we can show that, in any right triangle, the median drawn to the hypotenuse is equal to half the hypotenuse. To show it for the right triangle $AHB$, let us draw a line $ME$ starting from the midpoint $M$, parallel to the leg $AH$, and ending to the intersection point $E$ with the other leg $HB$. We know that the angle $AHB$ is right. The angles $MEB$ and $AHB$ are congruent, since they are corresponding angles if we consider the parallel lines $AH$ and $ME$, and the transverse line $HB$. So, the angle $MEB$ is a right angle.
Because the line $ME$ starts from the midpoint $M$ and is parallel to $AH$, it divides the leg $HB$ in two congruent segments $HE$ and $EB$ with equal length. These considerations show that the triangles $MEH$ and $MEB$ are right triangles, have congruent legs $HE$ and $EB$, and a common leg $ME$. This means that these triangles are congruent. We then get that the segments $MH$ and $MB$ are also congruent, since they are corresponding sides (hypothenuses) of these triangles. So we have shown that, in the right triangle $AHB$, the median $MH$ is equal to half the hypotenuse $AB$.
Applying the same procedure to the right triangle $BKA$, we get that $MK$ is congruent with $MA$, so that it also equals half of the hypothenuse $AB$.
We then conclude that $MH=MK$, and so the triangle $MHK$ is isosceles.
As the figure shows, denote the circumcenter as $O$, and the intersection point of $MP$ and $BC$ as $N$.
Accoring to the given conditions, it's clear that $A,O,P$ are collinear. Notice that $PB \perp AB$ and $CH \perp AB$. Hence $PB // CH$. Similarily, $PC \perp AC$ and $BH \perp AC $. Hence $PC // BC$. As a result, $BPCH$ is a parallelogram.
Thus, $N$ is the midpoint of $BC$. Hence, $MN // AB$. But $CH \perp AB$, therefore $CH \perp MP$.
Notice that $CH$ is the altitude to the hypotenuse $MP$ in the right triangle $\triangle MPC$. Hence,$$CH=\sqrt{HM \cdot HP}=4\sqrt{5}.$$ Moreover,$$HN=\frac{1}{2}HP=8.$$ Thus, $$CN=\sqrt{CH^2+HN^2}=12.$$ It follows that $$BC=2CN=24.$$
Best Answer
Let $\alpha:=\angle BAC$, $\beta:=\angle CBA$, and $\gamma:=\angle ACB$. Without loss of generality, assume that $\beta<\gamma$. Note that $$\angle AHO=\angle AHB+\angle BHO=(\pi-\gamma)+\beta=\pi-(\gamma-\beta).$$ Observe that $$\angle HOS=2\,\angle HCS=2\,\left(\angle HCB-\angle SCB\right)$$ and $$\angle HOS=2\,\angle HBS=2\,\left(\angle SBC-\angle HBC\right)\,.$$ Thus, $$\angle HOS=\left(\angle HCB-\angle SCB\right)+\left(\angle SBC-\angle HBC\right)=\angle HCB-\angle HBC\,,$$ as $\angle SBC=\angle SCB$. That is, $$\angle HOS=\left(\frac{\pi}{2}-\beta\right)-\left(\frac{\pi}{2}-\gamma\right)=\gamma-\beta\,.$$ Ergo, $$\angle OHS=\frac{\pi-(\gamma-\beta)}{2}=\frac{\angle AHO}{2}\,.$$ In other words, $HS$ is the internal angular bisector of $\angle AHO$.
If $AH=AS$, then $OA\perp HS$. Therefore, the triangle $AHO$ must be isosceles with $AH=HO$ because the internal angular bisector of $\angle AHO$ coincides with the altitude from $H$ to $AO$. In other words, $AH=R$, where $R$ is the radius of the circumcircle $\Gamma$ the triangle $ABC$ (noting that the triangles $ABC$ and $BHC$ have the same circumradius).
Let $D$ be the intersection of the line perpendicular to $BC$ at $B$ and the line perpendicular to $AC$ at $A$. Then, $D$ is on the circle $\Gamma$ and $CD$ is a diameter of $\Gamma$. It is easy to show that $AH=BD$ by noting that $AHBD$ is a parallelogram. Because the right triangle $BCD$ satisfies $\angle CBD=\dfrac{\pi}{2}$ and $$CD=2R=2\,AH=2\,BD\,,$$ we deduce that $$\alpha=\angle BAC=\angle BDC=\dfrac{\pi}{3}\,.$$
In fact, the converse also holds. That is, in an acute triangle $ABC$, $AS=AH$ where $H$ is the orthocenter of the triangle $ABC$ and $S$ is the midpoint of the circular arc $BHC$ if and only if $\angle BAC=\dfrac{\pi}{3}$.