# $AH=AS$ where $H$ is the orthocenter of $\triangle ABC$ and $S$ is the midpoint of the arc $BHC$ of the circumcircle of $\triangle BHC$

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 $$\angle BAC$$ . I found out two interesting equalities:

1. $$\angle BAC =\angle BOS$$

2. $$\angle HAS = \angle HOS$$

But apart from this I can't move further with the solution.

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 $$\angle BAC=\dfrac{\pi}{3}$$.

• 1. The fourth formula should end with $\angle HCB - \angle HBC$, shouldn't it? 2. Why $\triangle ABC$ and $\triangle BHC$ have the same circumradius? – Paweł Orliński Oct 23 '18 at 5:57
• @PawełOrliński Thanks. I've fixed the typo. About the circumradii, this is a well known result, but if you want to prove it, then you can use tigonometry, noting that the two triangles have a common side $BC$ with $\angle BHC=\pi-\angle BAC$. You can also use a geometric argument by reflecting $H$ about $BC$ to get $H_a$. Then, $ABCH_a$ is a cyclic quadrilateral. – Batominovski Oct 23 '18 at 6:40
• Ok I got it. But still I dont understand how point D should be drawn. Could you attach please some graph? – Paweł Orliński Oct 25 '18 at 19:53
• @PawełOrliński The point $D$ looks exactly like the one in the figure here: artofproblemsolving.com/community/…. – Batominovski Oct 25 '18 at 20:23
• Then "Let D be the intersection of the line perpendicular to BC at B and the line perpendicular to AC at A." instead of "Let D be the intersection of the line perpendicular to AB at B and the line perpendicular to AC at A." – Paweł Orliński Oct 26 '18 at 7:32

A picture is placed in the middle of the text.

• The Euler circle and the nine points are present in the figure. This is not necessary, but it gives a good guide for the relevant points.
• Let $$O$$ be the center of the cyclic (=inscriptible) quadrilateral $$BSHC$$.
• $$S$$ is on the side bisector of $$BC$$ by construction, so $$SO\perp BC\perp AH$$, so $$OS\|AH$$. The two triangles $$\Delta ASH$$ and $$OSH$$ are thus equal, being isosceles, with the same basis, and with same base angles (interior for the parallels $$AH$$ and $$OS$$. So $$ASOH$$ is a rhombus, its diagonals are perpendicular, the angles in $$A$$ and $$O$$ in it are equal.
• Let $$y$$ be measure of $$\angle HAC=\angle HBC$$.
• Let $$z$$ be measure of $$\angle HAB=\angle HCB$$.
• Assume $$z\ge y$$ without restriction.
• The angle $$\angle BOH=\overset \frown{BH}$$ is twice $$\angle HCB=z$$, so it is $$2z$$.
• The angle $$\angle HOC=\overset \frown{HC}$$ is twice $$\angle HBC=y$$, so it is $$2y$$.
• The angle $$\angle BOC$$ is thus $$2(y+z)$$, its half is $$\angle BOS=y+z$$, and for $$\angle SOH$$ there remain $$2z-(y+z)=z-y$$. This is also $$\angle SAH$$.
• Let us now consider as in the picture the heights / angle bisectors from $$O$$ in the triangles $$\Delta BOH$$, and $$\Delta HOC$$. They pass through Euler points, and further intersect $$AB$$ and $$AC$$ respectively in $$Q$$ and $$P$$.
• The triangles $$\Delta HPA$$, $$\Delta HPO$$ are equal, $$HP$$ common, two corresponding sides are sides of the rhombus $$AHOS$$, and the angles in $$A$$ and $$O$$ are $$y$$. So $$HP$$ is an angle bisector of $$\angle AHO$$, so $$S,H,P$$ are colinear.
• Similarly, $$S,H,Q$$ are also colinear. • $$QO$$ and $$AC$$ are both perpendicular on $$BH$$, so they are parallel.
• $$PO$$ and $$AB$$ are both perpendicular on $$CH$$, so they are parallel.
• So $$AQOP$$ is a parallelogram. Its diagonals are perpendicular, so it is a rhombus.
• Which are now the angles of $$\Delta OQP$$? The angle in $$O$$ is half of $$\angle BOC=2(y+z)$$, so it is $$y+z$$. The angle in $$P$$ is $$\angle QPO=\angle OPC=\angle BAC$$ (because of $$OP\| BA$$), so it is also $$y+z$$. The angle in $$Q$$ is $$\angle PQO=\angle OQB=\angle CAB$$ (because of $$OQ\| CA$$), so it is again $$y+z$$.
• The triangle $$\Delta OPQ$$ is thus equilateral, so $$\hat A=y+z= 60^\circ$$.

This answers the question in the original post.

$$\square$$

Bonus: Note that in case of an angle $$\angle A = 60^\circ$$, the triangles $$\Delta SBO$$ and $$\Delta SCO$$ are equilateral, so $$SB=SC=SO=SA$$, so $$S$$ is the center of the circumscribed circle $$(ABC)$$.

(The reciprocal is also true.)