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 $\angle BAC$ . Circle described on triangle 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.


2 Answers 2


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

  • $\begingroup$ 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? $\endgroup$ Oct 23, 2018 at 5:57
  • 2
    $\begingroup$ @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. $\endgroup$ Oct 23, 2018 at 6:40
  • $\begingroup$ Ok I got it. But still I dont understand how point D should be drawn. Could you attach please some graph? $\endgroup$ Oct 25, 2018 at 19:53
  • 1
    $\begingroup$ @PawełOrliński The point $D$ looks exactly like the one in the figure here: artofproblemsolving.com/community/…. $\endgroup$ Oct 25, 2018 at 20:23
  • $\begingroup$ 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." $\endgroup$ Oct 26, 2018 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.

Problem 2959781

  • $QO$ 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.


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.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .