# Extremely hard geometric problem

Given a triangle ABC. BL is the bisector of angle ABC, H is the orthocenter and P is the mid-point of AC. PH intersects BL at Q. If $\angle ABC= \beta$, find the ratio $PQ:HQ$.If $QR\perp BC$ and $QS \perp AB$, prove that the orthocenter lies on $RS$.

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What is $S$? A diagram would be nice. – robjohn Apr 8 '12 at 17:36
Unless there are further constraints on $R$ and $S$, the claim to be proven is not generally true. – Ilmari Karonen Apr 8 '12 at 17:48
Ah, I see... presumably $R$ is supposed to lie on $BC$ and $S$ on $AB$. – Ilmari Karonen Apr 8 '12 at 17:58
@robjohn: I added a picture I drew quickly in GeoGebra, using the assumptions stated in the previous comment. – Ilmari Karonen Apr 8 '12 at 18:10
The answer to the first question is: $HQ:PQ=\frac{2 cos \beta}{1-cos\beta}$ – Adam Apr 8 '12 at 20:15

In the figures below, I have added the circumcenter, $U$, and the centroid, $E$. I have also placed $L$ on the circumcircle.

$\hspace{8mm}$

Note that since both are perpendicular to $\overline{AC}$, we have $\overline{BH}\,||\,\overline{UP}$; furthermore, $|\overline{BH}|=2|\overline{UP}|$. The latter is because $\triangle PUE$ is similar to $\triangle BHE$ and $$P=\frac{A+C}{2}\text{ and }E=\frac{A+B+C}{3}\tag{1}$$ so that $$P-E=\frac{A-2B+C}{6}\text{ and }E-B=\frac{A-2B+C}{3}\tag{2}$$ Thus, $$|\overline{UP}|=R\cos(B)\text{ and }|\overline{BH}|=2R\cos(B)\tag{3}$$ where $R$ is the circumradius of $\triangle ABC$.

Since the line containing $\overline{UP}$ is the perpendicular bisector of $\overline{AC}$, the point at which $\overrightarrow{UP}$ intersects the circumcircle of $\triangle ABC$ splits the arc between $A$ and $C$ in half. Of course, the bisector of $\angle ABC$ also splits the arc between $A$ and $C$ in half. Thus, the perpindicular bisector of $\overline{AC}$ and the bisector of $\angle ABC$ meet on the circumcircle at $L$.

$\hspace{8mm}$

Note that $\triangle BHQ$ is similar to $\triangle LPQ$. Equation $(3)$ gives that $|\overline{UP}|=R\cos(B)$ so that $$|\overline{PL}|=R(1-\cos(B))\tag{4}$$ Therefore, $(3)$ and $(4)$ yield \begin{align} |\overline{HQ}|/|\overline{PQ}| &=|\overline{BQ}|/|\overline{LQ}|\\ &=|\overline{HB}|/|\overline{PL}|\\ &=\frac{2\cos(B)}{1-\cos(B)}\tag{5} \end{align} which answers the first part.

Because $\triangle BUL$ is isosceles with central angle $2A+B=\pi-(C-A)$, we have $$|\overline{BL}|=2R\sin\left(A+\frac{B}{2}\right)=2R\cos\left(\frac{C-A}{2}\right)\tag{6}$$ Equation $(5)$ yields that $|\overline{BQ}|/|\overline{BL}|=\frac{2\cos(B)}{1+\cos(B)}$. Thus, $(6)$ gives $$|\overline{BQ}|=2R\cos\left(\frac{C-A}{2}\right)\frac{2\cos(B)}{1+\cos(B)}\tag{7}$$ Let $X$ be the intersection of $\overline{BQ}$ and $\overline{RS}$. Since $X$ is on the angle bisector of $\angle ABC$, $\overline{RS}$ is perpendicular to $\overline{BQ}$ and $|\overline{BR}|=|\overline{BS}|$. Thus, $|\overline{BR}|/|\overline{BQ}|=|\overline{BX}|/|\overline{BR}|=\cos(B/2)$. Therefore, $$\frac{|\overline{BX}|}{|\overline{BQ}|}=\cos^2(B/2)=\frac{1+\cos(B)}{2}\tag{8}$$ Equations $(7)$ and $(8)$ yield $$|\overline{BX}|=2R\cos\left(\frac{C-A}{2}\right)\cos(B)\tag{9}$$ Since $\angle HBC=\frac\pi2-C$ and $\angle QBC=\frac{B}{2}$ we get that $\angle HBQ=\frac{C-A}{2}$. Using $(3)$, the orthogonal projection of $\overline{BH}$ onto $\overline{BQ}$ has length is $2R\cos(B)\cos\left(\frac{C-A}{2}\right)$. Thus, the orthogonal projection of $H$ onto $\overline{BQ}$ is $X$. Therefore, $H$ lies on $\overline{RS}$.

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 @AdamAndersson: This answer has been accepted and unaccepted twice. Is there something that you would like to see that is not here? – robjohn♦ May 8 '12 at 19:45 No, your answer is great. What do you use to draw graphics? – Adam Jun 24 '12 at 18:19 @AdamAndersson: I use Intaglio which is only available on the Mac. – robjohn♦ Jun 24 '12 at 18:26 Could please have a look at this geometry problem? math.stackexchange.com/questions/162165/… – Adam Jun 24 '12 at 18:44