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August 17th, 2018, 04:08 AM  #1 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2  Prove that P, Q and R are collinear
Straight line $\displaystyle p$ contains orthocenter $\displaystyle H$ of the acute triangle $\displaystyle ABC$ and intersects its edges $\displaystyle AB$ and $\displaystyle CA$, and edge $\displaystyle CA$ at point $\displaystyle P$. Straight line $\displaystyle q$ perpendicular to $\displaystyle p$ contains point $\displaystyle H$ also, intersects edges $\displaystyle AB$ and $\displaystyle BC$, and edge $\displaystyle BC$ at point $\displaystyle Q$. Straight line through $\displaystyle A$ parallel to $\displaystyle q$, and staight line through $\displaystyle B$ parallel to $\displaystyle p$ intersect at point $\displaystyle R$. Prove that points $\displaystyle P$, $\displaystyle Q$ and $\displaystyle R$ are collinear.

September 14th, 2018, 12:34 PM  #2 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
I tried to make some progress. See the picture. We can write: $\displaystyle DH=RE$, and $\displaystyle RD=HE$. $\displaystyle PH=RE+PD$ $\displaystyle QH=QE+RD$ After squaring the two above equations, we get: $\displaystyle PH^{2}=RE^{2}+2\cdot RE\cdot PD + PD^{2}$ and $\displaystyle QH^{2}=QE^{2}+2\cdot QE\cdot RD + RD^{2}$. After the addition of two previous equations, we get: $\displaystyle PH^{2}+QH^{2}=RE^{2}+QE^{2}+PD^{2}+RD^{2}+2\cdot(R E\cdot PD + QE\cdot RD)$ Since $\displaystyle RE^{2}+QE^{2}=RQ^{2}$ and $\displaystyle PD^{2}+RD^{2}=PR^{2}$, we have: $\displaystyle PH^{2}+QH^{2}=RQ^{2}+PR^{2}+2\cdot(RE\cdot PD + QE\cdot RD)$, i.e: $\displaystyle PH^{2}+QH^{2}=(RQ+PR)^{2}2\cdot PR\cdot RQ+2\cdot(RE\cdot PD + QE\cdot RD)$. If points $\displaystyle P$, $\displaystyle R$ and $\displaystyle Q$ are collinear, we would have: $\displaystyle PH^{2}+QH^{2}=PQ^{2}$. So, that $\displaystyle P$, $\displaystyle Q$ and $\displaystyle R$ would be collinear, we have to prove that $\displaystyle PR\cdot RQ= RE\cdot PD + QE\cdot RD$. Here I stuck. 
September 14th, 2018, 01:00 PM  #3 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
When I look better, last equation seems to be the same as $\displaystyle \cos^{2}\alpha + \sin^{2}\alpha =1$ (when the equation is divided by $\displaystyle PR\cdot RQ$), where $\displaystyle \alpha =\angle PRD$ which is equal to $\displaystyle \angle RQH$, when $\displaystyle P$, $\displaystyle R$ and $\displaystyle Q$ are collinear. In this moment, I'm feeling like I'm running my own tail. Last edited by skipjack; October 26th, 2018 at 07:09 AM. 
October 25th, 2018, 11:15 PM  #4 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
I managed to solve this problem ''with a little help from my friends''. Problem can be solved analytically, taking coordinates: $\displaystyle A(a,b), B(0,0)$ and $\displaystyle C(1,0)$. If someone is interested, I can write complete proof. 
October 26th, 2018, 07:22 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,751 Thanks: 2135 
I'm interested. My gut feeling was that this problem had more to do with angles than with Pythagoras.

October 29th, 2018, 02:19 AM  #6 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Equation of line AB is: $\displaystyle y_{AB}(x)=\frac{b}{a}x$, with note that besides $\displaystyle 0<a<1$, it is true that $\displaystyle b>\sqrt{a(1a)}$, i.e. point $\displaystyle A$ must be above the semicircle over the diameter $\displaystyle BC$ for $\displaystyle a$ given. Line $\displaystyle HC$ is perpendicular to line $\displaystyle AB$, so its equation is: $\displaystyle y_{HC}(x)=\frac{a}{b}(1x)$ from where it follows that the coordinates of the point $\displaystyle H$ are: $\displaystyle H(a, \frac{a}{b}(1a)) $. If slope of the line $\displaystyle p$ is $\displaystyle k$, then equations of lines $\displaystyle p$ and $\displaystyle q$ are: $\displaystyle y_{PH}(x)=k(xa)+\frac{a}{b}(1a)$ and $\displaystyle y_{PH}(x)=\frac{1}{k}(xa)+\frac{a}{b}(1a)$, respectively, where $\displaystyle \frac{a}{b}<k<\frac{a}{b}+\frac{1}{b}$. Equations of lines $\displaystyle AC$ and $\displaystyle BC$ are: $\displaystyle y_{AC}(x)=\frac{b}{1a}(1x)$ and $\displaystyle y_{BC}(x)=0$, respectively. From equation $\displaystyle y_{PH}(x_{P})=y_{AC}(x_{P})$ we get point $\displaystyle P$ coordinates: $\displaystyle x_{P}=\frac{b^{2}+a(1a)(bk(1a))}{b(b+(1a)k)}$, $\displaystyle y_{P}=\frac{(1a)(a+bk)}{b+(1a)k}$. From equation $\displaystyle y_{QH}(x_{Q})=y_{BC}(x_{Q})$ we get point $\displaystyle Q$ coordinates: $\displaystyle x_{Q}=\frac{a}{b}(b+(1a)k)$ and $\displaystyle y_{Q}=0$. Equations of lines $\displaystyle RB$ and $\displaystyle RA$ are: $\displaystyle y_{RB}(x)=kx$ and $\displaystyle y_{RA}(x)=\frac{1}{k}(xa)+b$, respectively. From equation $\displaystyle y_{RB}(x_{R})=y_{RA}(x_{R})$ we get point $\displaystyle R$ coordinates: $\displaystyle x_{R}=\frac{a+bk}{k^{2}+1}$ and $\displaystyle y_{R}=\frac{k}{k^{2}+1}(a+bk)$. Now, equation of the line $\displaystyle RQ$ is $\displaystyle y_{RQ}(x)=\frac{y_{R}}{x_{R}x_{Q}}(xx_{Q})$. If the coordinates of point $\displaystyle P$ satisfy this equation, then points $\displaystyle P$, $\displaystyle Q$ and $\displaystyle R$ are collinear. Indeed, after a little calculation, it can be shown that $\displaystyle y_{RQ}(x_{P})=y_{P}$, what should have been proved. Sorry for waiting. P.S. Few days ago I found another proof using right triangles similarity. I'll post it too, if you're interested. Last edited by lua; October 29th, 2018 at 02:24 AM. 
October 31st, 2018, 07:30 AM  #7  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 141  Quote:
Then we get $\displaystyle H(0,bc/a)$. From this we get lines p and q as $\displaystyle y=mx+bc/a$ and $\displaystyle y=x/m+bc/a$. Line p intersects $\displaystyle AC$ at $\displaystyle P \left(\dfrac{a^2cbc^2}{a^2+amc},\dfrac{amc+bc}{a+mc} \right)$. Line q intersects $\displaystyle BC$ at $\displaystyle Q \left(\dfrac{mbc}{a},0 \right)$. The lines through $\displaystyle A$ parallel to q and through $\displaystyle B$ parallel to p intersect at $\displaystyle R \left(\dfrac{ambm^2}{1+m^2},\dfrac{am^2+mb}{1+m^2} \right)$. Now calculate the slope of $\displaystyle PQ$ and the slope of $\displaystyle QR$. They both come out $\displaystyle \dfrac{a^2m+ab}{a^2ambbcbcm^2}$ Therefore $\displaystyle P,Q,R$ are collinear points. I've got to believe that there is a slightly less ugly solution!  

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