My Math Forum Prove that P, Q and R are collinear

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August 17th, 2018, 04:08 AM   #1
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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.
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September 14th, 2018, 12:34 PM   #2
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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.
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 Drz2016_VIII_z5_2.jpg (7.8 KB, 6 views)

 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 $\displaystyle 0\sqrt{a(1-a)}$, 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: $\displaystyle y_{HC}(x)=\frac{a}{b}(1-x)$ from where it follows that the coordinates of the point $\displaystyle H$ are: $\displaystyle H(a, \frac{a}{b}(1-a))$. If slope of the line $\displaystyle p$ is $\displaystyle k$, then equations of lines $\displaystyle p$ and $\displaystyle q$ are: $\displaystyle y_{PH}(x)=k(x-a)+\frac{a}{b}(1-a)$ and $\displaystyle y_{PH}(x)=-\frac{1}{k}(x-a)+\frac{a}{b}(1-a)$, respectively, where $\displaystyle -\frac{a}{b} October 31st, 2018, 07:30 AM #7 Senior Member Joined: Feb 2010 Posts: 706 Thanks: 141 Quote:  Originally Posted by lua Straight line$\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$are collinear. Establish a coordinate system with$\displaystyle A(0,a),B(-b,0),C(c,0)$and$\displaystyle a,b,c >0$. Then we get$\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^2c-bc^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{am-bm^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^2-amb-bc-bcm^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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