Geometry Geometry Math Forum

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.
Attached Images Drz2016_VIII_z5.jpg (9.2 KB, 36 views) 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.
Attached Images 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! Tags collinear, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gelatine1 Algebra 15 August 26th, 2013 05:17 AM Tooomy Algebra 1 December 28th, 2010 02:55 PM liaofei1128 Algebra 0 April 1st, 2010 10:18 PM qweiop90 Algebra 1 July 31st, 2008 06:27 AM Mazaheri Complex Analysis 9 March 6th, 2007 08:22 AM

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