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August 17th, 2018, 05:08 AM   #1
lua
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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, 01:34 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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September 14th, 2018, 02:00 PM   #3
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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 08:09 AM.
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October 26th, 2018, 12:15 AM   #4
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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.
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October 26th, 2018, 08:22 AM   #5
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I'm interested. My gut feeling was that this problem had more to do with angles than with Pythagoras.
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October 29th, 2018, 03:19 AM   #6
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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 $\displaystyle b>\sqrt{a(1-a)}$, 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 $\displaystyle -\frac{a}{b}<k<-\frac{a}{b}+\frac{1}{b}$.

Equations of lines $\displaystyle AC$ and $\displaystyle BC$ are:
$\displaystyle y_{AC}(x)=\frac{b}{1-a}(1-x)$ and
$\displaystyle y_{BC}(x)=0$, respectively.

From equation $\displaystyle y_{PH}(x_{P})=y_{AC}(x_{P})$ we get point $\displaystyle P$ coordinates:
$\displaystyle x_{P}=\frac{b^{2}+a(1-a)(bk-(1-a))}{b(b+(1-a)k)}$,
$\displaystyle y_{P}=\frac{(1-a)(a+bk)}{b+(1-a)k}$.

From equation $\displaystyle y_{QH}(x_{Q})=y_{BC}(x_{Q})$ we get point $\displaystyle Q$ coordinates:
$\displaystyle x_{Q}=\frac{a}{b}(b+(1-a)k)$ and
$\displaystyle y_{Q}=0$.

Equations of lines $\displaystyle RB$ and $\displaystyle RA$ are:
$\displaystyle y_{RB}(x)=kx$ and
$\displaystyle y_{RA}(x)=-\frac{1}{k}(x-a)+b$, 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 $\displaystyle y_{RQ}(x)=\frac{y_{R}}{x_{R}-x_{Q}}(x-x_{Q})$.
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 03:24 AM.
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October 31st, 2018, 08:30 AM   #7
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Quote:
Originally Posted by lua View Post
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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