non-equilateral triangle- concurrent.

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non-equilateral triangle- concurrent.

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Pirkuliyev Rovsen

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Pirkuliyev Rovsen

#1 h Dec 11, 2014, 8:59 AM • 2 Y

Y by Adventure10, Mango247

In a non-equilateral triangle $ABC$ it is given that $AB^2 + BC^2 = 2AC^2$ . Let $AT$ and $CP$ be the altitudes,
$H$ be the orthocenter, and $M$ be the intersection point of the medians of the triangle $ABC$ . Prove that the lines $AC$ , $PT$ and $HM$ are concurrent.

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#2 h Dec 11, 2014, 2:51 PM • 2 Y

Y by Pirkuliyev Rovsen, Adventure10

Hello.

Let $\bf \color{red} (1)$ be the given condition.

It's not difficult to see that the triangle is acute.Let $BI$ be the third altitude and $F$ the middle of $AC$ .

First step: The quadrilateral $HMFI$ is cyclic.

From the cyclic $CIHT$ we get $BT\cdot BC=BH\cdot BI$ .

The generalised pythagorean theorem gives $AC^2=AB^2+BC^2-2BC\cdot BT\overset{\bf \color{red} (1)}\Leftrightarrow \boxed{BC\cdot BT=\frac{AC^2}{2}}$

We have $BM\cdot BF=\frac{2BF^2}{3}$ .

Applying the thorem of medians gives $BM\cdot BF=\frac{2}{3}\cdot \frac{2(AB^2+BC^2)-AC^2}{4}\overset{\bf \color{red} (1)}=\frac{AC^2}{2}$

Thus, $BM\cdot BF=BT\cdot BT=BH\cdot BI$ whence we get that $HMFI$ is cyclic.

Second step: Since $HMFI$ is cyclic we have $\hat{HMB}=\hat{HIF}=90^{\circ}$ thus $\hat{HMB}=\hat{HTB}$ hence $HMBT$ is also cyclic.

Since $PHTB$ is also cyclic,the points $P,H,T,B,M$ are concyclic.

Third step: $HM$ is the common chord of the circles $(H,M,P,B,T)$ and $(H,M,F,I)$ thus their radical axis.

$FI$ is the common chord of the circles $(H,M,F,I)$ and the Euler circle of $\triangle{ABC}$ thus their radical axis.

$PT$ is the common chord of the circles $(H,M,P,B,T)$ and the Euler circle of $\triangle{ABC}$ thus their radical axis.

Thus $FI,PT,HM$ all pass from the radical center of the three circles Q.E.D.

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