$ABC$ and $XYZ$ have the same orthocenter.

Question

Let $ABC$ a nonequilateral triangle with $tg B\cdot tg C=3$ and $X\in BC$ s.t. $AX\perp BC$.

If $Y$ is the middle of $AB$ and $Z$ is the middle of $AC$ then $ABC$ and $XYZ$ have the same orthocenter.

I need a synthetic argument.

Answer 1

Let $\Delta ABC$ be the given triangle with $\tan B\cdot \tan C = 3$. Let $H,X$ be its orthocenter, respectively the projection of $A$ on $BC$. Then: $$ \begin{aligned} \frac {BX}{HX} &=\cot \widehat{HBX} =\tan \hat C =\frac{AX}{CX}\ ,\\ \frac {CX}{HX} &=\cot \widehat{HCX} =\tan \hat B =\frac{AX}{BX}\ ,\\ \end{aligned} $$ We take the product and get $$ \frac {AX^2}3 = BC\cdot CX =3HX^2\ . $$ This gives $AX^2 =9HX^2$, i.e. $AX=3HX$.

So the point $H$ is placed on the segment $AX$ at one third distance of it from $X$.

The same calculus applies also in the triangle $\Delta XYZ\sim \Delta ABC$. The homothety factor is $1/2$. This is because $YB=YA=YX=\frac 12 AB$ and correspondingly for $Z$ instead of $Y$. Let $W$ be the intersection of $AX$ with $YZ$.

Then with the same argument, the orthocenter of $\Delta XYZ$ is placed on $XW=\frac 12 AX$, at one third distance of $XW$ from $W$. We get the same point because of the relation considered on the half-line $XA$ $$\frac 13+\frac 13\cdot \frac 12=\frac 12\ .$$