# Lines through vertices of $\triangle ABC$ and a point $Q$ meet opposite sides at $M$, $N$, $P$. When is $Q$ the orthocenter of $\triangle MNP$?

Consider a point $$Q$$ inside the $$\triangle ABC$$ triangle, and $$M$$, $$N$$, $$P$$ the intersections of $$\overleftrightarrow{AQ}$$, $$\overleftrightarrow{BQ}$$, $$\overleftrightarrow{CQ}$$ with respective sides $$\overline{BC}$$, $$\overline{CA}$$, $$\overline{AB}$$.

What are the triangles $$\triangle ABC$$, respectively, which are the positions of $$Q$$, for which $$Q$$ is the intersection of the heights (aka, the orthocenter) in $$\triangle MNP$$?

I have not been able to solve the problem using synthetic geometry elements. In order to solve the problem with methods of analytical geometry, we considered $$A(0, a)$$, $$B(b, 0)$$, $$C(c, 0)$$, and $$Q(m, n)$$, where $$0. We calculated the coordinates of $$M$$, $$N$$, $$P$$ and then we set the conditions that the $$\overleftrightarrow{AQ}$$, $$\overleftrightarrow{BQ}$$, $$\overleftrightarrow{CQ}$$ are perpendicular to $$\overline{NP}$$, $$\overline{PM}$$, $$\overline{MN}$$. But the calculations became complicated, and I quit.

If I could have continued, I would have found three conditions (equalities) that I would have to satisfy $$m$$ and $$n$$ simultaneously. It follows that the $$\triangle ABC$$ triangle must be a particular triangle; and, within this triangle, $$Q$$ must have a particular position. One such case is the equilateral triangle and $$Q$$ is its center.

How can triangles be characterized with this property?

Any isosceles triangle can comply with your stated conditions. Consider the sketch below. The positions of $$E$$ and $$F$$ can be adjusted between $$E_1$$ and $$E_2$$ and $$F_1$$ and $$F_2$$ respectively until the lines $$BE$$ and $$AF$$ are perpendicular to $$DF$$ and $$DE$$.

$$EF$$ is parallel to $$AB$$ and symmetry plays a big part in this setup.

• We have already said that if the AQ, BQ, CQ lines are perpendicular to NP, PM, MN then the ABC triangle must be a particular one. Can we prove that under these conditions, the ABC triangle is isoscel? And the example given by you supports this supposition. – medicu Nov 10 '18 at 10:17
• I'm not fully understanding what you are saying but a summary of the proof for an isosceles triangle is that for a particular point Q along line CD, there is a position of E and F where EF is parallel to AB, where line AF is perpendicular to line ED and line BE is perpendicular to DF. The locations of E and F are between E1 , E2 and F!, F2 which span a continuous range of angles which includes 90 degrees of the aforementioned lines.. – Phil H Nov 10 '18 at 13:14
• Your example is clear. The question is whether such situations are only in the isosceles triangle, or there are triangles in the non-isosceles. – medicu Nov 10 '18 at 15:40

Changing notation a bit, we'll consider a triangle with circumdiameter $$1$$ and angles $$\alpha$$, $$\beta$$, $$\gamma$$. Duplicating OP's configuration with $$A$$ on the $$y$$-axis, we can write $$A = (0,\sin\beta\sin\gamma) \qquad B = (\cos\beta\sin\gamma,0) \qquad C = (-\sin\beta\cos\gamma,0) \tag{1}$$ Now, define our variable point, $$P$$, using barycentric coordinates: $$P = \frac{p A + q B + r C}{p + q + r} \qquad(p, q, r \neq 0) \tag{2}$$ Let $$\overleftrightarrow{AP}$$, $$\overleftrightarrow{BP}$$, $$\overleftrightarrow{CP}$$ meet $$\overleftrightarrow{BC}$$, $$\overleftrightarrow{CA}$$, $$\overleftrightarrow{AB}$$ at $$D$$, $$E$$, $$F$$. We find that $$D = \frac{qB+rC}{q+r} \qquad E = \frac{rC+pA}{r+p} \qquad F = \frac{pA+qB}{p+q} \tag{3}$$ For $$P$$ to be the orthocenter of $$\triangle DEF$$, we must have $$\overleftrightarrow{AP}\perp\overleftrightarrow{EF}$$, $$\overleftrightarrow{BP}\perp\overleftrightarrow{FD}$$, $$\overleftrightarrow{CP}\perp\overleftrightarrow{DE}$$, where any two guarantee the third. Considering the last two gives equations that reduce to \begin{align} (B-P)\cdot(F-D) &= 0 \quad\to\quad \overline{p}^2\left(\overline{q} + \overline{r}\cos\alpha \right) = \overline{r}^2\left(\overline{q} + \overline{p}\cos\gamma \right) \tag{4}\\ (C-P)\cdot(D-E) &= 0 \quad\to\quad \overline{q}^2 \left(\overline{r} + \overline{p} \cos\beta \right) = \overline{p}^2 \left(\overline{r} + \overline{q} \cos\alpha \right) \tag{5} \end{align} where $$\overline{p}:= p/\sin\alpha$$, $$\overline{q}:=q/\sin\beta$$, $$\overline{r}:=r/\sin\gamma$$.

Using the method of resultants (that is, the Resultant[] function in Mathematica), we can eliminate $$\overline{p}$$ from $$(4)$$ and $$(5)$$. The relation reduces to

\begin{align} 0 &= \phantom{2} \overline{q}^4\phantom{\overline{r}^4} \sin^2\beta \\[4pt] &+ \phantom{2} \overline{q}^3 \overline{r} \phantom{^4} \cos\beta \left(\cos\gamma - \cos\alpha \cos\beta \right) \\[4pt] &- 2 \overline{q}^2 \overline{r}^2 \left(1 - \cos\alpha \cos\beta \cos\gamma\right) \\[4pt] &+ \phantom{2}\overline{q}\phantom{^4} \overline{r}^3 \cos\gamma \left(\cos\beta - \cos\alpha\cos\gamma \right) \\[4pt] &+ \phantom{2}\phantom{\overline{q}^4}\overline{r}^4 \sin^2\gamma \end{align} \tag{\star}

To get an orthocenter configuration, we can take $$r=1$$ (so that $$\overline{r}=1/\sin\gamma$$), solve $$(\star)$$ for $$\overline{q}$$, then return to $$(4)$$ to solve for $$\overline{p}$$. These values in turn give $$p$$ and $$q$$, which determine the desired $$D$$, $$E$$, $$F$$. $$\square$$

There's no bias in $$(\star)$$ for isosceles triangles. Indeed, after admittedly scant experimentation, it appears that the four roots are valid for any triangle (barring unanticipated degeneracies), and that they correspond to one "internal" orthocenter and three "external" orthocenters (one per side). For example: