To solve the problem: Lines through vertices of $\triangle ABC$ and a point $Q$ meet opposite sides at $M$, $N$, $P$. When is $Q$ the orthocenter of $\triangle MNP$?, I tried to use methods of analytical geometry. I kept the notations used in the initial phase and found the following:
-$AQ$ perpendicular to $NP$:
$(1)$ $$ \frac{an(n-a)(2am+bn+cn-ab-ac)}{m(-a^2m^2 +bcn^2+a^2bm+a^2cm-a^2bc)} =1 $$
-$BQ$ perpendicular to $MP$:
$(2)$ $$\frac{a(c-b)(a-n)n^2}{(b-m)(-a^2m^2-2abmn-bcn^2+a^2bm+a^2cm+2abcn-a^2bc)}=1$$
-$CQ$ perpendicular to $MN$:
$(3)$ $$\frac{a(b-c)(a-n)n^2}{(c-m)(-a^2m^2-2acmn-bcn^2+a^2bm+a^2cm+2abcn-a^2bc)}=1$$
After making the calculations, we get the equalities:
$(1')$
$a^2m^3+2a^2mn^2-bcmn^2 +abn^3+acn^3-a^2bm^2-a^2cm^2-2a^2bn^2-2a^2cn^2-2a^3mn+a^2bcm+a^3bn +a^3cn=0$
$(2')$
$a^2m^3+2abm^2n+bcmn^2-acn^3+abn^3-a^2cm^2-2a^2bm^2-2abcmn-2ab^2mn+ a^2cn^2-a^2bn^2-b^2cn^2+2a^2bcm+a^2b^2m+2ab^2cn-a^2b^2c =0$
$(3')$
$a^2m^3+2acm^2n+bcmn^2-abn^3+acn^3-a^2bm^2-2a^2cm^2-2abcmn-2ac^2mn+ a^2bn^2-a^2cn^2-bc^2n^2+2a^2bcm+a^2c^2m+2abc^2n-a^2bc^2 =0$
Manipulating these relationships to characterize the ABC triangle is very difficult. We have noticed that by decreasing relations $(2)$ and $(3)$ a third-degree equality is achieved:
$(4)$ $$2m(b-m)(c-m)=(b+c-2m)n(n-a)$$
Does anyone know a way of eliminating the numbers $m$ and $n$ from relations $(1), (2), (3)$ or between relations $(1), (2), (4)$? Note that the data of the problem is known as $a, b, c, m, n$ are real numbers so $0<n<a, b<0<c, b<m<c.$