# How can I eliminate $m$ and $n$ the following equations?

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

• Using the method of resultants on your $(1)$, $(2)$, $(3)$, and ignoring extraneous factors, I get $a^2-bc=0$. However, there is a problem with regard to how these equations are supposed to be capturing the situation in your previous question. If two perpendicularity properties hold (say, $AQ\perp MP$ and $BQ\perp NP$), then we already know that $Q$ is the orthocenter of $\triangle MNP$. The third perpendicularity condition is redundant, so your equations should be dependent. We shouldn't be able to eliminate $m$ and $n$. Ergo, there must be an error in your calculations. (continued) – Blue Feb 11 at 13:58
• (continued) Working from your previous question, I confirm $(1)$. But I get a couple of different signs in $(2)$ & $(3)$, equivalent to changing both of your right-hand sides to $-1$. Using the method of resultants to remove $m$ from "my" $(1)$ & $(2)$, and then also from "my" $(1)$ & $(3)$, I get extraneous factors of $a$, $n$, $(a-n)$, $(b-c)$; I also get a common factor (quartic in $n$), confirming the dependency of the original equations. I haven't examined the quartic; if it only has viable roots for $b=-c$, then it confirms that the orthocenter property is limited to isosceles triangles. – Blue Feb 11 at 14:29
• @Blue: Thanks a lot for interesting observations. I checked the calculations carefully. I'll do more checks. – medicu Feb 11 at 18:18