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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 $0<n<a, b<0<c, b<m<c.$

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  • $\begingroup$ 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) $\endgroup$ – Blue Feb 11 at 13:58
  • $\begingroup$ (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. $\endgroup$ – Blue Feb 11 at 14:29
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    $\begingroup$ @Blue: Thanks a lot for interesting observations. I checked the calculations carefully. I'll do more checks. $\endgroup$ – medicu Feb 11 at 18:18

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