I am stuck on this problem -
Let $O$ and $H$ be the circumcenter and orthocenter of triangle $ABC$, respectively. Let $a$, $b$, and $c$ denote the side lengths.
Find $AH^2 + BH^2+CH^2$ if the circumradius is equal to $7$ and $a^2 + b^2 + c^2 = 432$.
So far, I've tried using vectors to tease out the solution, but nothing's working. I have found that $OH$ is $3$ but I'm not sure how that helps.
On
Orthocentre distance to triangle vertices as a function of triangle angles:
$\frac{AH}{|cos(α)|}=\frac{BH}{|cos(β)|}=\frac{CH}{|cos(γ)|}=2·R$
$AH^2+BH^2+CH^2=(2·R)^2·(cos^2(α)+cos^2(β)+cos^2(γ))$
$$ \begin{matrix} \text{replacing:}&cos(α)=\frac{-a^2+b^2+c^2}{2·b·c}&cos(β)=\frac{a^2-b^2+c^2}{2·a·c}&cos(γ)=\frac{a^2+b^2-c^2}{2·a·b}\\ \end{matrix} $$
$cos^2(α)+cos^2(β)+cos^2(γ)=3-\frac{a^2+b^2+c^2}{(2·R)^2}=3-\frac{432}{(2·7)^2}=\frac{39}{49}$
$AH^2+BH^2+CH^2=(2·7)^2·\frac{39}{49}=156$
Consider the triangle obtained by drawing at each vertex of $ABC$ a line parallel to its opposite side. These three new lines form a new triangle $A'B'C'$ that can be seen to be similar to the original triangle, with doubled sides. Also, notice that the circumcenter of $A'B'C'$ is the orthocenter of $ABC$, which we denoted $H$. Thus, by Pythagoras,
\begin{gather*} (HC')^2 = (AC')^2 + AH^2 = a^2 + AH^2\\ (HA')^2 = (BA')^2 + BH^2 = b^2 + BH^2\\ (HB')^2 = (CB')^2 + CH^2 = c^2 + CH^2. \end{gather*} Summing the three expressions above and using $HC' = HB' = HA' = 2 \cdot 7 = 14$ (by the similarity relation), we have $$ 3 \cdot 14^2 = a^2 + b^2 + c^2 + (CH^2 + AH^2 + BH^2) = 432 + (CH^2 + AH^2 + BH^2) $$ implying that $$ CH^2 + AH^2 + BH^2 = 588 - 432 = 156. $$