MHB Prove one of the angles of triangle is 60°

[anemone]

Given that $PA,\,QB,\,RC$ are the altitudes of the acute triangle $PQR$ such that $9\vec{PA}+4\vec{QB}+7\vec{RC}=0$.

Show that one of the angles of triangle $PQR$ is $60^{\circ}$.

 

[Opalg]

[sp]Denote the lengths of the sides of the triangle by $|\vec{QR}| = p$, $|\vec{RP}| = q$, $|\vec{PQ}| = r$, and the lengths of the altitudes by $|\vec{PA}| = a$, $|\vec{QB}| = b$, $|\vec{RC}| = c.$ Then the area of the triangle is $\frac12ap = \frac12bq = \frac12cr$, so that the ratios $\frac1a : \frac1b:\frac1c$ and $p:q:r$ are the same.

If $9\vec{PA}+4\vec{QB}+7\vec{RC}=0$, then the vectors $9\vec{PA}$, $4\vec{QB}$ and $7\vec{RC}$ form a triangle whose sides have lengths in the ratio $9a : 4b : 7c$. Also, each side of this triangle is in a direction perpendicular to the corresponding side of the triangle $PQR$ ($\vec{PA}$ is perpendicular to $\vec{BC}$ and so on), so the two triangles have the same angles and are therefore similar. Thus the lengths of their sides are in the same ratio, so that the ratios $9a : 4b : 7c$ and $\frac1a : \frac1b:\frac1c$ are the same. Therefore $9a^2 = 4b^2 = 7c^2$, or equivalently $p^2 = 9\lambda$, $q^2 = 4\lambda$, $r^2 = 7\lambda$ for some constant $\lambda.$

By the cosine rule in the triangle $PQR$, $\cos R = \dfrac{p^2 + q^2 - r^2}{2pq} = \dfrac{9+4-7}{2\sqrt9\sqrt4} = \dfrac6{12} = \dfrac12$ and hence $\angle R = 60^\circ.$[/sp]

 

[anemone]

Thank you Opalg for your neat solution!

Alternate solution that I saw somewhere online:

Spoiler

Let $\alpha=\angle QPR,\,\beta=\angle RQP$ and $\gamma=\angle PRQ$ and $a,\,b,\,c$ be the respective magnitudes of vectors $\vec{PA},\,\vec{QB},\,\vec{RC}$.

Taking the dot product of the vector equation with $\vec{QR}$ and note that $\angle BQR=90^{\circ}-\beta$, we find that $4b\sin \gamma=7c\sin \beta$.

Similarly, $9a\sin \gamma=7c\sin \alpha$ and $9a\sin \beta=4b\sin \alpha$.

Using the conventional notation for the sides of the triangle, we have that $p:q:r=\sin alpha:\sin \beta:\sin \gamma=9a:4b:7c$.

However, we also have that twice the area of triangle $PQR$ is equal to $pa=qb=rc$, so that $p:q:r=\dfrac{1}{a}:\dfrac{1}{b}:\dfrac{1}{c}$. Therefore $9a^2=4b^2=7c^2=k$ for some constant $k$.

$\begin{align*}\therefore \cos PRQ&=\dfrac{p^2+q^2-r^2}{2pq}\\&=\dfrac{81a^2+16b^2-49c^2}{72ab}\\&=\dfrac{9k+4k-7k}{12k}\\&=\dfrac{1}{2}\end{align*}$

from which it follows that $\angle R=60^{\circ}$.