Possible vector solution for synthetic geometry involving parallelogram and triangle sharing centroid?

Question

Let $ABC$ be a triangle with $AB=30$, $BC=51$, $CA=63$. Points $P$ and $Q$ lie on $\overline{BC}$, $R$ lies on $\overline{CA}$, and $S$ lies on $\overline{AB}$ such that $PQRS$ is a parallelogram, and the center of $PQRS$ coincides with the centroid of $\triangle ABC$. What is the area of parallelogram $PQRS$? (Source: CMC)

There is a synthetic solution involving half the height from $SR$ to $PQ$ being $1/3$ the height from $A$ to $BC$, but I'm looking for a solution involving vectors or complex geometry to learn how to apply them in this kind of problem.

I've tried some things that haven't lead very far. $\cos A = 3/5$ by the Law of Cosines and that seems to be the only nice angle, so I tried scaling the sides, then setting $w = \mathrm{cis}A$ and $|z| = 1$ s.t. $10z=AB$, $21zw=AC$ but that doesn't really help since we already know the side lengths. I tried doing the same instead with the parallelogram, but my angle chasing didn't get very far.

By the way, the answer is

336.

I would most appreciate an analytic solution not entirely depending on the insight that half the height from $SR$ to $PQ$ is $1/3$ the height from $A$ to $PQ$, since the easiest step to go from there would not be the proceed with an analytic solution, but just to use Heron's and complete the problem from there.

Answer 1

Yes, pure vector solution is possible. We let $c:=\overrightarrow{AC}$, $b:=\overrightarrow{AB}$, $\ b,c$ be the basis and $A$ be the origin meaning if I write $X=xb+yc$ it reads $\overrightarrow{AX}=xb+yc$ (for an arbitrary point $X$ and scalars $x,y$). The picture:

It's given that $$\begin{cases} S=yb\\R=xc\\R-S=t(b-c)\\ P=uc+(1-u)b\\Q=vc+(1-v)b\\ R-S=Q-P\\Q-R=P-S\\\frac{Q+S}{2}=\frac{0+b+c}{3} \end{cases}$$ It's a purely linear vector problem, meaning we don't take constraints $|b|=30,\,|c|=63,\,|b-c|=51$ at all for a while. Solving the system we get $$\begin{cases} x=\frac13\\y=\frac13\\u=\frac13\\v=\frac23 \end{cases}$$ Now, as $P,\,Q,\,R,\,S$ are known along with $b^2=30^2$, $c^2=63^2$ and $\langle b,c\rangle = \frac12\left(|b|^2+|c|^2-|b-c|^2\right)$ we can express the area of $PQRS$ as \begin{align*} \operatorname{Area}_{PQRS}&= |\overrightarrow{SR}|\cdot |\overrightarrow{SP}| \cdot\left|\sin\angle RSP\right|\\ &=|\overrightarrow{SR}|\cdot |\overrightarrow{SP}| \cdot\sqrt{1-\cos^2\angle RSP}\\ &=|\overrightarrow{SR}|\cdot |\overrightarrow{SP}| \cdot\sqrt{1-\left(\frac{ \langle\overrightarrow{SR},\overrightarrow{SP}\rangle}{ |\overrightarrow{SR}|\cdot |\overrightarrow{SP}|} \right)^2}\\ &=\sqrt{ \overrightarrow{SR}^2\cdot\overrightarrow{SP}^2- \left(\langle\overrightarrow{SR},\overrightarrow{SP}\rangle\right)^2 } \end{align*} hence we can find the answer of $336$.

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Edit: how do we solve the linear system. Okay, the answer on "how did I solve it" is "I didn't, WA did", but it should be solvable). I'll leave this $$\begin{cases} S=yb\\R=xc\\ P=uc+(1-u)b\\Q=vc+(1-v)b \end{cases}$$ rather as "definitions" of $P,\,Q,\,R,\,S$ and will work with the rest of the system. $$\begin{cases} R-S=t(b-c)\\ R-S=Q-P\\ Q-R=P-S\\ \frac{Q+S}{2}=\frac{0+b+c}{3} \end{cases}$$ $$\begin{cases} xc-yb=t(b-c)\\ xc-yb=vc+(1-v)b-(uc+(1-u)b)\\ vc+(1-v)b-xc=uc+(1-u)b-yb\\ 3(vc+(1-v)b+yb)=2(b+c) \end{cases}$$ As $b,\,c$ form a basis, each vector has the only representation as a linear combination of basis vectors, which means that the coefficients of $b$ and $c$ (seperately) of each LHS and RHS of the system are equal, i.e. $$\begin{cases} x=t\\ -y=-t\\ x=v-u\\ -y=(1-v)-(1-u)\\ v-x=u\\ 1-v=(1-u)-y\\ 3v=2\\ 3((1-v)+y)=2 \end{cases}$$ again we could feed it to WA (click to see the result) or perform Gaussian elimination.

Answer 2

Let's use $BC$ is the $x$ axis, say with $B$ being the origin. Then the coordinates of the points are $A=(x_a,y_a)$, $B=(0,0)$, $C=(x_c,0)$, $P=(x_p,0)$, $Q=(x_q,0)$, $R=(x_r,y_r)$, $S=(x_s,y_s)$. We know $PQRS$ is parallelogram, so $y_r=y_s$ and $x_r=x_s+(x_q-x_p)$. Then the center of the parallelogram along $x$ is at $$\frac12(x_q+x_s)=\frac12(x_r+x_p)$$ The $y$ position of the center of parallelogram is at $$\frac 12 (y_s+y_q)=\frac12 y_s$$ The center of the triangle is at $$\left(\frac{x_a+x_c}3, \frac{y_a}3\right)$$ From here you get $$y_s=\frac23 y_a$$ and since $S$ is on $BA$ you have $$\frac{y_s-y_b}{x_s-x_b}=\frac{y_a-y_b}{x_a-x_b}$$ or $$\frac{y_s}{x_s}=\frac{y_a}{x_a}$$ Therefore $x_s=\frac23 x_a$. Similarly $R$ is on $AC$ and therefore $x_r-x_c=\frac 23(x_a-x_c)$ or $$x_r=\frac23x_a+\frac13 x_c$$

Now the area of $\triangle ABC$ is $A_3=\frac 12 y_a\cdot x_c$, and the area of the parallelogram is $$A_4=y_s(x_r-x_s)=\frac 23y_a\left(\frac23 x_a+\frac 13 x_c-\frac 23 x_a\right)=\frac 29 y_a x_c$$ From here $$\frac{A_4}{A_3}=\frac49$$ This is valid for any triangle. Now use Heron or any method to calculate $A_3=756$, and you get $$A_4=\frac 49 756=336$$