areaeuclidean-geometrygeometryplane-geometrytriangles
For reference:
In the figure, calculate the area of the shaded region if $ML=LN, PM=a, NQ=b, AC=c.$ (Answer: $\frac{(a+b)c}{4}$)
My progress:
Let $ML=LN=BL=x.$
$S_{ALC} = S = S_{APQC}-S_{ALP}-S_{CLQ}\implies=\frac{(CQ+AP)(a+b+2x)}{2}-\frac{AP.(a+x)}{2}-\frac{CQ(b+x)}{2}$
$\therefore S =\frac{AP}{2}(b+x)+\frac{CQ}{2}(a+x)$
$S_{ABCL} = \frac{c\cdot x}{2}$
With the usual conventions:
\begin{aligned} \overrightarrow{GF}\times\overrightarrow{GB} &= (0,0,32)\\ \overrightarrow{GF}\times\overrightarrow{GC} &= (0,0,-18)\\ \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC} &= \overrightarrow{0} \end{aligned}
The area you search: \begin{multline*} \frac{1}{2}\big\lvert\overrightarrow{GF}\times\overrightarrow{GA}\big\rvert=\frac{1}{2}\Big\lvert\overrightarrow{GF}\times\big(-\overrightarrow{GB}-\overrightarrow{GC}\big)\Big\rvert =\frac{1}{2}\big\lvert-\overrightarrow{GF}\times\overrightarrow{GB}-\overrightarrow{GF}\times\overrightarrow{GC}\big\rvert=\frac{1}{2}\big\lvert(0,0,-14)\big\rvert = 7 \end{multline*}
Unbelievable setup! I had great joy of simply angle chasing and discovering and I wish to share with you.
$\triangle BFG$ is the reference triangle. $FP,GQ$ are given its external angle bisectors; let them meet at excenter $E$. Drop perpendiculars from $E$ onto $AB,BC$ and complete the square $ABCE$. Lets observe that $AEC$ is exact copy of $ABC$, so $\angle PEQ = 45^\circ$.
Now the fun begins. $\angle PEG=45^\circ=\angle PCG$, hence $PECG$ is cyclic. $\angle EPG$ being opposite to $\angle ECG$ is a right angle. Similarly $AFQE$ is cyclic, making $\angle FQE$ another right angle. Observe that $FG$ subtends $90^\circ$ at $B,P,Q$. As a result, $B,F,P,Q,G$ lie on same circle $!!$ Its center is $D$, midpoint of $FG$.
Now $DP=DF=$ radius of circle, $\angle DPF = \angle DFP = \angle PFA$ implying that $DP \parallel BA$. Hence $\triangle BPF$ and $\triangle BDF$ have same area. Take away their common area and we get $\triangle FRP$ and $\triangle BRD$ have same area. Similarly $DQ \parallel BC$ resulting in $\triangle GSQ$ and $\triangle BSD$ having same area. Adding the common area of $PRSQ$ to these, we see $FPQG$ and $\triangle BPQ$ have same area.
We do angle chasing one more time to find some lengths. $\angle BPC =$$ \angle PAB + \angle ABP = \angle PBQ + \angle ABP = \angle ABQ $. Therefore $\triangle ABQ \sim \triangle CPB$ by $AA$ similarity. So $$\frac{AB}{PC}=\frac{AQ}{BC} \Rightarrow AB^2=9\times 8 \Rightarrow AB=6\sqrt{2} \Rightarrow AC=12$$
From this it is found $AP=3, PQ=5, QC=4 \, (!)$ Consequently, $$[FPQG]=[BPQ]=\frac{PQ}{AC}\times [ABC]=\frac{5}{12}\times 36=15 \quad \square$$
Best Answer
First notice that $AB$ and $CB$ are angle bisectors of $\angle CAP$ and $\angle ACQ$ respectively. Next, drop a perp from $M$ to $AC$. Say it meets $AC$ at $H$.
As $L$ is midpoint of $MN$, $ \displaystyle LE = \frac{a+b}{2}$.
That gives area of the shaded region as $~~\dfrac{(a+b) \cdot c}{4}$
Why $AB$ and $CB$ are angle bisectors of $\angle CAP$ and $\angle ACQ$.