euclidean-geometrygeometryplane-geometry
For reference: Given the triangle below, where $AB = BC$, find the value of $x$ (Answer: $150^o$)
My progress:
$2\alpha+a+2\theta = 90^o\implies (\alpha+\theta) = 45^o – \frac{a}{2}\\
\alpha+b + 2\theta = 90^o\\\
\alpha+b = a+2\alpha \implies \alpha = b-a\\
x=180^0-(\alpha+\theta)=135^o +\frac{a}{2}\\
180-(3\theta+2\alpha)+180-(a+b)+x = 360 \implies x = 3\theta+2\alpha+a+b\\
$
Some detail is missing
It is not possible to change the solution and get the original figure
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$$
Best Answer
(I use $t,a$ instead of $\theta$, $\alpha$ for ease of typing.)
In the picture, $E$ is the mid point of $AC$. A prime denotes reflection w.r.t. $BE$, so for instance $A'=C$. Build $K'$, then $L=AK\cap A'K'\in BE$. Divide the $4t$ angle in $\hat B$ in four $t$-parts, the angles $\widehat{KAB}$, $\widehat{K'CB}$ in two $a$-parts. Draw all remaining angle bisectors in $\Delta BAL$ and $\Delta BCL$. Denote by $u$ the half angles built in $L$ in these triangles. The sum of angles in them is $180^\circ$, so $$ a+t+u=\frac 12180^\circ=90^\circ\ . $$ Since $KK'\|AA'=AC$ is perpendicular to $BLE$, we have two $(a+t)$-angles in $\Delta LKK'$, they are then copied in $\Delta LAC$. The sum of angles in $\Delta ABC$ is $180^\circ$ so: $$ 180^\circ=4t+(2a+(a+t))+(2a+(a+t))=6(a+t)\ . $$ This gives $a+t=30^\circ$, so the angle in $K$ in $\Delta BKC$ is the remaining amount of $180^\circ$.
$\square$