areaeuclidean-geometrygeometryplane-geometry
If $A$, $B$ and $T$ are points of tangency, and the radius of the quadrant measures $\sqrt2+1$, calculate the area of the shaded region.
(Answer:$\frac{\pi}{4}+\frac{\sqrt2}{2}-1)$
$HC = \sqrt2+1:\angle HDC = \angle HCD = 45^\circ$
$DH=CH \implies \triangle CHD{~\text{(isosceles)}}$
$CD^2=2(\sqrt2+1)^2=6+4\sqrt2 \implies CD = \sqrt{6+4\sqrt2}$
I couldn't finish
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$$
Best Answer
Here's your diagram, with a line, a point and couple of line lengths added:
As shown, have $\lvert KT\rvert = \lvert KB\rvert = \lvert HB\rvert = r$. Thus, $\lvert KH\rvert = \lvert HT\rvert - \lvert KT\rvert = (\sqrt{2} + 1) - r$. With the isosceles right-angled $\triangle KBH$, the hypotenuse length is $\sqrt{2}$ times the leg lengths, so
$$\begin{equation}\begin{aligned} \sqrt{2} + 1 - r & = \sqrt{2}r \\ (\sqrt{2} + 1)r & = \sqrt{2} + 1 \\ r & = 1 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Also as shown, draw from $V$ the line perpendicular to $HC$, with it meeting at $M$. Let $\lvert VM\rvert = h$, so $\lvert HM\rvert = h$ as well. Then since $\triangle AHC \sim \triangle VMC$, we have
$$\begin{equation}\begin{aligned} \frac{\lvert AH\rvert}{\lvert HC\rvert} & = \frac{\lvert VM\rvert}{\lvert MC\rvert} \\ \frac{1}{\sqrt{2} + 1} & = \frac{h}{\sqrt{2} + 1 - h} \\ \sqrt{2} + 1 - h & = h(\sqrt{2} + 1) \\ (2 + \sqrt{2})h & = \sqrt{2} + 1 \\ \sqrt{2}(\sqrt{2} + 1)h & = \sqrt{2} + 1 \\ h & = \frac{1}{\sqrt{2}} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$
This means
$$\lvert VH\rvert = \sqrt{2}h \;\;\to\;\; \lvert VH\rvert = 1 \tag{3}\label{eq3A}$$
Also, we have
$$\lvert VK\rvert = \lvert HK\rvert - \lvert VH\rvert = \sqrt{2} - 1 \tag{4}\label{eq4A}$$
Using $VK$ as the base, the height of $\triangle VKB$ is
$$\frac{1}{2}\lvert AB\rvert = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \tag{5}\label{eq5A}$$
With $A$ being the area of the darker shaded region, as it's the quarter circle area minus twice the area of $\triangle VKB$, we get using \eqref{eq1A}, \eqref{eq4A} and \eqref{eq5A} that
$$\begin{equation}\begin{aligned} A & = \frac{1}{4}(\pi(1^2)) - \frac{1}{\sqrt{2}}(\sqrt{2} - 1) \\ & = \frac{\pi}{4} - 1 + \frac{1}{\sqrt{2}} \\ & = \frac{\pi}{4} + \frac{\sqrt{2}}{2} - 1 \end{aligned}\end{equation}\tag{6}\label{eq6A}$$