areaeuclidean-geometrygeometryplane-geometry
In the figure the triangle $DCF$ is joined to the square $ABCD$ of side $2$, $M$ is middle of $CD$. So the gray area is:
My progress:
$S_{AMB}=\frac{4}2=2\implies S_{MCB}=S_{ADM}=1$
$\triangle CMF = \cong \triangle BMC\\ \therefore S_{CMF}=1$
$\frac{S_{MDG}}{S_{DFC}}=\frac{DG\cdot DM}{DF\cdot CF}=\frac{DG}{\sqrt2}$
$S_{DGM} = 1-S_{FGM}$
From Stweart's theorem; $FM =\sqrt5$
Using law of cosines in $\triangle ADF$,
$\angle ADG = 135^\circ\implies DF = 2\sqrt2$
…?
Sides of $\triangle DAB$ are in ratio $1:2:\sqrt5$
As $\triangle ARB \sim \triangle DAB$
$AR = \frac{2R}{\sqrt5}, BR = \frac{4R}{\sqrt5} $
As $\angle DAB = 90^\circ, $ and $AR \perp BD$, $ \displaystyle AR^2 = BR \cdot RD \implies RQ = RD = \frac{R}{\sqrt5}, BQ = \frac{3R}{\sqrt5}$.
So, $ \displaystyle AT = CT = \frac{R}{2}$ and that leads to $QT = \frac{3R}{2\sqrt5}, QS = \frac{R}{\sqrt5} ~ , ~ $ where $S$ is the foot of the perp from $C$ to $QT$.
By intersecting chords theorem,
$ \displaystyle BQ \cdot RQ = \frac{3R^2}{5} = PQ \cdot (2 PS - PQ) = x \left(x + \frac{2R}{\sqrt5}\right)$
Solving, $ \displaystyle x = \frac{R}{\sqrt5}$
Alternatively, once you show that $CT = \frac{R}{2}$, you know that perp from $P$ to $AB$ is $R$ (using similarity of triangle to $\triangle DAB$). As the perp is $R$, it must be $CP$ that is perp $AB$. That leads to an alternate solution using Pythagoras in $\triangle CPT$ or $\triangle PQD$.
I found this solution:
$BM=x$
$\angle BAC = \alpha\\ \angle BAM = \theta\\ \therefore \angle NMB = \theta +\alpha\\ \overset{\LARGE{\frown}}{PQ}= \overset{\LARGE{\frown}}{AQ}+\overset{\LARGE{\frown}}{BC}\implies \angle MBN = \theta+\alpha \therefore BN = MN = m\\ \overset{\LARGE{\frown}}{AP}\cong \overset{\LARGE{\frown}}{PC} \implies$
M is midpoint:$AM = MC = m + n (NC = n)$
Power point $N$: $m.PN= n(2m+n) \implies \boxed{PN = \frac{n(2m+n)}{m}}$
Drew $NK \perp MB\\ \angle MNK = (\alpha+\theta) \\ \triangle NMK \rightarrow \boxed{cos(\theta+\alpha) = \frac{x}{2m}}\\ \triangle PCB \sim \triangle NCP \therefore \boxed{PC^2 = PN.PB}\\ PN+m=: \boxed{PB=\frac{(m+n)^2}{m}}\\ \therefore \boxed{PC^2=\frac{n(2m+n)(m+n)^2}{m^2}}\\ \triangle PCH \sim \triangle MNP : \frac{a}{m}=\frac{PC}{PN}$
Replacing PN e PC, and squaring
$\frac{a^2n^2(2m+n)^2}{m^2}=\frac{n(2m+n)(m+n)^2}{m^2}\\ \boxed{a²=\frac{m^2(m+n)^2}{n(2m+n)}}\\ \triangle MNP: cos(2(\theta+\alpha))=\frac{m}{PN}\\ (cos2a =2cos^2a-1) \therefore 2*\frac{x^2}{4m^2}-1=\frac{m^2}{n(2m+n)}\\ \frac{x^2}{2}=1+\frac{m^2}{n(2m+n)} \implies \frac{x^2}{2}=\frac{m²(m+n)²}{n(2m+n)} \\ \therefore x^2 = 2a^2 \implies \boxed{\color{red}x=a\sqrt2} $
(Solution:by jvmago)
Best Answer
We have $S_{MAB} = \frac12 S_{ABCD} = 2$ and $S_{MCF} = \frac14 S_{ABCD} = 1$, so it's only $S_{MGD}$ that needs tricky calculation.
We have \begin{align} S_{MGD} : S_{MFG} &= GD : FG & \text{(triangles with same height, different base)} \\ &= S_{BGD} : S_{BFG} & \text{(triangles with same height, different base)} \\ &= S_{BMD} : S_{BFM} & \text{(subtract equal ratios)} \\ &= 1 : 2. \end{align} (To explain the penultimate step: since $S_{MGD} : S_{MFG} = S_{BGD} : S_{BFG}$, it is also equal to the difference $(S_{BGD} - S_{MGD}) : (S_{BFG} - S_{MFG})$, which simplifies to $S_{BMD} : S_{BFM}$.)
Since $S_{MGD} + S_{MFG} = S_{MFD} = 1$, we conclude that $S_{MGD} = \frac13$, so the total area is $\frac13 + 1 + 2 = \frac{10}{3}$.