euclidean-geometrygeometryplane-geometry
For reference: In triangle ABC, $S_1$ and $S_2$ are areas of the shaded regions. If $S_1 \cdot{S}_2=16 cm^4$, calculate $MN$.
My progress:
$\frac{AM.DM}{2}.\frac{CN.FN}{2}=16 \implies AM.DM.CN.FN=64\\
\frac{S1}{S2} = \frac{AM.MD}{CN.FN}\\
\frac{S1}{\frac{MI.DM}{2}}=\frac{AM}{MI}\implies S1 = \frac{AM.DM}{2}\\
\frac{S2}{\frac{NI.FN}{2}}=\frac{CN}{NI}\implies S2 = \frac{CN.FN}{2}$
…..????
Draw segment $QR = QP$ then,
$\angle PRQ = \angle PCA \implies AC \parallel QR$. Also, $CQ$ is angle bisector of $\angle ACB$.
So, $ \displaystyle \frac{QR}{AC} = \frac{BQ}{BA} = \frac{BC}{BC+AC} = \frac{a}{a+b}$
$ \displaystyle \implies PQ = QR = \frac{ab}{a+b}$
The key is to recognize that since $\overline{AC}$ bisects $\angle DAB$, the altitudes from $C$ to $\overline{DE}$ and $\overline{AB}$ are congruent. Then since $AB$ is $5/8$ times $DE$, the area of $\triangle ABC$ must be $5/8$ times the area of $\triangle CDE$. You already have the the area of $\triangle CDE$, so the area of $\triangle ABC$ follows.
Best Answer
$$\angle A=\angle IDM = \angle DIB = \angle IBF = \angle FIN = \angle CFN$$
$$\frac{DM}{AM}=\frac{IM}{DM}=\frac{FI}{BF}=\frac{FN}{IN}=\frac{CN}{FN}=x$$
$$DM=x AM, IM=x DM = x^2 AM,$$ $$BF=ID=\sqrt{IM^2+DM^2}=x AM \sqrt{1+x^2}, FI=x BF=x^2 AM \sqrt{1+x^2}$$ $$FI^2=IN^2+FN^2=IN^2+(x IN)^2=(1+x^2) IN^2 \Rightarrow IN=\frac{FI}{\sqrt{1+x^2}}=x^2 AM$$ $$FN=x IN=x^3 AM, CN=x FN=x^4 AM$$
$$S_1=AM\cdot DM / 2 = x AM^2 / 2, S_2=FN\cdot CN / 2= x^7 AM^2 /2$$ $$S_1 S_2 = x^8 AM^4 / 4 \Rightarrow x^2 AM = \sqrt[4]{4S_1 S_2}$$ $$MN=IM+IN=2x^2 AM=\sqrt[4]{64 S_1 S_2}=4\sqrt{2} \rm{\ cm}$$