A semicircle combined with a rectangle with 1:2 ratios for arcs, segments and areas

Question

A semicircle and a rectangle are combined together as shown. AB =2. P is on the semicircle with arc ratio AP:BP = 1:2. Q is on AB with AQ:BQ = 1:2. The extension of PQ intersects DC at E and PE separates the whole combined shape into two parts, with the area ratio $S_{ADEP}:S_{PECB}=$ 1:2. What is DE?

I've been given this question for homework for an online class and I can't seem to solve it. I know the lengths of $AQ$, $BQ$, arc $AP$ and arc $PB$, but I can't reduce the number of variables enough to be able to get an equation using the information about the areas to solve for DE.

Hints would be appreciated so much!

Answer 1

We know from the given that $\angle POQ = 60$, OP = 1, AQ = $\frac 23$, QO = $\frac13$ and the areas [OPQ] = $\frac12$OP$\cdot$OQ$\>\sin60 = \frac{\sqrt3}{12}$,

$$[APQ]= [APO] -[OPQ] =\frac\pi6- \frac{\sqrt3}{12}$$ $$[AQDE]=\frac12\left(\frac23+x\right)y,\>\>\>\>\>\>\>[APBCD]=\frac\pi2+2y$$

Use the given [APDE]=$\frac12$[PECB] to establish the equation below for the unknowns $x$ and $y$,

$$\frac\pi6- \frac{\sqrt3}{12}+\frac12\left(\frac23+x\right)y=\frac13 \left(\frac\pi2+2y\right)$$

Simplify to get

$$6xy-\sqrt3=4y\tag{1}$$

Then, apply the cosine and sine rules for the triangle PQO to obtain $PQ=\frac{\sqrt7}{3}$ and then $\sin\alpha= \frac32\sqrt{\frac37}$. Recognize that $\tan\alpha =3\sqrt3 = \frac{y}{x-\frac23}$ to establish another equation for $x$ and $y$,

$$3\sqrt3 x-2\sqrt3=y\tag{2}$$

Now, solve (1) and (2) to obtain,

$$DE=x=\frac{4+\sqrt2}{6}$$

Answer 2

The arc ratio $|\stackrel{\frown}{AP}|:|\stackrel{\frown}{PB}|=1:2$ implies that $\angle AOP=60^\circ$, so that $\triangle OAP$ is equilateral. For simplicity, define $m:=1/3$ so that we can conveniently write $|AB|=12m$ and still have the desired length $|AP|=2$.

Now, define $R$ as the area of one-third the rectangle, $S$ as the area of one-third the semicircle. Also, define $T := |\triangle OPQ|$ and $U:=|\triangle EQQ'|$. Drop perpendiculars from $P$ and $Q$ to $P'$ and $Q'$, and define $V:=|\triangle QPP'|$ and $n:=|Q'E|$.

Note that, because $|PP'|$ is the altitude of equilateral $\triangle OAP$, we have $P'$ is the midpoint of $\overline{OA}$, so that $|QP'|=m$ and $|OQ|=2m$ (that's why $|AB|=12m$ was so convenient), and therefore, most-importantly, $T=2V$.

Now to the business of actually solving the problem: $$\begin{align} |ADEP|:|PECB|=1:2 &\quad\to\quad 2(R+U+S-T)=2R-U+2S+T \\[4pt] &\quad\to\quad U = T = 2V \\[4pt] &\quad\to\quad \frac{U}{V}=2 \end{align}\tag{1}$$ But, $\triangle QPP'\sim\triangle EQQ'$, so that $$\frac{U}{V}=\left(\frac{n}{m}\right)^2\tag{2}$$ Consequently, $n=m\sqrt{2}$, and we have

$$|DE| = m\left(4 + \sqrt{2}\right)=\frac13(4+\sqrt{2}) \tag{$\star$}$$