points $R$ and $T$ lie on the side $CD$ of the parallelogram $ABCD$ such that $DR= RT= TC$ what is the area, in $cm^2$ , of the shaded region?

Question

points $R$ and $T$ lie on the side $CD$ of the parallelogram $ABCD$ such that $DR= RT= TC$ . Lines $AR$ and $AT$ intersect the extension of $BC$ at points $M$ and $L$ respectively, and the lines $BT$ and $BR$ intersect the extension of $AD$ at points $S$ and $P$ respectively. If the area of the parallelogram $ABCD$ is $48$ cm$^2$ , then what is the area, in cm$^2$ , of the shaded region? Im pretty sure ML=SP but im not sure how to prove it also I think $SMCD$ is a paralleogram but im not sure how to prove that either, hints or suggestions would be appreciated aswell as solutions

from the 2019 SAIMC

Answer 1

Let $AL\cap BP=\{X\}$.

Thus, $$\frac{LC}{AD}=\frac{CT}{DT}=\frac{1}{2}=\frac{DR}{RC}=\frac{DP}{BC}=\frac{DP}{AD},$$ which gives $$LC=DP,$$ $$AP=BL,$$ which gives $APLB$ is a parallelogram.

By the similar way we obtain $$MC=2BC=2AD=DS,$$ which gives that $ASMB$ is a parallelogram.

Now, let $h$ be an altitude of $ABCD$ from $B$ to $AD$.

Thus, $$S_{shaded}=\frac{3}{4}S_{ASMB}-S_{\Delta APR}-S_{\Delta BLT}-S_{\Delta AXB}=$$ $$=\frac{3}{4}\cdot3AD\cdot h-2\cdot\frac{\frac{3}{2}AD\cdot\frac{1}{3}h}{2}-\frac{1}{4}S_{APLB}=$$ $$=\left(\frac{9}{4}-\frac{1}{2}\right)S_{ABCD}-\frac{1}{4}\cdot\frac{3}{2}AD\cdot h=\left(\frac{9}{4}-\frac{1}{2}-\frac{3}{8}\right)\cdot48=66.$$

Answer 2

Let $AD$ have the length of $2$ (units). Then $PD=1$ and $SP=3$. Same proportions on the segment $MB$. This is because we have: $PD:PA=DR:AB=1:3$ and $SD:SA=DT:AB=2:3$.

This implies: $$ \frac {\operatorname{Area}(SPB)} {\operatorname{Area}(DAB)} = \frac{SP}{DA} = \frac 32\ . $$ The same also on the other side.

Now $\operatorname{Area}(DAB)$ is half of the area of the parallelogram.

---

Later edit: Thanks to the comment of

ganeshie8

the above is not enough. The "common shaded area" is taken twice. So let us compute it. Let $X$ be the intersection of the diagonals of the parallelogram $SABM$. Let $Y$ be the intersection of the diagonals of the parallelogram $PABL$. Then $X$ is the mid point of the diagonals $SB$ and $MA$, as $P$, $L$ are also the mid points of the sides $SA$ and $MB$. So $XR:XA=PD:PA$ or $XR:XA=RT:AB$ and we get $1:3$ as a common value for $XR:XA=XT:XB$. We may need also the position of $Y$ on $RB$ and/or $AT$, for this we use the theorem of Menelaos for the triangle $RXB$ w.r.t. the line AYT: $$ \frac{AR}{AX}\cdot \frac{TX}{TB}\cdot \frac{YB}{YR} =1\ . $$ This is: $$ \frac23\cdot \frac12\cdot \frac{YB}{YR} =1\ . $$ So $YB:YR=3$.

Let $S$ be the area of the parallelogram $ABCD$. We can finally write: $$ \begin{aligned} \operatorname{Area}(XAB) &= \frac 14 \operatorname{Area}(SABM) = \frac 14\cdot 3S = \frac 34S\ , \\ \operatorname{Area}(ARB) &= \operatorname{Area}(ARB) =\frac 12S\ , \\ \operatorname{Area}(AXT) &= \operatorname{Area}(BXR) \\ &= \operatorname{Area}(AXB) - \operatorname{Area}(ATB) =\left(\frac 34-\frac 12\right)S =\frac 14S\ , \\ \frac {\operatorname{Area}(AYB)} {\operatorname{Area}(ARB)} &= \frac{YB}{RB} =\frac 34\ , \\ \operatorname{Area}(AYB) &= \frac 34 \operatorname{Area}(ARB) = \frac 34\cdot\frac 12S =\frac 38S\ , \\ \operatorname{Area}(XRYT) &= \operatorname{Area}(AYB) + \operatorname{Area}(AXT) + \operatorname{Area}(BXR) - \operatorname{Area}(AXB) \\ &=\left(\frac 14+\frac 14+\frac 38-\frac 34\right)S \\ &=\frac 18S\ . \end{aligned} $$

---

Putting all together, the gray are is: $$ \begin{aligned} \operatorname{Area}(SPB) + \operatorname{Area}(MLA) - \operatorname{Area}(XRYT) &= \left(\frac 34+\frac 34-\frac 18\right)S = \frac {11}8S\ . \end{aligned} $$