, 2021

Solution 1

We begin by drawing and labelling a diagram, as shown.

[[IMAGE0]]

The diagonals of a rectangle intersect at the centre of the rectangle. That is, $E$ is the midpoint of $AC$. Thus, the $x$-coordinate of $E$ is the average of the $x$-coordinates of $A$ and $C$, or $\frac{0+6}{2}=3$.

The $y$-coordinate of $E$ is the average of the $y$-coordinates of $A$ and $C$, or $\frac{0+12}{2}=6$, and so the coordinates of $E$ are $(3,6)$.

Consider base $AD=6$ of $\triangle ADE$, then its height is equal to the distance from $E$ to the $x$-axis, which is 6.

The area of $\triangle ADE$ is $\frac12(6)(6)=18$.

Solution 2

The diagonals of a rectangle divide the rectangle into 4 non-overlapping triangles having equal area. (You should consider why this is true before reading on.)

Thus, the area of $\triangle ADE$ is equal to $\frac14$ of the area of rectangle $ABCD$ or $\frac 14(6)(12)=18$.
Solution 1

We begin by drawing and labelling a diagram, as shown.

[[IMAGE1]]

The area of rectangle $ABCD$ is equal to the area of trapezoid $BCDP$ plus the area of $\triangle PAD$.

Since the area of trapezoid $BCDP$ is twice the area of $\triangle PAD$, then the area of $\triangle PAD$ is $\frac13$ the area of $ABCD$ (and the area of trapezoid $BCDP$ is $\frac 23$ the area of $ABCD$).

The area of rectangle $ABCD$ is $6\times12=72$, and so the area of $\triangle PAD$ is $\frac13\times72=24$.

The area of $\triangle PAD$ is $\frac12(AD)(AP)=\frac12(6)(p)=3p$, and so $3p=24$ or $p=8$.

Solution 2

Point $P$ has coordinates $(0,p)$ and so $AP=p$ and $BP=12-p$.

The area of $\triangle PAD$ is $\frac12(AD)(AP)=\frac12(6)(p)=3p$.

The area of trapezoid $BCDP$ is $\frac12(BC)(BP+CD)=\frac12(6)(12-p+12)=3(24-p)$.

The area of trapezoid $BCDP$ is twice the area of $\triangle PAD$, and so $3(24-p)=2(3p)$ or $24-p=2p$, and so $3p=24$ or $p=8$.
The area of rectangle $ABCD$ is $6\times12=72$.

The sum of the areas of the two trapezoids is equal to the area of rectangle $ABCD$.

Since the ratio of the areas of these two trapezoids is $5:3$, then the areas of the two trapezoids are $\frac58\times72=45$ and $\frac38\times72=27$.

(We may check that $45:27=5:3$ and $45+27=72$.)

Let $\ell$ be the line that passes through $U$, $V$ and $W$.

Begin by assuming $\ell$ does not pass through a vertex of $ABCD$. In this case, $\ell$ either intersects opposite sides of $ABCD$, or it intersects adjacent sides of $ABCD$.

If $\ell$ intersects opposite sides of $ABCD$, then $\ell$ divides $ABCD$ into two trapezoids, as required.

If $\ell$ intersects adjacent sides of $ABCD$, then $\ell$ divides $ABCD$ into a triangle and a pentagon. This is not possible.

Assume $\ell$ passes through at least one vertex of $ABCD$.

In this case, $\ell$ divides $ABCD$ into two figures, at least one of which is a triangle. This is not also possible.

Thus, $\ell$ intersects opposite sides of $ABCD$ and does not pass through $A$, $B$, $C$, or $D$.

That is, line $\ell$ can intersect opposite sides of $ABCD$ in the two different ways shown below.

[[IMAGE2]]
Hide/Reveal Description of the Graph

Case 1 shows line $\ell$ intersecting the sides AB and CD of rectangle ABCD, so that point U lies between points A and B,
and point W lies between points C and D. Case 2 shows line $\ell$
intersecting sides AD and BC of rectangle ABCD, so that point U lies on AB extended, outside of side AB and point W lies on CD extended, outside of side CD.

In each case, since $\ell$ is a straight line passing through $U$, $V$ and $W$, then the slope of $UV$ is equal to the slope of $VW$.

That is, $$\begin{aligned} \frac{4-u}{2-0}& = \frac{w-4}{6-2}\\ 4(4-u)& = 2(w-4)\\ 2(4-u)& = w-4\\ 8-2u& = w-4\\ w& = 12-2u \end{aligned}$$

Case 1: Line $\ell$ intersects sides $AB$ and $CD$.

That is, $U$ lies between $A$ and $B$, and $W$ lies between $C$ and $D$.

[[IMAGE3]]

In this case, $0<u<12$, $0<w<12$, $AU=u$, and $DW=w$.

The area of trapezoid $ADWU$ is $$\tfrac12(AD)(DW+AU)=\tfrac12(6)(w+u)=3(w+u).$$ Since $w=12-2u$, the area of trapezoid $ADWU$ becomes $3(12-u)$.

We consider each of two possibilities: the area of trapezoid $ADWU$ is equal to 27, or the area is equal to 45.

If the area of trapezoid $ADWU$ is equal to 27, then $$\begin{aligned} 3(12-u)& = 27\\ 12-u& = 9\\ u& = 3\end{aligned}$$ Substituting $u=3$ into $w=12-2u$, we get $w=12-6=6$.

The Case 1 conditions that $0<u<12$ and $0<w<12$ are satisfied and thus the ratio of the areas of the two trapezoids is $5:3$ for the pair of points $U(0,3)$ and $W(6,6)$.

If the area of trapezoid $ADWU$ is equal to 45, then $$\begin{aligned} 3(12-u)& = 45\\ 12-u& = 15\\ u& = -3\end{aligned}$$ Here, the condition that $0<u<12$ is not satisfied and so there is no pair of points $U$ and $W$ for which the ratio of the areas of the two trapezoids is $5:3$.

Case 2: Line $\ell$ intersects sides $AD$ and $BC$.

That is, $U$ lies on $AB$ extended, outside of side $AB$, and $W$ lies on $CD$ extended, outside of side $CD$.

We begin by drawing and labelling a diagram, including $E(e,0)$ and $F(f,12)$, the points where $\ell$ intersects sides $AD$ and $BC$ respectively, as shown.

[[IMAGE4]]

In this case, $u<0$ and $w>12$ (as in the diagram shown), or $u>12$ and $w<0$ (when $U$ lies above $B$ and $W$ lies below $D$). We note that what follows is true for each of these two cases, and thus we need not consider them separately.

In this case, we require that $0<e<6$, $0<f<6$, and so we get $BF=f$ and $AE=e$.

The area of trapezoid $BFEA$ is $$\tfrac12(AB)(BF+AE)=\tfrac12(12)(f+e)=6(f+e).$$

Further, since $\ell$ is a straight line passing through $E$, $V$ and $F$, then the slope of $EV$ is equal to the slope of $FV.$
That is, $$\begin{aligned} \frac{4-0}{2-e}& = \frac{12-4}{f-2}\\ \frac{4}{2-e}& = \frac{8}{f-2}\\ 4(f-2)& = 8(2-e)\\ f-2& = 2(2-e)\\ f& = 6-2e \end{aligned}$$ Since $f=6-2e$, the area of trapezoid $BFEA$ becomes $6(6-e)$.

We consider each of two possibilities: the area of trapezoid $BFEA$ is equal to 27, or the area is equal to 45.

If the area of trapezoid $BFEA$ is equal to 27, then $$\begin{aligned} 6(6-e)& = 27\\ 6-e& = \tfrac92\\ e& = \tfrac32\end{aligned}$$ Substituting $e=\frac32$ into $f=6-2e$, we get $f=3$, and these values satisfy the Case 2

conditions $0<e<6$ and $0<f<6$.

Here, we get $E(\frac32,0)$ and $F(3,12)$ and use these points to determine $U$ and $W$.

The slope of $FV$ is $\dfrac{12-4}{3-2}=8$ and so the slope of $WV$ is also 8, which gives $\dfrac{w-4}{4}=8$,

and solving we get $w=36$.

Similarly, the slope of $VU$ is also 8, which gives $\dfrac{4-u}{2}=8$, and solving we get $u=-12$.

We note that $w=36$ and $u=-12$ satisfy the conditions $w>12$ and $u<0$ and so the ratio of the areas of the two trapezoids is $5:3$ for the points $U(0,-12)$ and $W(6,36)$.

If the area of trapezoid $BFEA$ is equal to 45, then $$\begin{aligned} 6(6-e)& = 45\\ 6-e& = \tfrac{15}{2}\\ e& = -\tfrac32\end{aligned}$$ Here, the condition that $0<e<6$ is not satisfied and so there is no pair of points $E$ and $F$ and thus no pair of points $U$ and $W$ for which the ratio of the areas of the two trapezoids is $5:3$.

Thus, there are two pairs of points $U$ and $W$ for which the ratio of the areas of the two trapezoids is $5:3$. These are $U(0,3)$, $W(6,6)$, and $U(0,-12)$, $W(6,36)$.