, 2023

The line $x=a$ intersects the
line $y=x$ at the point $(a,a)$.

Thus, the length of the base and the height of the triangle are each
equal to $a$, and so the area of the
triangle is $\$\dfrac12\times a\times$
a$.

Solving $\dfrac12 a^2=32$, we get
$a^2=64$, and so $a=8$ (since $a>0$).
Solution 1

The line $x=10$ intersects the
line $y=2x$ at the point $(10,20)$.

The line $x=4$ intersects the line
$y=2x$ at the point $(4,8)$.

Thus, the trapezoid has parallel sides of length 20 and 8, and the
distance between the parallel sides is $10-4=6$.

The area of the trapezoid is $\dfrac{6}{2}(20+8)=3(28)$ which is equal
to 84.

Solution 2

If the area of the trapezoid is $T$, the area of the new unshaded triangle
is $B$, and the area of the original
triangle is $A$, then $T=A-B$.

The line $x=4$ intersects the line
$y=2x$ at the point $(4,8)$.

Thus, the unshaded triangle has base length 4 and height 8, and so $B=\dfrac12\times 4\times 8=16$.

The line $x=10$ intersects the line
$y=2x$ at the point $(10,20)$.

Thus, the original triangle has base length 10 and height 20, and so
$\$A=\dfrac12\times 10\times$
20=100$.

The area of the trapezoid is $T=A-B$ or $T=100-16$ which is 84.
Solution 1

We begin by determining the area of the trapezoid.

The line $x=21$ intersects the line
$y=3x$ at the point $(21,63)$.

The line $x=c$ intersects the line
$y=3x$ at the point $(c,3c)$.

Thus, the trapezoid has parallel sides of length 63 and $3c$, and the distance between the parallel
sides is $21-c$ (since $0<c<21$).

The area of the trapezoid is $\dfrac{21-c}{2}(63+3c)$.

Next, we determine the area of the new triangle.

If the length of its base is $c$,
then its height is $3c$, and so the
area of the new triangle is $\$\dfrac12\times c\times 3c=\dfrac12\times$ 3c^2$.

The area of the trapezoid is 8 times the area of the new
triangle.

Solving, we get $$\begin{align*} \dfrac{21-c}{2}(63+3c)&=8\times\dfrac12\times 3c^2 \\ (21-c)(63+3c)&=8\times3c^2 \\ (21-c)(21+c)&=8\times c^2 \\ 441-c^2&=8c^2 \\ 441&=9c^2 \\ c^2&=49\end{align*}$$ and so $c=7$ (since $c>0$).

Solution 2

If the area of the trapezoid is $T$, the area of the new triangle is $B$, and the area of the original triangle
is $A$, then $T=A-B$.

The area of the trapezoid is 8 times the area of the new triangle, or
$T=8B$.

Substituting, we get $8B=A-B$ or
$9B=A$.

The line $x=21$ intersects the
line $y=3x$ at the point $(21,63)$.

Thus, $\$A=\dfrac12\times 21\times 63=\dfrac{1323}{2}\$.$

The line $x=c$ intersects the line
$y=3x$ at the point $(c,3c)$.

Thus, $\$B=\dfrac12\times c\times 3c=\dfrac{3c^2}{2}\$.$

Substituting into $9B=A$ and solving,
we get $$\begin{align*} 9\times\dfrac{3c^2}{2}&=\dfrac{1323}{2} \\ 27c^2&=1323 \\ c^2&=49\end{align*}$$ and so $c=7$ (since $c>0$).
Solution 1

As was shown in parts (b) and (c), the vertical line drawn at $x=p$ divides the original triangle into a
trapezoid and a new triangle.

We begin by determining the area of the trapezoid.

The line $x=1$ intersects the line
$y=4x$ at the point $(1,4)$.

The line $x=p$ intersects the line
$y=4x$ at the point $(p,4p)$.

Thus, the trapezoid has parallel sides of length 4 and $4p$, and the distance between the parallel
sides is $1-p$ (since $0<p<1$).

The area of the trapezoid is $\dfrac{1-p}{2}(4+4p)$.

Next, we determine the area of the new triangle.

If the length of its base is $p$,
then its height is $4p$, and so the
area of the new triangle is $\$\dfrac12\times p\times 4p=\dfrac12\times$ 4p^2$.

The line $x=p$ divides the area of
the original triangle in half, and so the area of the trapezoid is equal
to the area of the new triangle.

Solving, we get $$\begin{align*} \dfrac{1-p}{2}(4+4p)&=\dfrac12\times 4p^2 \\ (1-p)(4+4p)&=4p^2 \\ (1-p)(1+p)&=p^2 \\ 1-p^2&=p^2 \\ 1&=2p^2 \\ p^2&=\dfrac12\end{align*}$$ and so $p=\dfrac{1}{\sqrt{2}}$ (since $p>0$).

Ahmed repeats the process by drawing a second vertical line at $x=q$, where $0<q<p$.

We wish to determine the value of $q$
in terms of $p$, so that we may use
this relationship to determine the position of the 12th vertical line
(without needing to repeat these calculations 12 times).

That is, we will repeat the above process without substituting $p=\dfrac{1}{\sqrt{2}}$ so that we may
determine the value of $q$ in terms
of $p$.

The vertical line drawn at $x=q$
divides the triangle bounded by the $x$-axis, the line $y=4x$, and the line $x=p$ into a new trapezoid and a new
triangle.

We begin by determining the area of the trapezoid.

The line $x=p$ intersects the line
$y=4x$ at the point $(p,4p)$.

The line $x=q$ intersects the line
$y=4x$ at the point $(q,4q)$.

Thus, the trapezoid has parallel sides of length $4p$ and $4q$, and the distance between the parallel
sides is $p-q$ (since $0<q<p$).

The area of the trapezoid is $\dfrac{p-q}{2}(4p+4q)$.

Next, we determine the area of the triangle.

If the length of its base is $q$,
then its height is $4q$, and so the
area of the triangle is $\$\dfrac12\times q\times 4q=\dfrac12\times$ 4q^2$.

The line $x=q$ divides the area of
the previous triangle in half, and so the area of the trapezoid is equal
to the area of the new triangle.

Solving, we get $$\begin{align*} \dfrac{p-q}{2}(4p+4q)&=\dfrac12\times 4q^2 \\ (p-q)(4p+4q)&=4q^2 \\ (p-q)(p+q)&=q^2 \\ p^2-q^2&=q^2 \\ p^2&=2q^2 \\ q^2&=\dfrac12\times p^2\end{align*}$$ and so $q=\dfrac{1}{\sqrt{2}}\times p$ (since
$q>0$).

This tells us that if Ahmed draws a vertical line at $x=n$ (where $n>0$ and $n$ is less than the $x$-intercept of the vertical line
previously drawn), then the next vertical line is drawn at $x=\dfrac{1}{\sqrt{2}}\times n$ (since the
process repeats).

Since the original vertical line is at $x=1$, then the 12th vertical line drawn by
Ahmed is at $x=1\times\left(\dfrac{1}{\sqrt{2}}\right)^{12}$
or $x=\left(\left(\dfrac{1}{\sqrt{2}}\right)^{2}\right)^{\!\!6}$
or $x=\left(\dfrac{1}{2}\right)^{6}$,
and so $k=\dfrac{1}{64}$.

Solution 2

As was shown in parts (b) and (c), the vertical line drawn at $x=p$ divides the original triangle into a
trapezoid and a new triangle.

The line $x=1$ intersects the line
$y=4x$ at the point $(1,4)$, and so the area of the original
triangle is $\$\dfrac12\times 1\times$
4=2$.

The line $x=p$ intersects the line
$y=4x$ at the point $(p,4p)$, and so the area of the new
triangle is $\$\dfrac12\times p\times$
4p=2p^2$.

The area of the new triangle is half of the area of the original
triangle, and so $2p^2=1$ or $p^2=\dfrac{1}{2}$, and so $p=\dfrac{1}{\sqrt{2}}$ (since $p>0$).

Ahmed repeats the process by drawing a second vertical line at $x=q$, where $0<q<p$.

We wish to determine the value of $q$
in terms of $p$, so that we may use
this relationship to determine the position of the 12th vertical line
(without needing to repeat these calculations 12 times).

That is, we will repeat the above process without substituting $p=\dfrac{1}{\sqrt{2}}$ so that we may
determine the value of $q$ in terms
of $p$.

The vertical line drawn at $x=q$
divides the triangle bounded by the $x$-axis, the line $y=4x$, and the line $x=p$ into a new trapezoid and a new
triangle.

As was determined above, the triangle bounded by the $x$-axis, the line $y=4x$, and the line $x=p$ has area $2p^2$.

The line $x=q$ intersects the line
$y=4x$ at the point $(q,4q)$, and so the area of the new
triangle is $\$\dfrac12\times q\times$
4q=2q^2$.

The area of the new triangle is half of the area of the previous
triangle, and so $2q^2=\dfrac{2p^2}{2}$ or $q^2=\dfrac12\times p^2$, and so $q=\dfrac{1}{\sqrt{2}}\times p$ (since
$q>0$).

This tells us that if Ahmed draws a vertical line at $x=n$ (where $n>0$ and $n$ is less than the $x$-intercept of the vertical line
previously drawn), then the next vertical line is drawn at $x=\dfrac{1}{\sqrt{2}}\times n$ (since the
process repeats).

Since the original vertical line is at $x=1$, then the 12th vertical line drawn by
Ahmed is at $x=1\times\left(\dfrac{1}{\sqrt{2}}\right)^{\!\!12}$
or $x=\left(\left(\dfrac{1}{\sqrt{2}}\right)^{\!\!2}\right)^{\!\!6}$
or $x=\left(\dfrac{1}{2}\right)^{\!\!6}$, and
so $k=\dfrac{1}{64}$.

Solution 3

Ahmed draws the 12th vertical line at $x=k$.

The line $x=k$ intersects the line
$y=4x$ at the point $(k,4k)$, and so the area of the new

triangle to the left of this line is $\dfrac12\times k\times 4k=2k^2$.

Since the area of each new triangle is half of the area of the
previous triangle, then the

triangle with area $2k^2$ has
$\left(\dfrac12\right)^{\!\!12}$ of
the area of the original triangle.

The line $x=1$ intersects the line
$y=4x$ at the point $(1,4)$, and so the area of the original
triangle is $\$\dfrac12\times 1\times$
4=2$.

Equating the areas and solving for $k$, we get $$\begin{align*} 2k^2&=\left(\dfrac12\right)^{12}\times 2 \\ k^2&=\left(\dfrac12\right)^{12} \\ k&=\left(\dfrac12\right)^{6}\end{align*}$$ and so $k=\dfrac{1}{64}$.