, 2025

When a wheel on a bicycle makes $1$ complete revolution, the bicycle moves
a distance forward equal to the circumference of the wheel.

The front wheel on Terry’s bicycle has a radius of $15 \text{ cm}$, so its circumference is
$\$2\pi \cdot (15\text{}$ cm}) = $30\pi\text{}$
cm}$.

The rear wheel on Terry’s bicycle has a radius of $9 \text{ cm}$, so its circumference is
equal to $\$2\pi \cdot (9\text{}$ cm}) =
$18\pi\text{}$ cm}$.

After $1$ complete revolution of the
front wheel, the total number of revolutions of the rear wheel is not a
whole number, since $30\pi$ is not an
integer multiple of $18\pi$.

After $2$ complete revolutions of the
front wheel, the total number of revolutions of the rear wheel is not a
whole number, since $60\pi$ is not an
integer multiple of $18\pi$.

After $3$ complete revolutions of the
front wheel, the rear wheel will have made $5$ complete revolutions, since $90\pi = 5 \cdot 18\pi$. (Note that $90$ is the least common multiple of $30$ and $18$.)

Therefore, after Terry has travelled $90\pi\text{ cm}$, both wheels have made a
whole number of revolutions for the first time.

Thus, the smallest possible value of $d$ is $d =
$90\pi \approx$ 282.7$ which means that the integer closest to the
smallest possible value of $d$ is
$283$.
Solution 1:

Since $ABCD$ is a rectangle, then
$DC = AB = 24$ and $\angle ADC = 90\degree$.

By the Pythagorean Theorem, $AC^2 = AD^2 +
DC^2 = 18^2 + 24^2 = 324 + 576 = 900$.

Since $AC>0$, then $AC = \sqrt{900} = 30$.

Consider $\triangle ADF$ and $\triangle CEF$.

Since $AD$ is parallel to $BC$, then $\angle DAF = \angle ECF$ and $\angle ADF = \angle CEF$.

This means that $\triangle ADF$ and
$\triangle CEF$ are similar.

Since $AD = 18$ and $CE = 6$, then $\$\dfrac{CF}{AF} = \dfrac{CE}{AD} = \dfrac{6}{18}\$.$

This means that $CF:AF = 1:3$.

Since $AC = CF + AF = 30$, then $CF = $\frac{1}{4}AC = \frac{30}{4} = \frac{15}{2}\$.$

Solution 2:

We put the diagram on a coordinate grid with $D$ at the origin $(0,0)$, $A$ on the positive $y$-axis, and $C$ on the positive $x$-axis.

Since $AD = 18$, then $A$ has coordinates $(0, 18)$.

Since $DC = AB = 24$, then $C$ has coordinates $(24, 0)$.

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Since $BC$ is vertical and $EC = 6$, then $E$ has coordinates $(24, 6)$.

Line segment $DE$ passes through the
origin and through $(24, 6)$ and so
has slope $\frac{6}{24}$ which means
that its equation is $y =
$\frac{1}{4}x\$.$

Line segment $AC$ passes through the
points $(0, 18)$ and $(24, 0)$.

Its slope is thus $\text{18}$ - 0}{0 - 24} =
$-\frac{3}{4}$, which means that its equation is y = $-\frac{3}{4}x$ + 18$.

Point $F$ is the point of
intersection of line segments $AC$
and $DE$.

Equating expressions for $y$, we
obtain $\$\frac{1}{4}x = -\frac{3}{4}x$ +
18$, which gives$x =
18$.

Substituting into $y = \frac{1}{4}x$,
we obtain $y = $\frac{18}{4} = \frac{9}{2}\$.$

Thus, $F$ has coordinates $\left(18, \frac{9}{2}\right)$.

Finally, $$CF = \sqrt{(24 - 18)^2 + \left(0 - \tfrac{9}{2}\right)^2} = \sqrt{36 + \tfrac{81}{4}} = \sqrt{\tfrac{225}{4}}$$ and so $CF =
$\frac{15}{2}\$.$