, 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 $(24, 0)$. [[IMAGE0]] 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 = $\frac{15}{2}\$.$