Rectangle $T E H F$ has dimensions $15 \mathrm{~m}$ by $30 \mathrm{~m}$, as shown. Tom the Cat begins at $T$, and Jerry the Mouse begins at $J$, the midpoint of $T E$. Jerry runs at $3 \mathrm{~m} / \mathrm{s}$ in a straight line towards $H$. Tom starts at the same time as Jerry, and, running at $5 \mathrm{~m} / \mathrm{s}$ in a straight line, arrives at point $C$ at the same time as Jerry. The time, in seconds, that it takes Tom to catch Jerry is closest to

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-107.jpg?height=277&width=458&top_left_y=257&top_left_x=1294)
(A) 5.4
(B) 5.6
(C) 5.8
(D) 6.0
(E) 6.2
Solution 1

Let $t$ be the amount of time in seconds that it takes Tom to catch Jerry.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-211.jpg?height=274&width=461&top_left_y=1018&top_left_x=889)

Then $T C=5 t$, since Tom runs at $5 \mathrm{~m} / \mathrm{s}$ for $t$ seconds.

Also, $J C=3 t$, since Jerry runs at $3 \mathrm{~m} / \mathrm{s}$.

We know as well that $J$ is the midpoint of $T E$, so $T J=15=J E$.

Since $H E=15$, then $\angle H J E=45^{\circ}$, so $\angle T J C=135^{\circ}$.

By the cosine law,

$$
\begin{aligned}
T C^{2} & =T J^{2}+J C^{2}-2(T J)(J C) \cos (\angle T J C) \\
(5 t)^{2} & =15^{2}+(3 t)^{2}-2(15)(3 t) \cos \left(135^{\circ}\right) \\
25 t^{2} & =225+9 t^{2}-90 t\left(-\frac{1}{\sqrt{2}}\right) \\
16 t^{2}-45 \sqrt{2} t-225 & =0
\end{aligned}
$$

Using the quadratic formula, since $t$ must be positive, we obtain

$$
t=\frac{45 \sqrt{2}+\sqrt{(45 \sqrt{2})^{2}-4(16)(-225)}}{2(16)}=\frac{45 \sqrt{2}+\sqrt{18450}}{32} \approx 6.23 \text { seconds }
$$

so it takes Tom about 6.2 seconds to catch Jerry.

## Solution 2

Let $t$ be the amount of time in seconds that it takes Tom to catch Jerry.

Then $T C=5 t$, since Tom runs at $5 \mathrm{~m} / \mathrm{s}$ for $t$ seconds.

Also, $J C=3 t$, since Jerry runs at $3 \mathrm{~m} / \mathrm{s}$.

We know as well that $J$ is the midpoint of $T E$, so $T J=15=J E$.

Since $H E=15$, then $\angle H J E=45^{\circ}$.

Drop a perpendicular from $C$ to $P$ on $J E$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-212.jpg?height=282&width=461&top_left_y=146&top_left_x=889)

Since $C J=3 t, \angle C J E=45^{\circ}$ and $\angle C P J=90^{\circ}$, then $J P=C P=\frac{1}{\sqrt{2}}(3 t)$. By the Pythagorean Theorem in $\triangle C P T$, we obtain

$$
\begin{aligned}
T C^{2} & =(T J+J P)^{2}+C P^{2} \\
(5 t)^{2} & =\left(15+\frac{1}{\sqrt{2}}(3 t)\right)^{2}+\left(\frac{1}{\sqrt{2}}(3 t)\right)^{2} \\
25 t^{2} & =225+2(15)\left(\frac{1}{\sqrt{2}}(3 t)\right)+\frac{1}{2}\left(9 t^{2}\right)+\frac{1}{2}\left(9 t^{2}\right) \\
25 t^{2} & =225+45 \sqrt{2} t+9 t^{2} \\
16 t^{2}-45 \sqrt{2} t-225 & =0
\end{aligned}
$$

Proceeding exactly as in Solution 1, we obtain that the time is closest to 6.2 seconds.

ANswer: (E)