2025 AMC 10B Problem 7

Set the fence along the $x$-axis with the locked gate at the origin, east positive. Then

• Frances is at $(0, -15)$,
• the box is at $(x, 0)$,
• the eastern unlocked gate is at $(x + 9, 0)$,
• the western unlocked gate is at $(-8, 0)$.

Each route is a straight walk to a gate followed by a straight walk along the fence to the box.

Western route. From $(0,-15)$ to $(-8,0)$ is $\sqrt{8^2 + 15^2} = \sqrt{289} = 17$ (the $8$-$15$-$17$ triple), and from there to the box is $x + 8$. Total: $17 + (x + 8) = 25 + x$.

Eastern route. From $(0,-15)$ to $(x+9, 0)$ is $\sqrt{15^2 + (x+9)^2}$, and from there back to the box is $9$. Total: $\sqrt{225 + (x+9)^2} + 9$.

Equating and isolating the radical,
$$\sqrt{225 + (x+9)^2} = 16 + x.$$
Both sides are positive, so square:
$$225 + x^2 + 18x + 81 = x^2 + 32x + 256,$$
$$306 + 18x = 256 + 32x \quad\Longrightarrow\quad 50 = 14x \quad\Longrightarrow\quad x = \frac{25}{7} = 3\tfrac47.$$

The answer is $\boxed{\textbf{(C)}\ 3\frac47}$.

Coordinates convert a wordy layout into two distances, and the only geometry needed is the Pythagorean theorem applied twice. The $x^2$ terms cancel when the equation is squared, so what looks like a quadratic is really linear. Spotting $8$-$15$-$17$ on the western side saves a surd and keeps the algebra on one side of the equation. The general lesson is to draw the axis along the fence and measure everything from one fixed landmark, here the locked gate, so that every position is a single expression in $x$.

• Comparing only the diagonal walks to the two gates and omitting the walks along the fence, 9 metres in the east and x + 8 metres in the west.
• Reading the description as "the box is $9$ metres east of the unlocked gate", which puts that gate at $(x-9,0)$ and yields $x = 1$.
• Squaring $\sqrt{225 + (x+9)^2} + 9 = 25 + x$ without first isolating the radical, which produces stray cross terms.