2025 AMC 10B Problem 7

Question

Frances stands

15

meters directly south of a locked gate in a fence that runs east-west. Immediately behind the fence is a box of chocolates, located

x

meters east of the locked gate. An unlocked gate lies

9

meters east of the box, and another unlocked gate lies

8

meters west of the locked gate. Frances can reach the box by walking toward an unlocked gate, passing through it, and walking toward the box. It happens that the total distance Frances would travel would be the same via either unlocked gate. What is the value of

x?

3\tfrac{2}{7}

3\tfrac{3}{7}

3\tfrac{4}{7}

3\tfrac{5}{7}

3\tfrac{6}{7}

Answer: C

Concepts:coordinate geometry Pythagorean Theorem radical

Difficulty rating: 1500

Small hint:

Put the locked gate at the origin and Frances at

(0, -15);

each route is a straight walk to a gate plus a walk along the fence to the box

Big hint:

The west route totals

17 + (x + 8);

set it equal to

\sqrt{(x + 9)^2 + 225} + 9

and square

Solution:

Put the fence on the

x

-axis, the locked gate at the origin, and Frances at

(0, -15).

Then the box is at

(x, 0),

the east gate at

(x + 9, 0),

and the west gate at

(-8, 0).

The east route is

\sqrt{(x + 9)^2 + 15^2} + 9;

the west route is

\sqrt{8^2 + 15^2}

+ (x + 8) = 17 + x + 8.

Set them equal:

\sqrt{(x + 9)^2 + 225} = x + 16.

Square and simplify to get

18x + 306 = 32x + 256,

so

x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}.

Thus, C is the correct answer.