Question

Frances 站在一扇上锁的门正南方 $15$ 米处，这扇门位于一条东西走向的栅栏上。栅栏正后方有一盒巧克力，位于上锁的门以东 $x$ 米处。另有一扇未上锁的门位于巧克力盒以东 $9$ 米处，还有一扇未上锁的门位于上锁的门以西 $8$ 米处。Frances 可以走向一扇未上锁的门，穿过它，再走向巧克力盒。碰巧无论经由哪一扇未上锁的门，她走的总距离都相同。求 $x$ ？

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 $x?$

$3\tfrac{2}{7}$

$3\tfrac{3}{7}$

$3\tfrac{4}{7}$

$3\tfrac{5}{7}$

$3\tfrac{6}{7}$

答案：C

知识点：坐标几何勾股定理根式

难度评级：1500

解答：

把栅栏放在 $x$ -轴上，上锁的门在原点，Frances 在 $(0, -15)$ 。则盒子在 $(x, 0)$ ，东门在 $(x + 9, 0)$ ，西门在 $(-8, 0)$ 。东侧路线为 $\sqrt{(x + 9)^2 + 15^2} + 9$ ；西侧路线为 $\sqrt{8^2 + 15^2}$ $+ (x + 8) = 17 + x + 8$ 。令二者相等，得 $\sqrt{(x + 9)^2 + 225} = x + 16$ 。平方并化简： $18x + 306 = 32x + 256$ ，所以 $x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}$ 。因此正确答案是 C。

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.

2025 AMC 10B 第 7 题

答案：C

难度评级：1500

解答：

其他年份的第 7 题