AMC10 2025 B

AMC10 2025 B · Q7

AMC10 2025 B · Q7. It mainly tests Quadratic equations, Pythagorean theorem.

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 is the same via either unlocked gate. What is the value of $x$?

Frances 站在一条东西走向的围栏中一个上锁大门正南方 15 米处。围栏紧后面有一个巧克力盒，位于上锁大门东边 $x$ 米处。一个未上锁大门位于巧克力盒东边 9 米处，另一个未上锁大门位于上锁大门西边 8 米处。Frances 可以走向一个未上锁大门，通过它，然后走向巧克力盒。恰好通过任一未上锁大门的总距离相同。求 $x$ 的值。

(A) \ 3 \frac{2}{7} \ 3 \frac{2}{7}

(B) \ 3 \frac{3}{7} \ 3 \frac{3}{7}

(D) \ 3 \frac{5}{7} \ 3 \frac{5}{7}

(E) \ 3 \frac{6}{7} \ 3 \frac{6}{7}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let Frances's location be $F$, the locked gate be $L$, the box be $B$, the unlocked gate on the west be $U$, and the unlocked gate on the east be $V$. We are given that $FL = 15$, $LB = x$, $BV$ = 9, and $LU = 8$. Because $FL$ is vertical and the fence is horizontal, $m \angle FLU = m \angle FLV = 90^\circ$. If Frances takes the western path, she must first travel the length of $FV$, and then $x + 8$ meters. By the Pythagorean theorem, $FV = \sqrt{8^2 + 15^2} = 17$. The remaining distance from Francis and the box of chocolates after she walks that distance is $8 + x$ meters. Thus, the distance Francis travels if she takes the western path is $25 + x$. If Frances takes the eastern path, she first travels $\sqrt{(x + 9)^2 + 15^2} = \sqrt{(x + 9)^2 + 225}$ meters, and then $9$ more meters. Her total distance on the eastern path is $\sqrt{(x + 9)^2 + 225} + 9$ meters. Now, we set the two distances equal to each other: \[x + 25 = \sqrt{(x + 9)^2 + 225} + 9\] \[x + 16 = \sqrt{(x + 9)^2 + 225}\] \[x^2 + 32x + 256 = (x + 9)^2 + 225\] \[x^2 + 32x + 256 = x^2 + 18x + 81 + 225\] \[14x = 50\] \[x = \frac{50}{14} = \frac{25}{7}.\] This corresponds with the answer $\boxed{\textbf{(C)}\hspace{3pt}3\frac{4}{7}}$.

设 Frances 的位置为 $F$，上锁大门为 $L$，巧克力盒为 $B$，西边未上锁大门为 $U$，东边未上锁大门为 $V$。给定 $FL = 15$，$LB = x$，$BV = 9$，$LU = 8$。因为 $FL$ 是垂直的且围栏是水平的，$m \angle FLU = m \angle FLV = 90^\circ$。如果 Frances 走西边路径，她首先走 $FV$ 的长度，然后 $x + 8$ 米。根据勾股定理，$FV = \sqrt{8^2 + 15^2} = 17$。走完这段距离后 Frances 与巧克力盒的剩余距离是 $8 + x$ 米。因此，走西边路径的总距离是 $25 + x$。如果 Frances 走东边路径，她首先走 $\sqrt{(x + 9)^2 + 15^2} = \sqrt{(x + 9)^2 + 225}$ 米，然后再走 9 米。东边路径总距离是 $\sqrt{(x + 9)^2 + 225} + 9$ 米。现在，将两条路径距离设相等： \[x + 25 = \sqrt{(x + 9)^2 + 225} + 9\] \[x + 16 = \sqrt{(x + 9)^2 + 225}\] \[x^2 + 32x + 256 = (x + 9)^2 + 225\] \[x^2 + 32x + 256 = x^2 + 18x + 81 + 225\] \[14x = 50\] \[x = \frac{50}{14} = \frac{25}{7}.\] 这对应答案 $\boxed{\textbf{(C)}\hspace{3pt}3\frac{4}{7}}$。

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