Question

一个直圆柱形筒仓直径为 $20$ 米，立在一片田地中。MacDonald 位于筒仓中心以西 $20$ 米、以南 $15$ 米处。McGregor 位于筒仓中心以东 $20$ 米、以南 $g \gt 0$ 米处。MacDonald 和 McGregor 之间的视线与筒仓相切。 $g$ 的值可写为 $\dfrac{a\sqrt{b} - c}{d}$ ，其中 $a, b, c$ 和 $d$ 是正整数， $b$ 不被任何质数的平方整除，且 $d$ 与 $a$ 和 $c$ 的最大公因数互质。 $a + b + c + d$ 等于多少？

A silo (right circular cylinder) with diameter $20$ meters stands in a field. MacDonald is located $20$ meters west and $15$ meters south of the center of the silo. McGregor is located $20$ meters east and $g \gt 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of $g$ can be written as $\dfrac{a\sqrt{b} - c}{d},$ where $a, b, c,$ and $d$ are positive integers, $b$ is not divisible by the square of any prime, and $d$ is relatively prime to the greatest common divisor of $a$ and $c.$ What is $a + b + c + d?$

$119$

$120$

$121$

$122$

$123$

答案：A

知识点：切线坐标几何二次方程

难度评级：2080

解答：

将筒仓中心放在原点，半径为 $10$ 。则 MacDonald 在 $D = (-20, -15)$ ，McGregor 在 $G = (20, -g)$ 。从 $D$ 到切点的切线长为 $DT = \sqrt{DS^2 - 10^2}$ $= \sqrt{25^2 - 100}$ $= \sqrt{525}$ ，从 $G$ 到切点的切线长为 $TG = \sqrt{g^2 + 20^2 - 10^2}$ 。切点 $T$ 在两人之间，所以 $DG = DT + TG$ 。另一方面 $DG = \sqrt{40^2 + (15 - g)^2}$ 。两次平方并化简，得到 $3g^2 + 150g - 925 = 0$ ，因而 $g = \frac{20\sqrt{21} - 75}{3}$ 。所以 $a + b + c + d$ $= 20 + 21 + 75 + 3$ $= 119$ 。因此正确答案是 A。

Put the silo's center at the origin with radius $10.$ Then MacDonald is at $D = (-20, -15)$ and McGregor at $G = (20, -g).$ The tangent length from $D$ is $DT = \sqrt{DS^2 - 10^2}$ $= \sqrt{25^2 - 100}$ $= \sqrt{525},$ and from $G$ it is $TG = \sqrt{g^2 + 20^2 - 10^2}.$ The tangent point $T$ sits between the two men, so $DG = DT + TG.$ But also $DG = \sqrt{40^2 + (15 - g)^2}.$ Squaring twice and simplifying gives $3g^2 + 150g - 925 = 0.$ Its positive root, which satisfies the original tangent-length equation, is $g = \frac{20\sqrt{21} - 75}{3}.$ Therefore $a + b + c + d$ $= 20 + 21 + 75 + 3$ $= 119.$ Thus, A is the correct answer.

2025 AMC 10A 第 20 题

答案：A

难度评级：2080

解答：

其他年份的第 20 题