Question

一个高为 $24$ 的等边三角形的一边在直线 $\ell$ 上。一个半径为 $12$ 的圆与 $\ell$ 相切，并且与该三角形外切。位于三角形和圆外部、并由三角形、圆和直线 $\ell$ 围成的区域面积可写成 $a\sqrt{b} - c\pi$ ，其中 $a$ 、 $b$ 和 $c$ 为正整数，且 $b$ 不被任何质数的平方整除。求 $a + b + c$ 。

One side of an equilateral triangle of height $24$ lies on line $\ell.$ A circle of radius $12$ is tangent to $\ell$ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line $\ell$ can be written as $a\sqrt{b} - c\pi,$ where $a,$ $b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is $a + b + c?$

$72$

$73$

$74$

$75$

$76$

答案：D

知识点：切线特殊直角三角形扇形筝形

难度评级：1660

解答：

等边三角形边长为 $16\sqrt3$ 。将 $\ell$ 放在 $x$ 轴上，并取底角顶点 $V = (16\sqrt3, 0)$ ；斜边所在直线为 $\sqrt3\,x + y = 48$ 。圆与 $\ell$ 相切，圆心高度为 $12$ ，且从外侧与该边相切。设圆心为 $O = (20\sqrt3, 12)$ 。令 $T = (20\sqrt3, 0)$ 为圆在 $\ell$ 上的切点， $P$ 为圆在斜边上的切点。所求区域由从 $V$ 引出的两条切线段和靠近的圆弧围成。切线长为 $VT = VP = 4\sqrt3$ ，因此风筝 $VTOP$ 的面积为 $4\sqrt3 \cdot 12 = 48\sqrt3$ 。顶点 $V$ 处夹角为 $120^\circ$ ，所以要减去的圆心角为 $60^\circ$ 的扇形，面积为 $\tfrac16 \pi (12)^2 = 24\pi$ 。区域面积为 $48\sqrt3 - 24\pi$ ，因此 $a + b + c = 48 + 3 + 24 = 75$ ，正确答案是 D。

The equilateral triangle has side $16\sqrt3.$ Put $\ell$ on the $x$ -axis with base vertex $V = (16\sqrt3, 0)$ ; the slanted side lies on $\sqrt3\,x + y = 48.$ The circle sits on $\ell,$ has radius $12,$ and touches that side externally, so its center is $O = (20\sqrt3, 12).$ Let $T = (20\sqrt3, 0)$ be its tangency point on $\ell,$ and let $P$ be the tangency point on the slanted side. The two tangent lengths from $V$ satisfy $VT = VP = 4\sqrt3,$ so kite $VTOP$ has area $4\sqrt3 \cdot 12 = 48\sqrt3.$ The angle at $V$ is $120^\circ,$ so the removed sector has angle $60^\circ$ and area $\tfrac16 \pi (12)^2 = 24\pi.$ The region has area $48\sqrt3 - 24\pi,$ giving $a + b + c = 48 + 3 + 24 = 75.$ Therefore, the answer is D.

2024 AMC 10A 第 14 题

答案：D

难度评级：1660

解答：

其他年份的第 14 题