AMC10 2024 A

AMC10 2024 A · Q14

AMC10 2024 A · Q14. It mainly tests Triangles (properties), Circle theorems.

One side of an equilateral triangle of height $24$ lies on line $\ell$. A circle of radius $12$ is tangent to line $\ l$ 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 $a + b + c$?

一个高度为 $24$ 的等边三角形的一条边位于直线 $\ell$ 上。一个半径为 $12$ 的圆与直线 $\ell$ 相切，并与三角形外切。位于三角形和圆外部、由三角形、圆和直线 $\ell$ 包围的区域面积可以写成 $a \sqrt{b} - c \pi$，其中 $a$、$b$ 和 $c$ 是正整数，且 $b$ 不能被任何质数的平方整除。$a + b + c$ 等于多少？

(A) 72 72

(B) 73 73

(D) 75 75

(E) 76 76

Answer

Correct choice: (D)

正确答案：（D）

Solution

Call the bottom vertices of the triangle $B$ and $C$ (the one closer to the circle is $C$) and the top vertex $A$. The tangency point between the circle and the side of the triangle is $D$, and the tangency point on line $\ell$ $E$, and the center of the circle is $O$. Draw radii to the tangency points, the arc is $60$ degrees because $\angle ACB$ is $60$, and since $\angle DCE$ is supplementary, it's $120^{\circ}$.(Using Angle of Intersecting Secants Theorem) The sum of the angles in a quadrilateral is $360$, which means $\angle DOE$ is $60^{\circ}$ Triangle $ODC$ is $30$-$60$-$90$ triangle so CD is $4\sqrt{3}$. Since we have $2$ congruent triangles ($\triangle ODC$ and $\triangle OEC$), the combined area of both is $48\sqrt{3}$. The area of the arc is $144 \cdot \frac{60}{360} \cdot \pi$ which is $24\pi$, so the area of the region is $48\sqrt{3}-24\pi$ $a+b+c$ is $48+3+24$ which is $\boxed{\textbf{(D)}~75}$

称三角形底边顶点为 $B$ 和 $C$（靠近圆的为 $C$），顶点为 $A$。圆与三角形边相切的点为 $D$，与直线 $\ell$ 相切的点为 $E$，圆心为 $O$。画出到相切点的半径，弧为 $60$ 度，因为 $\angle ACB$ 为 $60$ 度，且 $\angle DCE$ 互补，为 $120^{\circ}$（使用相交割线定理）。四边形内角和为 $360$ 度，故 $\angle DOE$ 为 $60^{\circ}$。三角形 $ODC$ 是 $30$-$60$-$90$ 三角形，故 $CD$ 为 $4\sqrt{3}$。有两个全等三角形（$\triangle ODC$ 和 $\triangle OEC$），两者面积和为 $48\sqrt{3}$。扇形面积为 $144 \cdot \frac{60}{360} \cdot \pi = 24\pi$，故区域面积为 $48\sqrt{3}-24\pi$。 $a+b+c=48+3+24=\boxed{\textbf{(D)}~75}$。

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