AMC12 2005 B

AMC12 2005 B · Q18

AMC12 2005 B · Q18. It mainly tests Triangles (properties), Circle theorems.

Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$?

设平面上的点$A(2,2)$和$B(7,7)$。定义$R$为第一象限中那些点$C$组成的区域，使得$\triangle ABC$为锐角三角形。求区域$R$面积的最接近整数。

(A) 25 25

(B) 39 39

(D) 60 60

(E) 80 80

Answer

Correct choice: (C)

正确答案：（C）

Solution

For angle $A$ and $B$ to be acute, $C$ must be between the two lines that are perpendicular to $\overline{AB}$ and contain points $A$ and $B$. For angle $C$ to be acute, first draw a $45-45-90$ triangle with $\overline{AB}$ as the hypotenuse. Note $C$ cannot be inside this triangle's circumscribed circle or else $\angle C > 90^\circ$. Hence, the area of $R$ is the area of the large triangle minus the area of the small triangle minus the area of the circle, which is $\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}$, which is approximately $51$. The answer is $\boxed{\mathrm{C}}$. Note: When doing this problem, it is important to double check that the circle with diameter $AB$ lies fully in the first quadrant, which luckily it does.

要使$\angle A$与$\angle B$为锐角，点$C$必须位于两条与$\overline{AB}$垂直且分别经过$A$与$B$的直线之间。要使$\angle C$为锐角，先作一个以$\overline{AB}$为斜边的$45-45-90$三角形。注意若$C$在该三角形外接圆内，则$\angle C > 90^\circ$。因此，$R$的面积等于大三角形面积减去小三角形面积再减去圆的面积，即 $\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}$，约为$51$。答案为$\boxed{\mathrm{C}}$。注：做此题时需确认以$AB$为直径的圆完全位于第一象限内，所幸确实如此。

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