Question

连接一个三角形内切圆与三边的三个切点，所得三角形的各角：

If the three points of contact of a circle inscribed in a triangle are joined, the angles of the resulting triangle:

总是等于 $60^\circ$

are always equal to $60^\circ$

总是一个钝角和两个不等的锐角

are always one obtuse angle and two unequal acute angles

总是一个钝角和两个相等的锐角

are always one obtuse angle and two equal acute angles

总是锐角

are always acute angles

总是互不相等

are always unequal to each other

答案：D

知识点：内切圆、内心与内切圆半径圆周角圆

难度评级：1780

小提示：

连接内心与两个相邻切点

Join the incenter to two adjacent points of tangency

大提示：

接触三角形中与顶角 $A$ 相对的角为 $90^\circ-\frac{A}{2}$

The angle of the contact triangle opposite vertex angle $A$ is $90^\circ-\frac{A}{2}$

解答：

通向两个相邻切点的半径分别垂直于对应边。这两条半径所夹的圆心角为 $180^\circ-A$ 。接触三角形中相应的角等于所截圆弧度数的一半，即 $90^\circ-\frac A2\text{。}$ 另外两个角分别为 $90^\circ-\frac{B}{2}$ 和 $90^\circ-\frac{C}{2}$ 。因为三角形的每个角都严格介于 $0^\circ$ 与 $180^\circ$ 之间，所以这三个角都是锐角。

因此，正确答案是 D。

The radii to two adjacent tangency points are perpendicular to the corresponding sides. The central angle between those radii is $180^\circ-A.$ The relevant angle of the contact triangle is half the intercepted arc, giving $90^\circ-\frac A2.$ The other two angles are $90^\circ-\frac{B}{2}$ and $90^\circ-\frac{C}{2}.$ Since every triangle angle lies strictly between $0^\circ$ and $180^\circ,$ all three are acute.

Thus, the correct answer is D.

1954 AMC 12 第 19 题

答案：D

难度评级：1780

小提示：

大提示：

解答：

其他年份的第 19 题