1954 AMC 12 第 19 题

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

连接内心与两个相邻切点

Join the incenter to two adjacent points of tangency

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}

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.

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