Question

坐标平面上的三点 $(\cos 40^\circ, \sin 40^\circ),\$ $(\cos 60^\circ, \sin 60^\circ),$ 和 $\ \ (\cos t^\circ, \sin t^\circ)$ 构成等腰三角形。求 $0$ 到 $360$ 之间所有可能的 $t$ 之和。

What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are $(\cos 40^\circ, \sin 40^\circ),\$ $(\cos 60^\circ, \sin 60^\circ),$ and $\ \ (\cos t^\circ, \sin t^\circ)$ is isosceles?

$100$

$150$

$330$

$360$

$380$

答案：E

知识点：圆弦等腰三角形

难度评级：1820

解答：

三个点位于单位圆上，对应角分别为 $40^\circ$ ， $60^\circ$ ， $t^\circ$ 。

若第三点到其余两点的距离相等，它必在这两点连线的垂直平分线上，故 $t = 50$ 或 $t = 230$ 。

若它到 $40^\circ$ 所对点的距离等于两个已知点间的弦长（角度差为 $20^\circ$ ），则 $t = 20$ （ $t = 60$ 时三角形退化）。同理，若它到 $60^\circ$ 所对点的距离相等，则 $t = 80$ （ $t = 40$ 时退化）。

有效值为 $50, 230, 20, 80$ ，总和为 $380$ 。

所以正确答案是 E。

The three points lie on the unit circle at angles $40^\circ,$ $60^\circ,$ and $t^\circ.$ A chord's length depends only on the angular separation of its endpoints.

If the third point is equidistant from the other two, it lies on the perpendicular bisector: $t = 50$ or $t = 230.$

If its distance to $40^\circ$ equals the fixed chord (separation $20^\circ$ ), then $t = 20$ (since $t = 60$ is degenerate). If its distance to $60^\circ$ matches, then $t = 80$ (since $t = 40$ is degenerate).

The valid values are $50, 230, 20, 80,$ summing to $380.$

Thus, the correct answer is E.

2021 AMC 12B Fall 第 10 题

答案：E

难度评级：1820

解答：

其他年份的第 10 题