Question

四边形 $ABCD$ 的内角成等差数列。三角形 $ABD$ 与 $DCB$ 相似，且 $\angle DBA = \angle DCB$ 、 $\angle ADB = \angle CBD$ 。此外，这两个三角形各自的角也都成等差数列。 $ABCD$ 中最大的两个角之和最大可能是多少度？

The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD.$ Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD?$

$210$

$220$

$230$

$240$

$250$

答案：D

知识点：等差数列导角分类讨论

难度评级：1700

解答：

三角形的三个角成等差数列，当且仅当中间的角为 $60^\circ$ 。令 $\angle DBA = x$ 、 $\angle ADB = y$ ，则四边形 $ABCD$ 的四个角为 $x, y, 180 - y, 180-x$ ，它们也必须成等差数列。再结合两个三角形中有一个角为 $60^\circ$ ，可知 $x,y,180-y,180-x$ 或 $x,180-y,y,180-x$ 。逐一讨论可得，两组可能的角为 $3y=x+180$ 和 $3y=360-x$ 。因此最大的两个角之和至多为 $105 + 135 = 240$ 。所以正确答案是 D。 $x,y,180-x-y$ $(60,80,100,120)$ $(45,75,105,135)$ 。

The angles of a triangle form an arithmetic progression exactly when the middle one is $60^\circ.$ With $\angle DBA = x$ and $\angle ADB = y,$ the four angles of $ABCD$ are $x, y, 180 - y, 180-x,$ which must itself be an arithmetic progression. In increasing order they are either $x,y,180-y,180-x$ or $x,180-y,y,180-x,$ giving $3y=x+180$ or $3y=360-x.$ One of the triangle angles $x,y,180-x-y$ is $60^\circ.$ Substitution leaves the angle sets $(60,80,100,120)$ and $(45,75,105,135).$ The two largest angles sum to at most $105 + 135 = 240.$ Thus, the correct answer is D.

2013 AMC 12B 第 13 题

答案：D

难度评级：1700

解答：

其他年份的第 13 题