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2013 AMC 12B 第 13 题

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所有题目均经美国数学协会（MAA）官方合法授权使用。

13.

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

The internal angles of quadrilateral ABCDABCD form an arithmetic progression. Triangles ABDABD and DCBDCB are similar with ∠DBA=∠DCB\angle DBA = \angle DCB and ∠ADB=∠CBD.\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?ABCD?

210210

220220

230230

240240

250250

答案：D
知识点：等差数列导角分类讨论
难度评级：1700
小提示：

一个三角形的三个角成等差数列，当且仅当中间角为 60∘60^\circ。

A triangle’s angles form an arithmetic progression exactly when its middle angle is 60∘60^\circ

大提示：

令 ∠DBA=x\angle DBA = x、∠ADB=y\angle ADB = y，则四边形 ABCDABCD 的四个角为 x,y,180−y,180−xx, y, 180-y, 180-x。它们必须成等差数列，并且三角形中有一个角等于 60∘60^\circ。

Let ∠DBA=x\angle DBA = x and ∠ADB=y;\angle ADB = y; the four angles of ABCDABCD become x,y,180−y,180−x,x, y, 180-y, 180-x, which must be an arithmetic progression with one triangle angle equal to 60∘60^\circ

解答：

三角形的三个角成等差数列，当且仅当中间的角为 60∘60^\circ。令 ∠DBA=x\angle DBA = x、∠ADB=y\angle ADB = y，则四边形 ABCDABCD 的四个角为 x,y,180−y,180−xx, y, 180 - y, 180-x，它们本身也必须成等差数列。按递增顺序排列，它们只能是 x,y,180−y,180−xx,y,180-y,180-x 或 x,180−y,y,180−xx,180-y,y,180-x，于是分别得到 3y=x+1803y=x+180 或 3y=360−x3y=360-x。而三角形的三个角 x,y,180−x−yx,y,180-x-y 中有一个等于 60∘60^\circ。代入后只剩下两组角 (60,80,100,120)(60,80,100,120) 和 (45,75,105,135)(45,75,105,135)。最大的两个角之和至多为 105+135=240105 + 135 = 240。所以正确答案是 D。

The angles of a triangle form an arithmetic progression exactly when the middle one is 60∘.60^\circ. With ∠DBA=x\angle DBA = x and ∠ADB=y,\angle ADB = y, the four angles of ABCDABCD are x,y,180−y,180−x,x, y, 180 - y, 180-x, which must itself be an arithmetic progression. In increasing order they are either x,y,180−y,180−xx,y,180-y,180-x or x,180−y,y,180−x,x,180-y,y,180-x, giving 3y=x+1803y=x+180 or 3y=360−x.3y=360-x. One of the triangle angles x,y,180−x−yx,y,180-x-y is 60∘.60^\circ. Substitution leaves the angle sets (60,80,100,120)(60,80,100,120) and (45,75,105,135).(45,75,105,135). The two largest angles sum to at most 105+135=240.105 + 135 = 240. Thus, the correct answer is D.

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