2013 AMC 12B Problem 13

Attempt Problem 13 of the 2013 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12B solutions, or check the answer key.

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13.

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
Answer: D
Concepts:arithmetic sequenceangle chasingcasework
Small hint:
A triangle’s angles form an arithmetic progression exactly when its middle angle is 60∘60^\circ
Big hint:
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
Solution:
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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