Problem:
The internal angles of quadrilateral ABCD form an arithmetic progression. Triangles ABD and DCB are similar with ∠DBA=∠DCB and ∠ADB=∠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 ?
Answer Choices:
A. 210
B. 220
C. 230
D. 240
E. 250
Solution:
Let the degree measures of the angles be as shown in the figure. The angles of a triangle form an arithmetic progression if and only if the median angle is 60∘, so one of x,y, or 180−x−y must be equal to 60 . By symmetry of the role of the triangles ABD and DCB, assume that x≤y. Because x≤y<180−x and x<180−y≤180−x, it follows that the arithmetic progression of the angles in ABCD from smallest to largest must be either x,y,180−y,180−x or x,180−y,y,180−x. Thus either x+180−y=2y, in which case 3y=x+180; or x+y=2(180−y), in which case 3y=360−x. Neither of these is compatible with y=60 (the former forces x=0 and the latter forces x=180 ), so either x=60 or x+y=120.
First suppose that x=60. If 3y=x+180, then y=80, and the sequence of angles in ABCD is (x,y,180−y,180−x)=(60,80,100,120). If 3y=360−x, then y=100, and the sequence of angles in ABCD is (x,180−y,y,180−x)= (60,80,100,120). Finally, suppose that x+y=120. If 3y=x+180, then y=75, and the sequence of angles in ABCD is (x,y,180−y,180−x)=(45,75,105,135). If 3y=360−x, then y=120 and x=0, which is impossible.
Therefore, the sum in degrees of the two largest possible angles is 105+135= 240​.
The problems on this page are the property of the MAA's American Mathematics Competitions