Problem:
The internal angles of quadrilateral ABCD form an arithmetic progression. Triangles ABD and DCB are similar with β DBA=β DCB and β ADB=β 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 60β, so one of x,y, or 180βxβy must be equal to 60 . By symmetry of the role of the triangles ABD and DCB, assume that xβ€y. Because xβ€y<180βx and x<180βyβ€180βx, it follows that the arithmetic progression of the angles in ABCD from smallest to largest must be either x,y,180βy,180βx or x,180βy,y,180βx. Thus either x+180βy=2y, in which case 3y=x+180; or x+y=2(180βy), in which case 3y=360βx. 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,180βy,180βx)=(60,80,100,120). If 3y=360βx, then y=100, and the sequence of angles in ABCD is (x,180βy,y,180βx)= (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,180βy,180βx)=(45,75,105,135). If 3y=360βx, then y=120 and x=0, which is impossible.
Therefore, the sum in degrees of the two largest possible angles is 105+135= 240β.
The problems on this page are the property of the MAA's American Mathematics Competitions