TMUA practice: Arithmetic Series, problem 1

Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ , $a+d$ , $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$ .

For the sake of simplicity, lets rename the angles of each similar triangle. Let $\angle ADB = \angle CBD = \alpha$ , $\angle DBA = \angle DCB = \beta$ , $\angle CDB = \angle BAD = \gamma$ .

Now the four angles of $ABCD$ are $\beta$ , $\alpha + \beta$ , $\gamma$ , and $\alpha + \gamma$ .

As for the similar triangles, we can name their angles $y$ , $y+b$ , and $y+2b$ . Therefore $y+y+b+y+2b=180$ and $y+b=60$ . Because these 3 angles are each equal to one of $\alpha, \beta, \gamma$ , we know that one of these three angles is equal to 60 degrees.

Now we we use trial and error. Let $\alpha = 60^{\circ}$ . Then the angles of ABCD are $\beta$ , $60^{\circ} + \beta$ , $\gamma$ , and $60^{\circ} + \gamma$ . Since these four angles add up to 360, then $\beta + \gamma= 120$ . If we list them in increasing value, we get $\beta$ , $\gamma$ , $60^{\circ} + \beta$ , $60^{\circ} + \gamma$ . Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, $\alpha, \beta, \gamma$ also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.

If we apply the same reasoning to $\beta$ and $\gamma$ , we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, $\boxed{\textbf{(D) }240}$ is the correct answer.