Following Through With Casework (Fall 2021 AMC 12A #24)

Not following through with a problem may be the most difficult pitfall to avoid. Try the problem below (it's casework) and try to follow through with all cases.

Fall 2021 AMC 12A #24

Problem: Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}$. In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a$. What is the sum of all possible values of $a$?

Solution: We see that the sides of $ABCD$ must have side lengths $x, x+d, x+2d, x+3d$ to form an arithmetic sequence with difference $d \geq 0$. Since $\overline{AB}$ has the maximum length, we have that $AB = x+3d = 18$, with $3! = 6$ ways to arrange the other sides of length $x, x+d$, and $x+2d$.

From the diagram, we can relate the length of $AB$ to the length of $CD$ because the distance between the two parallel sides is the same at all points. So, our equation becomes $AB = \frac{1}{2}AD + CD \pm \sqrt{CB^2-\frac{3}{4}AD^2}$ depending on the location of $C$, which can be rewritten by isolating $\pm \sqrt{CB^2-\frac{3}{4}AD^2}$ and squaring, giving us

$$\left(AB - \left(\frac{1}{2}AD + CD\right)\right)^2 = CB^2-\frac{3}{4}AD^2$$ Using this, we have six cases below.

Case 1: $AD = x, CD = x+d, BC = x+2d$. We have

$$\left((x+3d)-(\frac{x}{2} + (x+d))\right)^2 = (x+2d)^2-\frac{3}{4}x^2$$ $$(2d-\frac{x}{2})^2 = x^2 + 4dx + 4d^2 - \frac{3x^2}{4}$$ $$4d^2 -2dx + \frac{x^2}{4} = x^2 + 4xd + 4d^2 - \frac{3x^2}{4}$$ This leaves us with $6dx = 0$, which gives us the possibility of $d = 0$, where all sides of $AB = BC = CD = AD = 18$. So, we get $a = 18$ is possible from this case. (I made the mistake of ignoring the result, forgetting that $d=0$ is possible).

Case 2: $AD = x, CD = x+2d, BC = x+d$. We get

$$\left((x+3d)-(\frac{x}{2} + (x+2d))\right)^2 = (x+d)^2-\frac{3}{4}x^2$$ $$(d-\frac{x}{2})^2 = x^2 + 2dx + d^2 - \frac{3x^2}{4}$$ $$d^2 -dx + \frac{x^2}{4} = x^2 + 2dx + d^2 - \frac{3x^2}{4}$$ This simplifies to $3dx = 0$, which gives us $d=0$ just like before, so no new values of $a$ are possible in this case.

Case 3: $AD = x+d, CD = x, BC = x+2d$. We have

$$\left((x+3d)-(\frac{x+d}{2} + x)\right)^2 = (x+2d)^2-\frac{3}{4}(x+d)^2$$ $$\left(\frac{1}{2}(5d-x)\right)^2 = x^2 + 4dx + 4d^2 - \frac{3}{4}(x^2+2dx+d^2)$$ $$\frac{1}{4} (25d^2 - 10dx + x^2) = \frac{1}{4}x^2 + \frac{5}{2}dx + \frac{13}{4}d^2$$ This can be simplified to $3d = 5x$, and since $x+3d=18$, we get $x=3$ and $d=5$ in this case. This gives us the possible side lengths $a = 3, 8, 13$, which adds $24$ to sum of possible values of $a$.

Case 4: $AD = x+d, CD = x+2d, BC = x$. We get

$$\left((x+3d)-(\frac{x+d}{2} + (x+2d))\right)^2 = x^2-\frac{3}{4}(x+d)^2$$ $$\left(\frac{1}{2}(d-x)\right)^2 = x^2 - \frac{3}{4}(x^2+2dx+d^2)$$ $$\frac{1}{4} (d^2 - 2dx + x^2) = \frac{1}{4}x^2 - \frac{3}{2}dx - \frac{3}{4}d^2$$ This leaves us with $d^2 + xd = d(d+x) = 0$, which doesn't add any more possible side lengths $a$ since we have already accounted for $d = 0$ and $d = -x$ is not possible.

Case 5: $AD = x+2d, CD = x, BC = x+d$. We have

$$\left((x+3d)-(\frac{x+2d}{2} + x)\right)^2 = (x+d)^2-\frac{3}{4}(x+2d)^2$$ $$\left(2d-\frac{x}{2}\right)^2 = x^2 +2dx +d^2 - \frac{3}{4}(x^2+4dx+4d^2)$$ $$4d^2 - 2dx + \frac{x^2}{4}  = \frac{x^2}{4} - dx - 2d^2$$ We can simplify this to $6d^2 - dx = d(6d-x) = 0$, so we see that $d=0$ or $x=6d$. Since we have already counted $d=0$, we can use $x=6d$ and $x+3d = 18$ to see that $x=12$ and $d=2$ is possible, giving us $a= 12, 14, 16$ which contributes $42$ to the sum of possible values of $a$. 

Case 6: $AD = x+2d, CD = x+d, BC = x$. We get

$$\left((x+3d)-(\frac{x+2d}{2} + (x+d))\right)^2 = x^2-\frac{3}{4}(x+2d)^2$$ $$\left(d-\frac{x}{2}\right)^2 = x^2 - \frac{3}{4}(x^2+4dx+4d^2)$$ $$d^2 - dx + \frac{x^2}{4}  = \frac{x^2}{4} - 3dx - 3d^2$$ We are left with $4d^2+2dx=d(4d+2x)=0$, so we have $d=0$, which has been covered, and $x=-2d$, which is not possible.

So, in all, we have the sum of possible values of $a = 18 + 3 + 8 + 13 + 12+14+16 = 84$.

So yes, definitely a tedious problem, but it's worth working through and most importantly, following through.