Harvard-MIT Mathematics Tournament

Solution:

Answer:

| $\frac{216}{\sqrt{145}}$ |
| :---: |

The locus of points $E$ such that $[AED] = [AEB]$ forms a line, since area is a linear function of the coordinates of $E$; setting the areas equal gives a linear equation in the coordinates $E$. Note that $A$ and $M$, the midpoint of $\overline{DB}$, are on this line; $A$ because both areas are $0$, and $M$ because the triangles share an altitude, and bases $\overline{DM}$ and $\overline{MB}$ are equal in length. Then $\overline{AM}$ is the set of points satisfying the area condition. The point $E$, then, is such that $\triangle AEC$ is a right angle (to make the distance minimal) and $E$ lies on $\overline{AM}$.

Let $X$ be the point of intersection of $\overline{AM}$ and $\overline{CD}$. Then $\triangle AMB \sim \triangle XMD$, and since $MD = BM$, they are in fact congruent. Thus $DX = AB = 10$, and $XC = 18$. Similarly, $BX = 15$, so $ABXD$ is a parallelogram. Let $Y$ be the foot of the perpendicular from $A$ to $\overline{DC}$, so that $DY = \frac{DC - AB}{2} = 9$. Then

$AY = \sqrt{AD^2 - DY^2} = \sqrt{225 - 81} = 12$.

Then $YX = DX - DY = 1$ and $AX = \sqrt{AY^2 + YX^2} = \sqrt{144 + 1} = \sqrt{145}$.

Since both $\triangle AXY$ and $\triangle CXE$ have a right angle, and $\angle EXC$ and $\angle YXA$ are congruent because they are vertical angles, $\triangle AXY \sim \triangle CXE$. Then $\frac{CE}{AY} = \frac{CX}{AX}$, so $CE = 12 \cdot \frac{18}{\sqrt{145}} = \frac{216}{\sqrt{145}}$.