AIME 1989 · 第 12 题

Solution

Call the midpoint of $\overline{AB}$ $M$ and the midpoint of $\overline{CD}$ $N$. $d$ is the median of triangle $\triangle CDM$. The formula for the length of a median is $m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}$, where $a$, $b$, and $c$ are the side lengths of triangle, and $c$ is the side that is bisected by median $m$. The formula is a direct result of the Law of Cosines applied twice with the angles formed by the median (Stewart's Theorem). We can also get this formula from the parallelogram law, that the sum of the squares of the diagonals is equal to the squares of the sides of a parallelogram (https://en.wikipedia.org/wiki/Parallelogram_law).

We first find $CM$, which is the median of $\triangle CAB$.

$$CM=\sqrt{\frac{98+2592-1681}{4}}=\frac{\sqrt{1009}}{2}$$ Now we must find $DM$, which is the median of $\triangle DAB$.

$$DM=\frac{\sqrt{425}}{2}$$ Now that we know the sides of $\triangle CDM$, we proceed to find the length of $d$.

$$d=\frac{\sqrt{548}}{2} \Longrightarrow d^2=\frac{548}{4}=\boxed{137}$$

Solution 2

Let M be the midpoint of AB. DM is a median of triangle ABD and CM is a median of triangle ABC We find that CM^2 = 1009/4 and DM^2 = 425/4 We can then find the median of triangle CDM squared from the 3 side lengths of the triangle which gives us 137. Therefore $d^2 = \boxed{137}.$

~ Ethan50083