# Parallelogram Contest Problem

In parallelogram $ABCD, AB = 1, BC = 4$, and $\angle ABC = 60^{\circ}$ . Suppose that $AC$ is extended from $A$ to a point $E$ beyond $C$ so that triangle $ADE$ has the same area as the parallelogram. Find the length of $DE$.

I have already gotten the area of the parallelogram using its $2$ diagonals and the cosine law.

$\sqrt{13} \sqrt{21} = \sqrt{273}/2$

I also see that $CDE$ must be half of the are of the parallelogram, but I cant seem to find any angles and side lengths to compute for $CD$. Some hints or maybe I overlooked some properties?

$E$ is just the symmetric of $A$ with respect to $C$:

By the cosine theorem, $$DE^2 = DF^2+FE^2-2\cdot DE\cdot EF\cdot \cos 60^\circ = 16+4-8=12$$ hence $DE=\color{red}{2\sqrt{3}}$.

I don't know too much about computing lengths in geometry. But I think it should be easy after a proper construction of $E$:

Draw $p$ parallel to $AB$ and $CD$ so that the distance between $AB$ and $CD$ is equal to the distance between $CD$ and $p$.

$AC \cap p = \{E\}$

From that construction we know:

The Triangle $DCE$ has the same area as $ABC$

length of $DC$ is $1$

length of $CE$ is equal to length of $AC$

That should be enough to compute $DE$

• Sorry but what does the 3rd line mean? – SuperMage1 Jan 18 '18 at 12:50
• well, you don't need it for your computations. It's just the simplest way (at least for me) to prove my observations by costruction. Now I know that length of CE is equal to length of AC and that DCE has the same area as ABC. – Max Punck-Institut Jan 18 '18 at 13:15

Reflect $E$ and $D$ in $C$, obtaining $A$ and $D'$. If $M$ is the midpoint of $BC$ then $$|MD'|^2=2^2+1^2-{1\over2}\cdot2\cdot 1=3\ .$$ It follows that $$|DE|=|AD'|=2\,|MD'|=2\sqrt{3}\ .$$