CEMC Fermat, 2024

Solution 1:

Suppose that $BE = AC = x$ and
$DE = y$.

Extend $BC$ to point $F$ so that $BC = DE = y$.

Since $BC + DE = 288$, then $BF = BC + CF = BC + DE = 288$.

Also, $\triangle BED$ is congruent to
$\triangle ACF$ by
side-angle-side.

Therefore, $$\angle BAF = \angle BAC + \angle ACF = \angle BDE + \angle DBE = 90\degree$$ since $DE$ and $AC$ are parallel.

Next, $\triangle BED$ is similar to
$\triangle BAF$ since both are
right-angled and they share an angle at $B$.

Therefore, $\dfrac{DE}{BD} = \dfrac{FA}{BF}$ and so $\dfrac{DE}{120} = \dfrac{120}{288}$, which
gives $DE = \dfrac{120 \cdot 120}{288} = 50$, as required.

Solution 2:

Suppose that $BE = AC = x$ and
$DE = y$.

Since $DE + BC = 288$, then $BC = 288 - y$.

We note that $\triangle BED$ is
similar to $\triangle BCA$ because
each is right-angled and their angles at $B$ are common.

Therefore, $\dfrac{BE}{DE} = \dfrac{BC}{AC}$ and so $\dfrac{x}{y} = \dfrac{288 - y}{x}$.

Manipulating, we obtain $x^2 = y(288-y)$ and so $x^2 = 288y - y^2$ or $x^2 + y^2 = 288y$.

Also, using the Pythagorean Theorem in $\triangle BED$ gives $x^2 + y^2 = 120^2$.

Since $x^2 + y^2 = 288y$ and $x^2 + y^2 = 120^2$, then $288y = 120^2$ which gives $2 \cdot 12 \cdot 12 \cdot y = 120 \cdot 120$ and so $2y = 10 \cdot 10$ or $y = 50$.

Therefore, $DE = 50$.