Rectangular cake divided into three parts by cuts parallel to the diagonal

Text

A rectangular cake $ABCD$ has sides $AB=8$ and $BC=4$. You want to cut it with two cuts parallel to the same diagonal, obtaining three equal-area slices: two triangular and one hexagonal (irregular). Determine the points $E$ on $AD$ and $F$ on $CD$ so that $EF$ is a suitable cut, and likewise $G$ on $AB$ and $H$ on $BC$ for the second cut $GH$.

Solution

Place $A(0;0)$, $B(8;0)$, $C(8;4)$, $D(0;4)$; the cuts are parallel to diagonal $AC$ (slope $\tfrac12$). The total area is $8\cdot4=32$, so each slice is $\dfrac{32}{3}$. Triangle $DEF$ at corner $D$. $E$ on $AD$ ($DE$ vertical) and $F$ on $CD$ ($DF$ horizontal). For $EF\parallel AC$ we need $\dfrac{DE}{DF}=\dfrac12$, i.e. $DF=2\,DE$. The right triangle $DEF$ has area $\frac12\,DE\cdot DF=DE^2=\frac{32}{3}\ \Longrightarrow\ DE=\frac{4\sqrt6}{3}\approx3{,}27,$ hence $DF=\dfrac{8\sqrt6}{3}\approx6{,}53$ (both within sides $AD=4$ and $CD=8$). Triangle $BGH$ at corner $B$. By symmetry, with $G$ on $AB$, $H$ on $BC$ and $GH\parallel AC$: $\;BG=2\,BH$ and $BH^2=\dfrac{32}{3}$, so $BH=\frac{4\sqrt6}{3}\approx3{,}27,\qquad BG=\frac{8\sqrt6}{3}\approx6{,}53.$ $\boxed{DE=BH=\tfrac{4\sqrt6}{3}\approx3{,}27,\qquad DF=BG=\tfrac{8\sqrt6}{3}\approx6{,}53.}$