Triangular cake cut into three slices of given area

Text

The triangular cake in the figure has $AB=6$, $AC=4$, $BC=7$. It has been cut into three slices with cuts parallel to $AB$: $DE$ (near $C$) and $FG$ (near $AB$), with $D,F$ on $AC$ and $E,G$ on $BC$. Each slice shows its area, in $\mathrm{dm}^2$. Determine the lengths of $AF$, $FD$, $CD$, $CE$, $EG$, $GB$.

Solution

The slices have areas $1,5,2$, so the whole triangle $ABC$ has area $1+5+2=8$. The cuts are parallel to $AB$: the triangles with vertex $C$ are similar to $ABC$.

• Triangle $CDE$ (above $DE$): area $1$, so $\dfrac{CDE}{CAB}=\dfrac18$ and the linear ratio is $\dfrac{CD}{CA}=\dfrac{1}{2\sqrt2}$.
• Triangle $CFG$ (above $FG$): area $1+5=6$, so $\dfrac{CFG}{CAB}=\dfrac34$ and $\dfrac{CF}{CA}=\dfrac{\sqrt3}{2}$.

On $AC$ ($CA=4$): $\;CD=\sqrt2\approx1{,}41$; $\;CF=2\sqrt3\approx3{,}46$. Hence $FD=2\sqrt3-\sqrt2\approx2{,}05,\qquad AF=4-2\sqrt3\approx0{,}54.$ On $BC$ ($CB=7$): $\;CE=\dfrac{7\sqrt2}{4}\approx2{,}48$; $\;CG=\dfrac{7\sqrt3}{2}\approx6{,}06$. Hence $EG=\frac{7(2\sqrt3-\sqrt2)}{4}\approx3{,}59,\qquad GB=\frac{7(2-\sqrt3)}{2}\approx0{,}94.$ $\boxed{CD=\sqrt2,\ FD=2\sqrt3-\sqrt2,\ AF=4-2\sqrt3,\ CE=\tfrac{7\sqrt2}{4},\ EG=\tfrac{7(2\sqrt3-\sqrt2)}{4},\ GB=\tfrac{7(2-\sqrt3)}{2}.}$