Rhombus-shaped cake divided into three equal parts

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A rhombus-shaped cake is cut with two symmetric cuts, parallel to the minor diagonal, one above and one below it. At which heights along the major diagonal (as a percentage) should the two cuts be made, if the cake’s area is to be split into three equal parts?

Solution

Place the major diagonal vertically, with centre $O$ at the middle. The upper half of the rhombus is a triangle with apex the top vertex and base the minor diagonal (at $O$); its area is $\dfrac{A}{2}$. The part above the upper cut is a triangle similar to it, with the same apex, and must have area $\dfrac{A}{3}$. The area ratio is $\frac{A/3}{A/2}=\frac23\ \Longrightarrow\ \text{linear ratio }k=\sqrt{\tfrac23}=\frac{\sqrt6}{3}\approx0{,}8165.$ The cut is at distance $k\cdot\dfrac{D}{2}$ from the top vertex (with $D$ the major diagonal), i.e. at height $1-\frac{k}{2}=0{,}5918\quad\text{of the major diagonal.}$ By symmetry the lower cut is at $0{,}4082$. As percentages: $\boxed{\text{cuts at }\approx40{,}8\%\ \text{and}\ \approx59{,}2\%\ \text{of the major diagonal.}}$