Following on from the earlier Cake problem...
Solution
Focus on the triangle with base length \(y\). It has a perpendicular height of \(\frac{1}{2}(x+y)\) therefore an area of \(\frac{y(x+y)}{4}\).
The entire cake has an area of \(\frac{3}{4}(x+y)^2\) so this triangle is a quarter of that.
\(\frac{y(x+y)}{4}=\frac{1}{4} \times \frac{3}{4}(x+y)^2\)
\(y(x+y)=\frac{3}{4}(x+y)^2\)
\(y=3x\)
So the ratio is 1:3.
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