# Ratio between the areas of two rectangles inscribed in a triangle and the area of the triangle

Let's denote $$A(X)$$ the area of the polygon $$X$$. Let $$S$$,$$R$$ be rectangles inside a triangle $$T$$.

Find the maximum value of: $$\frac{A(R)+A(S)}{A(T)}$$

My try

The only property i know (from a problem that i solved in the past) is that the maximum area of a rectangle inscribed in a triangle is $$1/2$$ the area of the triangle, but actually i don't know how to use it here.

Any hints?

The ratio will be maintained under affine transformations, so we can solve the problem for our favorite triangle. I will use the one with vertices $$(0,0),(1,0),(0,1)$$. Let $$R$$ have height $$h$$. The area of the triangle above $$R$$ is $$\frac 12(1-h)^2$$, so the area of $$S$$ is half that or $$\frac 14(1-h)^2$$. The area of $$R$$ is $$h(1-h)$$, so the area of $$R+S$$ is $$A(R)+A(S)=h-h^2+\frac 14(1-h)^2$$. Taking the derivative and setting to $$0$$ we have $$1-2h-\frac 12(1-h)=0\\ \frac 12-\frac 32h=0\\ h=\frac 13\\ A(R)+A(S)=\frac 13\cdot \frac 23+\frac 14(1-\frac 13)^2=\frac 13\\ \frac {A(R)+A(S)}{A(T)}=\dfrac {\frac 13}{\frac 12}=\frac 23$$
• I am using OP's result that $S$ is half the area of the triangle it is in. In fact the area of the triangle above $R$ is four times the area of the triangle above $S$, but I only know that at the end when I find both rectangles have height $\frac 13$ – Ross Millikan Mar 27 '19 at 4:07