Q:

Rectangle in a Triangle

There is a rectangle inside a triangle, with one of the rectangle's sides up against one of the triangle's sides, and the other two corners touching the triangle's other two sides.

Prove that the area of the rectangle is at most $1/2$ the area of the triangle.

In other words, prove that the blue region is no bigger than the red region.

I heard this puzzle from Solomon Golomb, a brilliant mathematician and lover of puzzles who  recently passed away. 

A:

Graphically...

...where H is the height of the triangle

A:

Any such arrangement may be rotated to have the adjoining edge at the base (as pictured in the question).

Say that the rectangle is $n$ high and $m$ wide and the large triangle is $p$ high and $q$ wide.
The large triangle will have area $A=\frac{pq}2$.

 The upper of the smaller triangles is a similar triangle to the large triangle and as such:
 $m=q\frac{p-n}p$,
 so the area of the rectangle is:
 $nm=qn\frac{p-n}p=q\frac{(p/2)^2-(n-p/2)^2}{p}$
 which may be maximised by letting $(n-p/2)^2=0\rightarrow n=p/2$,
 whereupon
 $nm=q\frac{(p/2)^2}{p}=\frac{pq}{4}=\frac12A$

Alternatively

 one could take the derivative of $n(p-n)$ with respect to $n$ and set it to zero for the same result.

A:

I'm not sure if this needs spoiler marking, but I'll play it safe:

 Look at it as three separate triangles. Flip each of the three red triangles on its edge with the blue rectangle. If the rectangle's height is exactly half the total height, the triangles perfectly cover it. If the rectangle's height is less than or greater than half the total height, the triangles more than cover it.

