Notes
square in a triangle in a square solution

Solution to the Square in a Triangle in a Square Puzzle

Solution by Dissection and Area of a Triangle

Consider the diagram as labelled above. The line from $E$ to $A$ means that the blue region consists of four triangles, each of which can be thought of as having a base along one of the edges of the inner square.

The vertical height of $I$ above $G F$ is the same as the length of $B L$. Similarly, the vertical height of $A$ to $H E$ is the same as the length of $A K$. The length of $L K$ is that of the side of the inner square, so the sum of the lengths of $B L$ and $A K$ is $2$. This means that the sum of the areas of $I G F$ and $A H E$ is $\frac{1}{2} \times 1 \times 2 = 1$.

A similar argument says that the areas of the other two blue triangles also sum to $1$. So the total area of the blue regions is $2$.

The yellow region therefore has area $3 \times 3 - 2 - 1 = 6$. So the ratio of the blue area to the yellow is $2 : 6 = 1 : 3$.

Solution by Invariance Principle

There is a range of positions for the small square within the larger square, and at one extreme is the configuration below where it is in the middle against the left-hand edge.

In this configuration, the top right corner of the small square lies on the diagonal of the outer square.

The large yellow triangle has area $\frac{1}{2} \times 3 \times 3 = \frac{9}{2}$. The smaller yellow triangle has area $\frac{1}{2} \times 3 \times 1= \frac{3}{2}$. So the yellow area is $\frac{9}{2} + \frac{3}{2} = 6$.

The total area of the square is $3 \times 3 = 9$, so the total blue region has area $9 - 6 - 1 = 2$.

Therefore the ratio of the blue area to the yellow is $2 : 6 = 1 : 3$.