Notes
stacked triangles solution

Solution to the Stacked Triangles Puzzle

Solution by Pythagoras' Theorem and Equilateral Triangles

With the points labelled as above, let $a$ be the length of $B C$, $b$ of $A C$, and $c$ of $A B$. Then these are also the lengths of the blue, purple, and yellow triangles respectively. The area of an equilateral triangle is $\frac{\sqrt{3}}{4}$ times the square of its side length, so considering the areas of the blue and yellow triangles gives:

Line segments $A E$ and $B C$ are parallel so the “height” of $A$ above $B C$ is the same as the “height” of $B$ above $A E$, which is $\frac{\sqrt{3}}{2} a$. The green area is therefore

Considering the right-angled triangle $A D C$, the length of $A D$ is $\frac{a + c}{2}$ and the length of $C D$ is $\frac{\sqrt{3}}{2}c - \frac{\sqrt{3}}{2}a$. Pythagoras' theorem therefore gives the following formula for $b^2$:

The area of the purple triangle is $\frac{\sqrt{3}}{4} b^2$ so the combined purple and green regions have total area:

Solution by Cosine Rule and Area of a Triangle

With $a$, $b$, and $c$ as above, the cosine rule applied to the green triangle gives:

So $b^2 + a c = a^2 + c^2$.

The sine rule for the area of a triangle shows that the areas of the three triangles are:

So the purple and green areas are

Solution by Invariance Principle

If the yellow and blue triangles are the same size, the green triangle becomes equilateral and all four are in fact congruent, whereupon the area of the green and purple is the same as that of the yellow and blue.