Triangular Thirds

The area of the whole triangle is $\frac{1}{2} \cdot 2 \cdot 1 = 1$ , so we expect each section to have an area of $\frac{1}{3}$ . The leftmost section is a right-angled triangle similar to the large one, so it has a base of $x_1$ and a height of $\frac{1}{2}x_1$ . If we label the area of this section $A_1$ , we have: $\begin{aligned} A_1 = \frac{1}{2} \cdot x_1 \cdot \frac{1}{2}x_1 &= \frac{1}{3} \\ \left( \frac{1}{2}x_1 \right)^2 &= \frac{1}{3} \\ \frac{1}{2}x_1 &= \frac{1}{ \sqrt{3} } \\ x_1 &= \frac{2}{ \sqrt{3} } \end{aligned}$ The middle section is not a right-angled triangle, but its area can be expressed as $A_2 = \frac{1}{2} \cdot x_2 \cdot \frac{1}{2}x_2 - A_1$ , so we have: $\begin{aligned} A_2 = \frac{1}{2} \cdot x_2 \cdot \frac{1}{2}x_2 - A_1 &= \frac{1}{3} \\ \left( \frac{1}{2}x_2 \right)^2 - \frac{1}{3} &= \frac{1}{3} \\ \left( \frac{1}{2}x_2 \right)^2 &= \frac{2}{3} \\ \frac{1}{2}x_2 &= \frac{ \sqrt{2} }{ \sqrt{3} } \\ x_2 &= \frac{2}{ \sqrt{3} } \cdot \sqrt{2} \end{aligned}$ Thus, we have $\dfrac{x_2}{x_1} = \boxed{ \sqrt{2} }$

For figures of similar shape, in the case a right triangle with legs of $1:2$ . the area of the figure is directly proportional to the square of the linear dimension. That is $A \propto x^2$ $\implies \dfrac {x_2^2}{x_1^2} = \dfrac {A_2}{A_1} = 2$ $\implies \dfrac {x_2}{x_1} = \boxed{\sqrt 2}$ .

The triangle with base $x_2$ is similar to, and has twice the area of the triangle with base $x_1$ . The scale factor of the areas is the square of the scale factor of the sides for similar shapes; hence $\frac{x_2}{x_1}=\boxed{\sqrt2}$ .

In fact, by comparing with the triangle of base $2$ , we can work out explicitly that $x_1=\frac{2\sqrt3}{3}$ and $x_2=\frac{2\sqrt6}{3}$ .