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\begin{document}

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\section*{two equilateral triangles solution}

\hypertarget{solutions_to_the_two_equilateral_triangles_problem}{}\section*{{Solutions to the Two Equilateral Triangles Problem}}\label{solutions_to_the_two_equilateral_triangles_problem}

\hypertarget{using_trigonometry}{}\subsection*{{Using Trigonometry}}\label{using_trigonometry}

From the formula for the [[sine formula for the area of a triangle|area of a triangle]] we can write the area as:

\begin{displaymath}
\frac{1}{2} a b \sin C
\end{displaymath}
where $a$ is the side of the black triangle, $b$ of the blue, and $C$ is the angle at the apex.

Let us write $x$ for the small angle between the triangles, so that $C = 120^\circ - x$. Then the angle at the bottom between the triangles is $60^\circ + x$ so the [[sine rule]] tells us that:

\begin{displaymath}
\frac{\sin(60^\circ + x)}{a} = \frac{\sin(60^\circ)}{b}
\end{displaymath}
which rearranges to

\begin{displaymath}
b \sin(60^\circ + x) = a \sin(60^\circ)
\end{displaymath}
Now the sine function satisfies $\sin(180^\circ - y) = \sin(y)$ so $\sin(60^\circ + x) = \sin(120^\circ - x)$ and so

\begin{displaymath}
\frac{1}{2} a b \sin C = \frac{1}{2} a b \sin(60^\circ + x) = \frac{1}{2} a^2 \sin(60^\circ)
\end{displaymath}
This is the area of the black triangle, and so the total area is $8$.

\hypertarget{using_the_agg_invariance_principle}{}\subsection*{{Using the Agg Invariance Principle}}\label{using_the_agg_invariance_principle}

Using the [[Agg Invariance Principle]], we can consider two extreme examples of the diagram. In one, the triangles overlap completely and so the blue area is coincidental with the black triangle. In the second, the triangles are adjacent but do not overlap and so are the same size and the blue area occupies half of each. In either case, the blue area is easily seen to be the same as that of the black triangle and so is $8$.

\hypertarget{using_angles_in_a_circle}{}\subsection*{{Using Angles in a Circle}}\label{using_angles_in_a_circle}

Knowing that the answer is $8$, we can see that the far point of the blue triangle must lie on a line parallel to the right hand edge of the black triangle. To demonstrate that this is the case we need to show that the angle it makes is the same as the small angle at the apex of the black triangle. One way to show this is to use the fact that [[angles in the same segment|angles in the same segment in a circle are equal]].

Draw the circumcircle of the blue triangle. The right hand edge of the blue triangle is a chord in this circle and the angle subtended at the left hand vertex of the blue triangle is $60^\circ$. The left hand vertex of the black triangle can also be viewed as being subtended from this chord and it also has angle $60^\circ$. It therefore also lies on the circumcircle.

Now we turn our attention to the small angles. These are both subtended from the same chord on this circumcircle and hence are equal. This is enough to show that the line segment joining the left hand vertices of the triangles is parallel to the right hand edge of the black triangle.

Using the right hand edge of the black triangle as the ``base'' of the blue area, its height is then the distance between these two parallel lines, which is independent of where the vertex of the blue triangle is along this line. The blue area is therefore the same area as the black triangle.



\end{document}