3420.71 – Hard Triangle Area

If GRAWGRAW is a square, and the triangle △GMR\triangle GMR is equilateral, what is the area of triangle △RAC\triangle RAC?

Solution

If the side of the square is aa and xx is the height of the △RAC\triangle RAC, then the area of △RAC=ax2\triangle RAC = \dfrac{ax}{2}.

Now, AD=xAD = x and RD=x3RD = x\sqrt3, so a=x+x3a = x + x\sqrt 3 and x=a3+1x = \dfrac{a}{\sqrt3 +1}. This leads to

Area(△RAC)=a2⋅a3+1=a22(3+1)=a2(3−1)2(3+1)(3−1)=a2(3−1)2(3−1)=a2(3−1)4\begin{align*}\text{Area}(\triangle\text{RAC}) &= \dfrac{a}{2}\cdot\dfrac{a}{\sqrt3 +1} \\[1em]&= \dfrac{a^2}{2(\sqrt 3 +1)} \\[1em]&= \dfrac{a^2(\sqrt 3 - 1)}{2(\sqrt 3+1)(\sqrt 3-1)} \\[1em]&= \dfrac{a^2(\sqrt 3-1)}{2(3-1)} \\[1em]&= \dfrac{a^2(\sqrt 3-1)}{4}\end{align*}

And now, how about the area of △GCM\triangle GCM?