3420.71 – Hard Triangle Area

If GRAW is a square, and the triangle GMR is equilateral, what is the area of triangle RAC?

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Solution

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

Now, AD = xx and RD = x3x\sqrt3, so a=x+x3a = x + x\sqrt 3 and x=a3+1.x = \frac{a}{\sqrt3 +1}. This leads to

△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.\triangle\text{RAC} = \frac{a}{2}\cdot\frac{a}{\sqrt3 +1} = \frac{a^2}{2(\sqrt 3 +1)}=\frac{a^2(\sqrt 3 - 1)}{2(\sqrt 3+1)(\sqrt 3-1)}=\frac{a^2(\sqrt 3-1)}{2(3-1)}=\frac{a^2(\sqrt 3-1)}{4}.

And now, how about the area of △\triangleGCM?

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