Find the area of an equilateral triangle

Question

Point $M$ lies on the side $AB$ of an equilateral triangle $ABC$ such that $AM:BM=1:2$. If $CM=m$, find the area of the triangle $ABC$.

The answer given in my book is $\dfrac{9m^2\sqrt3}{28}$.

So we have $$S_{ABC}=S_{AMC}+S_{BMC}\iff\\\iff\dfrac{AB\cdot CH}{2}=\dfrac{AM\cdot CH}{2}+\dfrac{BM\cdot CH}{2}\\AB\cdot CH=AM\cdot CH+BM\cdot CH.$$ This doesn't seem to help very much. Any help would be appreciated.

Answer 1

The cosine rule on triangle $MBC$ gives $$m^2=\dfrac49a^2+a^2-2\cdot\dfrac23 a\cdot a\cdot\cos60^\circ\\\Rightarrow m=\dfrac{a\sqrt7}{3}.$$ Now using $S=\dfrac{1}{2}a\cdot a\cdot\sin60^\circ=\dfrac{a^2\sqrt3}{4}...$

Answer 2

Hint. Let $AM=x.$ Then $MB=2x.$ Apply Heron to the equilateral triangle to find its area $$\sqrt{s(s-3x)^3},$$ where $s=9x/2.$ Then calculate the areas of the component triangles to find the area of the equilateral triangle in another form, to get $$\sqrt{t(t-x)(t-m)(t-3x)}+\sqrt{u(u-2x)(u-3x)(u-m)},$$ where $t=\frac{x+m+3x}{2}$ and $u=\frac{2x+3x+m}{2}.$ Then you can find the area of the equilateral triangle in terms of $m$ alone.

Answer 3

I like your approach of looking at the areas of smaller triangles. You nearly showed that $S_{BMC}=\frac23 S_{ABC}$ , which means if we find $S_{BMC}$ , we can simply multiply it by $\frac32$ to get the desired $S_{ABC}$ . Let's see how we can do that! If we choose point $N$ on $BC$ such that $BN : CN = 1 : 2$ then $S_{MNC} = \frac23 S_{BMC}$. Besides, $\triangle MNB$ is a right triangle (why?). Now, this helps us calculate side length $a$ of $\triangle ABC$ in terms of $m$. Can you take it from here?