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Let $ABC$ be a triangle. Construct points $B'$ and $C'$ such that $ACB'$ and $ABC'$ are equilateral triangles that have no overlap with $\triangle ABC$. Let $BB'$ and $CC'$ intersect at $X$. If $AX = 3, BC = 4,$ and $CX = 5$, find the area of quadrilateral $BCB'C'$.

We know that $X$ is the Fermat point. From the cosine rule, we can work out $CA$, but I'm not sure how to proceed.

Original source: 2014 Berkeley Math Tournament G9

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As you said, $X$ is the Fermat point.

And also with having three equilateral triangles on the sides, we have an interesting fact that $AA'=BB'=CC'=L$. (Here's a good link where you may find the proof and some other interesting things)

Now consider $\Delta CXA'$, we could see $$CX=L-5$$$$XA'=L-3$$$$CA'=4$$ And the calculations will show us $L=7.6$, now we are interested in calculating $BX$ and $B'X$

$\dfrac{C'X}{A'B}=\dfrac{2.6}{4}=\dfrac{KC}{KA'}=\dfrac{KX}{KB}$

$\Rightarrow \begin{cases} \dfrac{4}{2.6}XK+CK=4 \\ \dfrac{4}{2.6}CK+XK=4.6 \end{cases} $

By solving this system of equations and using $$\dfrac{CK}{XK}=\dfrac{CA'}{BX}=\dfrac{4}{BX}$$ We can find out that $BX=2$, therefore $BX'=5.6$.

Now by calculating areas of $\Delta B'XC'$, $\Delta BXC$, $\Delta BXC'$ and $\Delta B'XC$ we may find the area of $BCB'C'$ ( forgive me if I skipped the calculation part in my solution! I hope the total vision is right)

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