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


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 $B'X$


$\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)

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.