# Point $B$ lies on line segment $\overline{AC}$ with $AB = 16$ , $BC = 4$ .

Point $$B$$ lies on line segment $$\overline{AC}$$ with $$AB = 16$$ , $$BC = 4$$ . Points $$D$$ and $$E$$ lie on the same side of line $$AC$$ forming equilateral triangles $$\Delta ABD$$ and $$\Delta BCE$$ . Let $$M$$ be the midpoint of $$\overline{AE}$$, and $$N$$ be the midpoint of $$\overline{CD}$$ . The area of $$\Delta BMN$$ is $$x$$ . Find $$x^2$$ .
Source :- $$2015$$ AIME Problem $$4$$ .

What I Tried :- Ok I want to say that I don't know very much of Geometry and I am a little weak on this subject, but I tried my best and want some hints. Here is the whole figure of my picture in Geogebra :- I have noted all the angles which are equal with the same colour. However, not all angles are understandable why they are equal, but I found them so in Geogebra . For example $$\angle BAE = \angle BDC$$, which means that in some way $$\Delta CAH$$ is similar to $$\Delta BDC$$ , but I don't know how. This is $$1$$ way from which I cannot proceed.

Another is that surprisingly, $$\Delta BGF$$ (Green Triangle) , is equilateral everytime ; and that is what we need as the area . First, if it is equilateral, then $$\angle GBA = \angle EBF$$ . But why is it so?

I was able to deduce that as $$CE \parallel BD$$ , I can find that $$\angle ECD = \angle CDB$$ , and maybe if I take their values to be $$\theta$$ , maybe angle-chasing can help?

Can I get some hints for this problem?

Note :- This Problem already has a solution, but I am trying without checking it and rather solve geometry problems myself by hints, hence posting it here .

Since $$\Delta DBC$$ goes to $$\Delta ABE$$ after rotation around $$B$$ on $$60^{\circ},$$ we obtain: $$\Delta DBC\cong\Delta ABE,$$ which gives that $$\Delta MBN$$ is an equilateral triangle.

Thus, $$x=\frac{BN^2\sqrt3}{4}.$$ Now, $$DC^2=16^2+4^2+2\cdot16\cdot4\cdot\frac{1}{2}=336,$$ which gives $$BN=\frac{1}{2}\sqrt{2\cdot16^2+2\cdot4^2-336}=\sqrt{52},$$ $$x=\frac{52\sqrt3}{4}=13\sqrt3$$ and $$x^2=507.$$ For getting of $$BN$$ we can use the following reasoning.

$$BN$$ is a median of $$\Delta DBC$$, where $$DB=16$$, $$BC=4$$ and $$\measuredangle DBC=120^{\circ}.$$

Now, by law of cosines we got $$DC$$.

Also, in $$\Delta ABC$$ for a median $$m_a$$ we have: $$m_a=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}.$$

• Can I know what is $M$? (Actually I thought of getting some hints, you posted a solution :]) Sep 20, 2020 at 11:13
• @Anonymous See your given. Sep 20, 2020 at 11:16
• Oh I see , in the question I gave midpoint to be M Sep 20, 2020 at 11:16
• @Anonymous My post it's more a hint than a full solution. Try to understand, what I wrote. Good luck! Sep 20, 2020 at 11:18
• Okay @MichaelRozenburg I understood what you did everything right there , except how did you get $BN$ to be that expression? Sep 20, 2020 at 11:27