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 .


1 Answer 1


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: $$m_a=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}.$$

  • $\begingroup$ Can I know what is $M$? (Actually I thought of getting some hints, you posted a solution :]) $\endgroup$
    – Anonymous
    Sep 20, 2020 at 11:13
  • $\begingroup$ @Anonymous See your given. $\endgroup$ Sep 20, 2020 at 11:16
  • $\begingroup$ Oh I see , in the question I gave midpoint to be M $\endgroup$
    – Anonymous
    Sep 20, 2020 at 11:16
  • 1
    $\begingroup$ @Anonymous My post it's more a hint than a full solution. Try to understand, what I wrote. Good luck! $\endgroup$ Sep 20, 2020 at 11:18
  • $\begingroup$ Okay @MichaelRozenburg I understood what you did everything right there , except how did you get $BN$ to be that expression? $\endgroup$
    – Anonymous
    Sep 20, 2020 at 11:27

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