# 2015 AIME I Problems/Problem 4

## Problem

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

## Diagram $[asy] pair A = (0, 0), B = (16, 0), C = (20, 0), D = (8, 8*sqrt(3)), EE = (18, 2*sqrt(3)), M = (9, sqrt(3)), NN = (14, 4*sqrt(3)); draw(A--B--D--cycle); draw(B--C--EE--cycle); draw(A--EE); draw(C--D); draw(B--M--NN--cycle); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(M); dot(NN); label("A", A, SW); label("B", B, S); label("C", C, SE); label("D", D, N); label("E", EE, N); label("M", M, NW); label("N", NN, NE); [/asy]$

Diagram by RedFireTruck (talk) 18:52, 15 February 2021 (EST)

## Solution 1

Let $A$ be the origin, so $B=(16,0)$ and $C=(20,0).$ Using equilateral triangle properties tells us that $D=(8,8\sqrt3)$ and $E=(18,2\sqrt3)$ as well. Therefore, $M=(9,\sqrt3)$ and $N=(14,4\sqrt3).$ Applying the Shoelace Theorem to triangle $BMN$ gives $$x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt3,$$

so $x^2=\boxed{507}.$

## Solution 2

Note that $AB=DB=16$ and $BE=BC=4$. Also, $\angle ABE = \angle DBC = 120^{\circ}$. Thus, $\triangle ABE \cong \triangle DBC$ by SAS.

From this, it is clear that a $60^{\circ}$ rotation about $B$ will map $\triangle ABE$ to $\triangle DBC$. This rotation also maps $M$ to $N$. Thus, $BM=BN$ and $\angle MBN=60^{\circ}$. Thus, $\triangle BMN$ is equilateral.

Using the Law of Cosines on $\triangle ABE$, $$AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot\left(-\frac{1}{2}\right)$$ $$AE = 4\sqrt{21}$$ Thus, $AM=ME=2\sqrt{21}$.

Using Stewart's Theorem on $\triangle ABE$, $$AM\cdot ME\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME$$ $$BM = 2\sqrt{13}$$

Calculating the area of $\triangle BMN$, $$[BMN] = \frac{\sqrt{3}}{4} BM^2$$ $$[BMN] = 13\sqrt{3}$$ Thus, $x=13\sqrt{3}$, so $x^2 = 507$. Our final answer is $\boxed{507}$.

Admittedly, this is much more tedious than the coordinate solutions.

I also noticed that there are two more ways of showing that $\triangle BMN$ is equilateral:

One way is to show that $\triangle ADB$, $\triangle BMN$, and $\triangle ECB$ are related by a spiral similarity centered at $B$.

The other way is to use the Mean Geometry Theorem. Note that $\triangle BCE$ and $\triangle BDA$ are similar and have the same orientation. Note that $B$ is the weighted average of $B$ and $B$, $M$ is the weighted average of $E$ and $A$, and $N$ is the weighted average of $C$ and $D$. The weights are the same for all three averages. (The weights are actually just $\frac{1}{2}$ and $\frac{1}{2}$, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem, $\triangle BMN$ is similar to both $\triangle BAD$ and $\triangle BEC$, which means that $\triangle BMN$ is equilateral.

Note: A much easier way to go about finding $BM$ without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that $AY = 18 \implies AX = 9 \implies BX = 16-9 = 7$. Additionally, $MX = \frac{1}{2} (2\sqrt{3}) = \sqrt{3}$ from similar triangles meaning we can now just do pythagorean theorem on right triangle $MXB$ to get $MB = \sqrt{52}$ - SuperJJ

## Solution 3 $AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE \cong \triangle DBC$

Medians are equal, so $MB = MN, \angle ABM = \angle DBN \implies$ $\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies$ $\triangle MNB$ is equilateral triangle.

The height of $\triangle BCE$ is $2 \sqrt{3},$ distance from $A$ to midpoint $BC$ is $16 + 2 = 18 \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} = 81 + 3 = 84.$ $BM$ is the median of $\triangle ABE \implies$ $BM^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 - 84 = 52.$

The area of $\triangle BMN$ $$[BMN] = \frac{\sqrt{3}}{4} BM^2 =13 \sqrt{3} \implies \boxed{\textbf{507}}.$$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 