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2015 AIME I Problems/Problem 4

Revision as of 21:04, 25 March 2015 by Sujaykazi (talk | contribs) (→‎Solution 3)

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 $x^2$.

Solution

Let point $A$ be at $(0,0)$. Then, $B$ is at $(16,0)$, and $C$ is at $(20,0)$. Due to symmetry, it is allowed to assume $D$ and $E$ are in quadrant 1. By equilateral triangle calculations, Point $D$ is at $(8,8\sqrt{3})$, and Point $E$ is at $(18,2\sqrt{3})$. By Midpoint Formula, $M$ is at $(9,\sqrt{3})$, and $N$ is at $(14,4\sqrt{3})$. The distance formula shows that $BM=BN=MN=2\sqrt{13}$. Therefore, by equilateral triangle area formula, $x=13\sqrt{3}$, so $x^2$ is $\boxed{507}$.

Solution 2

Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle $BMN$ to solve for its area.

Solution 3

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 $\triangle ABE$, \[AE\cdot AM\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME\] \[BM = 2\sqrt{13}\]

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.

See Also

2015 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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