Difference between revisions of "2015 AIME I Problems/Problem 4"

Revision as of 21:03, 25 March 2015

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 actually just $\frac{1}{2}$ and $\frac{1}{2}$ , so they 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.

@@ Line 10: / Line 10: @@
 Note that <math>AB=DB=16</math> and <math>BE=BC=4</math>. Also, <math>\angle ABE = \angle DBC = 120^{\circ}</math>. Thus, <math>\triangle ABE \cong \triangle DBC</math> by SAS.
-From this, it is clear that a <math>60^{\circ}</math> rotation about B will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.
+From this, it is clear that a <math>60^{\circ}</math> rotation about <math>B</math> will map <math>\triangle ABE</math> to <math>\triangle DBC</math>.
 This rotation also maps <math>M</math> to <math>N</math>. Thus, <math>BM=BN</math> and <math>\angle MBN=60^{\circ}</math>. Thus, <math>\triangle BMN</math> is equilateral.
@@ Line 29: / Line 29: @@
 Admittedly, this is much more tedious than the coordinate solutions.
-I noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:
+I also noticed that there are two more ways of showing that <math>\triangle BMN</math> is equilateral:
 One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle ECB</math> are related by a spiral similarity centered at <math>B</math>.
-The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. (The weights are actually <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so they are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.
+The other way is to use the Mean Geometry Theorem. Note that <math>\triangle BCE</math> and <math>\triangle BDA</math> are similar and have the same orientation. Note that <math>B</math> is the weighted average of <math>B</math> and <math>B</math>, <math>M</math> is the weighted average of <math>E</math> and <math>A</math>, and <math>N</math> is the weighted average of <math>C</math> and <math>D</math>. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so they are also unweighted averages.) Thus, by the Mean Geometry Theorem, <math>\triangle BMN</math> is similar to both <math>\triangle BAD</math> and <math>\triangle BEC</math>, which means that <math>\triangle BMN</math> is equilateral.
 ==See Also==
 {{AIME box|year=2015|n=I|num-b=3|num-a=5}}
 {{MAA Notice}}

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

Page

Toolbox

Search