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

Revision as of 21:57, 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 $\triangle ABE$ , $AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot(-\frac{1}{2})$ $AE = 4\sqrt{21}$ 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 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$ (Error compiling LaTeX. Unknown error_msg)\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 $. (The weights are actually$ \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 6: / Line 6: @@
 ==Solution 2==
 Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle <math>BMN</math> to solve for its area.
+==Solution 3==
+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>.
+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.
+Using the Law of Cosines on <math>\triangle ABE</math>,
+<cmath>AE^2 = 16^2 + 4^2 - 2\cdot 16\cdot 4\cdot(-\frac{1}{2})</cmath>
+<cmath>AE = 4\sqrt{21}</cmath>
+Thus, <math>AM=ME=2\sqrt{21}</math>.
+Using Stewart's Theorem on <math>\triangle ABE</math>,
+<cmath>AE\cdot AM\cdot AE + AE\cdot BM^2 = BE^2\cdot AM + BA^2\cdot ME</cmath>
+<cmath>BM = 2\sqrt{13}</cmath>
+Calculating the area of <math>\triangle BMN</math>,
+<cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2</cmath>
+<cmath>[BMN] = 13\sqrt{3}</cmath>
+Thus, <math>x=13\sqrt{3}</math>, so <math>x^2 = 507</math>. Our final answer is <math>\boxed{507}</math>.
+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:
+One way is to show that <math>\triangle ADB</math>, <math>\triangle BMN</math>, and <math>\triangle </math>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$ is equilateral.
 ==See Also==
 {{AIME box|year=2015|n=I|num-b=3|num-a=5}}
 {{MAA Notice}}

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