Difference between revisions of "2015 AIME I Problems/Problem 4"
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==Solution 2== | ==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. | 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== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=3|num-a=5}} | {{AIME box|year=2015|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:57, 25 March 2015
Problem
Point lies on line segment
with
and
. Points
and
lie on the same side of line
forming equilateral triangles
and
. Let
be the midpoint of
, and
be the midpoint of
. The area of
is
. Find
.
Solution
Let point be at
. Then,
is at
, and
is at
. Due to symmetry, it is allowed to assume
and
are in quadrant 1. By equilateral triangle calculations, Point
is at
, and Point
is at
. By Midpoint Formula,
is at
, and
is at
. The distance formula shows that
. Therefore, by equilateral triangle area formula,
, so
is
.
Solution 2
Use the same coordinates as above for all points. Then use the Shoelace Formula/Method on triangle to solve for its area.
Solution 3
Note that and
. Also,
. Thus,
by SAS.
From this, it is clear that a rotation about B will map
to
.
This rotation also maps
to
. Thus,
and
. Thus,
is equilateral.
Using the Law of Cosines on ,
Thus,
.
Using Stewart's Theorem on ,
Calculating the area of ,
Thus,
, so
. Our final answer is
.
Admittedly, this is much more tedious than the coordinate solutions.
I noticed that there are two more ways of showing that is equilateral:
One way is to show that ,
, and
ECB
B$.
The other way is to use the Mean Geometry Theorem. Note that$ (Error compiling LaTeX. Unknown error_msg)\triangle BCE\triangle BDA
B
B
B
M
E
A
N
C
D
\frac{1}{2}
\frac{1}{2}
\triangle BMN
\triangle BAD
\triangle BEC
\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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