Difference between revisions of "2015 AIME I Problems/Problem 4"
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==Solution 3== | ==Solution 3== | ||
− | <math> AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE= \triangle DBC | + | |
− | + | [[File:2015 AIME I 4.png|430px|right]] | |
− | The height of BCE is <math>2 \sqrt{3} \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} | + | |
− | <math>BM</math> is the median of <math>\triangle ABE \implies MB^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 – 84 = 52.</math> | + | <math> AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE= \triangle DBC</math> |
− | The area of <math>\triangle BMN</math> | + | |
+ | Medians are equal, so <math>MB = MN, \angle ABM = \angle DBN \implies</math> | ||
+ | <math>\angle MBN = \angle ABD - \angle ABM + \angle DBN = 60^\circ \implies </math> | ||
+ | |||
+ | <math>\triangle MNB</math> is equilateral triangle. | ||
+ | |||
+ | The height of <math>\triangle BCE</math> is <math>2 \sqrt{3},</math> distance from <math>A</math> to midpoint <math>BC</math> is <math>16 + 2 = 18 \implies \frac {AE^2}{4} =\frac{ (16 + 2)^2 +2^2 \cdot 3}{4} = 81 + 3 = 84.</math> | ||
+ | |||
+ | <math>BM</math> is the median of <math>\triangle ABE \implies</math> | ||
+ | <math>MB^2 = \frac {AB^2}{2} + \frac {BE^2}{2} - \frac {AE^2}{4}=16 \cdot 8 + 4 \cdot 2 – 84 = 52.</math> | ||
+ | |||
+ | The area of <math>\triangle BMN</math> | ||
+ | |||
+ | <cmath>[BMN] = \frac{\sqrt{3}}{4} BM^2 =13 \sqrt{3} \implies \boxed{\textbf{507}}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==See Also== | ==See Also== |
Revision as of 09:40, 1 September 2022
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
.
Diagram
Diagram by RedFireTruck (talk) 18:52, 15 February 2021 (EST)
Solution 1 (fastest)
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
by Shoelace Theorem,
, so
is
.
Solution 2
Note that and
. Also,
. Thus,
by SAS.
From this, it is clear that a rotation about
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 also noticed that there are two more ways of showing that is equilateral:
One way is to show that ,
, and
are related by a spiral similarity centered at
.
The other way is to use the Mean Geometry Theorem. Note that and
are similar and have the same orientation. Note that
is the weighted average of
and
,
is the weighted average of
and
, and
is the weighted average of
and
. The weights are the same for all three averages. (The weights are actually just
and
, so these are also unweighted averages.) Thus, by the Mean Geometry Theorem,
is similar to both
and
, which means that
is equilateral.
Solution 3
Medians are equal, so
is equilateral triangle.
The height of is
distance from
to midpoint
is
is the median of
The area of
vladimir.shelomovskii@gmail.com, vvsss
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.