Difference between revisions of "2015 AIME I Problems/Problem 4"
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− | Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FFFFFF">talk</font>]] | + | Diagram by [[User:RedFireTruck|<font color="#FF0000">RedFireTruck</font>]] ([[User talk:RedFireTruck|<font color="#FFFFFF">talk</font>]]) |
− | ==Solution 1 | + | ==Solution 1== |
− | Let | + | Let <math>A</math> be the origin, so <math>B=(16,0)</math> and <math>C=(20,0).</math> Using equilateral triangle properties tells us that <math>D=(8,8\sqrt3)</math> and <math>E=(18,2\sqrt3)</math> as well. Therefore, <math>M=(9,\sqrt3)</math> and <math>N=(14,4\sqrt3).</math> Applying the Shoelace Theorem to triangle <math>BMN</math> gives |
+ | |||
+ | <cmath>x=\dfrac 1 2 |16\sqrt3+36\sqrt3+0-(0+14\sqrt3+64\sqrt3)| =13\sqrt3,</cmath> | ||
+ | |||
+ | so <math>x^2=\boxed{507}.</math> | ||
==Solution 2== | ==Solution 2== | ||
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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 the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these 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 the same for all three averages. (The weights are actually just <math>\frac{1}{2}</math> and <math>\frac{1}{2}</math>, so these 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. | ||
+ | |||
+ | |||
+ | Note: A much easier way to go about finding <math>BM</math> without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that <math>AY = 18 \implies AX = 9 \implies BX = 16-9 = 7</math>. Additionally, <math>MX = \frac{1}{2} (2\sqrt{3}) = \sqrt{3}</math> from similar triangles meaning we can now just do pythagorean theorem on right triangle <math>MXB</math> to get <math>MB = \sqrt{52}</math> - SuperJJ | ||
==Solution 3== | ==Solution 3== | ||
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[[File:2015 AIME I 4.png|430px|right]] | [[File:2015 AIME I 4.png|430px|right]] | ||
− | <math> AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE | + | <math> AB = BD, BE = BC, \angle ABE = \angle CBD \implies \triangle ABE \cong \triangle DBC</math> |
Medians are equal, so <math>MB = MN, \angle ABM = \angle DBN \implies</math> | Medians are equal, so <math>MB = MN, \angle ABM = \angle DBN \implies</math> | ||
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<math>BM</math> is the median of <math>\triangle ABE \implies</math> | <math>BM</math> is the median of <math>\triangle ABE \implies</math> | ||
− | <math> | + | <math>BM^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> | The area of <math>\triangle BMN</math> |
Latest revision as of 00:12, 8 May 2024
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)
Solution 1
Let be the origin, so
and
Using equilateral triangle properties tells us that
and
as well. Therefore,
and
Applying the Shoelace Theorem to triangle
gives
so
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.
Note: A much easier way to go about finding without having to use Stewart's Theorem is to simply drop the altitudes from M and E to AC, thus hitting AC at points X and Y. Then clearly AEY and AMX are similar with ratio 2. But we know that
. Additionally,
from similar triangles meaning we can now just do pythagorean theorem on right triangle
to get
- SuperJJ
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.