Difference between revisions of "2022 AMC 12B Problems/Problem 19"
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+ | ==Solution 3 (Law of Cosine)== | ||
+ | |||
+ | Let <math>AG = AE = GE = CE = 1</math>. Since <math>G</math> is the centroid, <math>DG = \frac12</math>, <math>BG = 2</math>. | ||
+ | |||
+ | <cmath>\angle BGD = \angle AGE = 60^{\circ}</cmath> | ||
+ | |||
+ | By the Law of Cosine in <math>\triangle BGD</math> | ||
+ | |||
+ | <cmath>BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD</cmath> | ||
+ | |||
+ | <cmath>BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angle BGD} = \frac{\sqrt{13}}{2}, \quad CD = \frac{\sqrt{13}}{2}</cmath> | ||
+ | |||
+ | By the Law of Cosine in <math>\triangle ACD</math> | ||
+ | |||
+ | <cmath>AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C</cmath> | ||
+ | |||
+ | <cmath>\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}</cmath> | ||
+ | |||
+ | <cmath> 5 + 13 + 26 = \boxed{\textbf{(A) }44}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 4 (Barycentric Coordinates)== | ||
+ | Using reference triangle <math>\triangle AGE</math>, we can let <cmath>A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).</cmath> If we move <math>A,B,C</math> each over by <math>(1,0,-2)</math>, leaving <math>\angle C</math> unchanged, we have <cmath>A=(2,0,-2),B=(1,3,-4),C=(0,0,0).</cmath> The angle <math>\theta</math> between vectors <math>\overrightarrow{CA}</math> and <math>\overrightarrow{CB}</math> satisfies <cmath>\cos\theta=\frac{(2)(1)+(0)(3)+(-2)(-4)}{\sqrt{\left[2^{2}+0^{2}+(-2)^{2}\right]\left[1^{2}+3^{2}+(-4)^{2}\right]}}=\frac{10}{\sqrt{8\cdot 26}}=\frac{10}{4\sqrt{13}}=\frac{5\sqrt{13}}{26},</cmath> giving the answer, <math>5+13+26=\boxed{\textbf{(A)}~44}</math>. | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/QNjvpYI1V5g | ||
+ | |||
+ | ~r00tsOfUnity | ||
==Video Solution (Just 3 min!)== | ==Video Solution (Just 3 min!)== |
Latest revision as of 03:53, 12 November 2023
Contents
Problem
In medians
and
intersect at
and
is equilateral. Then
can be written as
, where
and
are relatively prime positive integers and
is a positive integer not divisible by the square of any prime. What is
Diagram
Solution 1 (Law of Cosines)
Let . Since
is the midpoint of
, we must have
.
Since the centroid splits the median in a ratio,
and
.
Applying Law of Cosines on and
yields
and
. Finally, applying Law of Cosines on
yields
. The requested sum is
.
Solution 2 (Law of Cosines: One Fewer Step)
Let . Since
(as
is the centroid),
. Also,
and
. By the law of cosines (applied on
),
.
Applying the law of cosines again on gives
, so the answer is
.
Solution 3 (Law of Cosine)
Let . Since
is the centroid,
,
.
By the Law of Cosine in
By the Law of Cosine in
Solution 4 (Barycentric Coordinates)
Using reference triangle , we can let
If we move
each over by
, leaving
unchanged, we have
The angle
between vectors
and
satisfies
giving the answer,
.
~r00tsOfUnity
Video Solution by MOP 2024
~r00tsOfUnity
Video Solution (Just 3 min!)
~Education, the Study of Everything
Video Solution(Length & Angle Chasing)
~Hayabusa1
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.