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Difference between revisions of "2022 AMC 12B Problems/Problem 19"

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== Problem ==
 
== Problem ==
  
−
In <math>\triangle ABC</math> medians <math>\overline{\rm AD}</math> and <math>\overline{\rm BE}</math> intersect at <math>G</math> and <math>\triangle AGE</math> is equilateral. Then <math>\cos(C)</math> can be written as <math>\frac{m\sqrt{p}}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers and <math>p</math> is a positive integer not divisible by the square of any prime. What is <math>m+n+p</math>?
+
In <math>\triangle{ABC}</math> medians <math>\overline{AD}</math> and <math>\overline{BE}</math> intersect at <math>G</math> and <math>\triangle{AGE}</math> is equilateral. Then <math>\cos(C)</math> can be written as <math>\frac{m\sqrt p}n</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers and <math>p</math> is a positive integer not divisible by the square of any prime. What is <math>m+n+p?</math>
  
−
<math>\textbf{(A)}\ 44 \qquad
+
<math>\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60</math>
−
\textbf{(B)}\ 48 \qquad
 
−
\textbf{(C)}\ 52 \qquad
 
−
\textbf{(D)}\ 56 \qquad
 
−
\textbf{(E)}\ 60 \qquad</math>
 
  
−
==Solution 1: Law of Cosines==
+
== Diagram ==
−
Note: can someone add the diagram here please, I don't know how to do that
+
<asy>
 +
            import geometry;
 +
            unitsize(2cm);
  
−
Let <math>\overline{AG}=\overline{AE}=\overline{EG}=2x</math>. Since <math>E</math> is the midpoint of <math>\overline{AC}</math>, <math>\overline{EC}</math> must also be <math>2x</math>.
+
real arg(pair p) {
 +
              return atan2(p.y, p.x) * 180/pi;
 +
            }
  
−
Since the centroid splits the median in a <math>2:1</math> ratio, <math>\overline{GD}</math> must be equal to <math>x</math> and <math>\overline{BG}</math> must be equal to <math>4x</math>.
+
            pair G=(0,0),E=(1,0),A=(1/2,sqrt(3)/2),D=1.5*G-0.5*A,C=2*E-A,B=2*D-C;
  
−
Applying Law of Cosines on <math>\triangle{}ADC</math> and <math>\triangle{}AGB</math> yields <math>\overline{AB}=\sqrt{28}x</math> and <math>\overline{CD}=\overline{BD}=\sqrt{13}x</math>. Finally, applying Law of Cosines on <math>\triangle{}ABC</math> yields <math>\cos{C}=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}</math>. The requested sum is <math>5+13+26=44</math>.
+
            pair t(pair p) {
 +
                return rotate(-arg(dir(B--C)))*p;
 +
            }
 +
 
 +
 
 +
            path t(path p) {
 +
                return rotate(-arg(dir(B--C)))*p;
 +
            }
 +
 
 +
            void d(path p, pen q = black+linewidth(1.5)) {
 +
                draw(t(p),q);
 +
            }
 +
 
 +
            void o(pair p, pen q = 5+black) {
 +
                dot(t(p),q);
 +
            }
 +
 
 +
            void l(string s, pair p, pair d) {
 +
                label(s, t(p),d);
 +
            }
 +
           
 +
            d(A--B--C--cycle);
 +
            d(A--D);
 +
            d(B--E);
 +
            o(A);
 +
            o(B);
 +
            o(C);
 +
            o(D);
 +
            o(E);
 +
            o(G);
 +
            l("$A$",A,N);
 +
            l("$B$",B,SW);
 +
            l("$C$",C,SE);
 +
            l("$D$",D,S);
 +
            l("$E$",E,NE);
 +
            l("$G$",G,NW);
 +
</asy>
 +
 
 +
==Solution 1 (Law of Cosines)==
 +
 
 +
Let <math>AG=AE=EG=2x</math>. Since <math>E</math> is the midpoint of <math>\overline{AC}</math>, we must have <math>EC=2x</math>.
 +
 
 +
Since the centroid splits the median in a <math>2:1</math> ratio, <math>GD=x</math> and <math>BG=4x</math>.
 +
 
 +
Applying Law of Cosines on <math>\triangle ADC</math> and <math>\triangle{}AGB</math> yields <math>AB=\sqrt{28}x</math> and <math>CD=BD=\sqrt{13}x</math>. Finally, applying Law of Cosines on <math>\triangle ABC</math> yields <math>\cos(C)=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}</math>. The requested sum is <math>5+13+26=44</math>.
 +
 
 +
==Solution 2 (Law of Cosines: One Fewer Step) ==
 +
 
 +
Let <math>AG = 1</math>. Since <math>\frac{BG}{GE}=2</math> (as <math>G</math> is the centroid), <math>BE = 3</math>. Also, <math>EC = 1</math> and <math>\angle{BEC} = 120^{\circ}</math>. By the law of cosines (applied on <math>\triangle BEC</math>), <math>BC = \sqrt{13}</math>.
 +
 
 +
Applying the law of cosines again on <math>\triangle BEC</math> gives <math>\cos{\angle{C}} = \frac{1 + 13 - 9}{2\sqrt{13}} = \frac{5\sqrt{13}}{26}</math>, so the answer is <math>\fbox{\textbf{(A)}\ 44}</math>.
 +
 
 +
~[[User:Bxiao31415 | Bxiao31415]]
 +
 
 +
==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
 +
 
 +
==Solution 5 (Coordinate Bash)==
 +
 
 +
Let \(A\) be at \((0,0)\), with \(E\) at \((1,0)\) and \(G\) at \((\tfrac{1}{2},\tfrac{\sqrt{3}}{2})\). Because \(\triangle AGE\) is equilateral, the equation of line \(AD\) is \(y=x\sqrt{3}\), and the equation for \(BE\) is \(y=-x\sqrt{3}+\sqrt{3}\). Because \(BE\) is a median, we know that \(C\) is at \((2,0)\). 
 +
 
 +
Therefore, the equation for line \(BC\) is \(y=m(x-2)\), where \(m\) is the slope of the line. Since \(AD\) is also a median, \(D\) is the midpoint of \(\overline{BC}\). 
 +
 
 +
To find \(m\), solve for the coordinates of \(D\) in terms of \(m\): 
 +
 
 +
<cmath>x\sqrt{3}=mx-2m</cmath> 
 +
<cmath>x(m-\sqrt{3})=2m</cmath> 
 +
<cmath>x=\frac{2m}{m-\sqrt{3}}</cmath> 
 +
 
 +
The \(y\)-value of \(D\) is then \(y=x\sqrt{3}\): 
 +
 
 +
<cmath>y=\frac{2m\sqrt{3}}{m-\sqrt{3}}</cmath> 
 +
 
 +
Because \(D\) is the midpoint of \(\overline{BC}\) and \(y_C=0\), the \(y\)-value of \(B\) is double that of \(D\): 
 +
 
 +
<cmath>y_B=\frac{4m\sqrt{3}}{m-\sqrt{3}}</cmath> 
 +
 
 +
Now use the equation for line \(BE\) to express \(B\): 
 +
 
 +
<cmath>mx-2m=-x\sqrt{3}+\sqrt{3}</cmath> 
 +
<cmath>x(m+\sqrt{3})=2m+\sqrt{3}</cmath> 
 +
<cmath>x_B=\frac{2m+\sqrt{3}}{m+\sqrt{3}}</cmath> 
 +
 
 +
Plug this back into \(BE\) to find \(y_B\): 
 +
 
 +
<cmath>y_B=\frac{-2m-\sqrt{3}}{m+\sqrt{3}}+\sqrt{3}=\frac{-m\sqrt{3}}{m+\sqrt{3}}</cmath> 
 +
 
 +
Equating the two expressions for \(y_B\): 
 +
 
 +
<cmath>-\frac{m\sqrt{3}}{m+\sqrt{3}}=\frac{4m\sqrt{3}}{m-\sqrt{3}}</cmath> 
 +
<cmath>-\sqrt{3}m^2+3m=4\sqrt{3}m^2+12m</cmath> 
 +
<cmath>5\sqrt{3}m^2+9m=0</cmath> 
 +
<cmath>m(5m\sqrt{3}+9)=0</cmath> 
 +
 
 +
Since \(m\neq0\), we have 
 +
 
 +
<cmath>m=-\frac{3\sqrt{3}}{5}.</cmath> 
 +
 
 +
Now find \(B\): 
 +
 
 +
<cmath>-x\sqrt{3}+\sqrt{3}=-\frac{3\sqrt{3}}{5}x+\frac{6\sqrt{3}}{5}</cmath> 
 +
<cmath>\frac{-2\sqrt{3}}{5}x=\frac{\sqrt{3}}{5}</cmath> 
 +
<cmath>x_B=-\frac{1}{2}</cmath> 
 +
 
 +
Plug this into \(BE\) to get 
 +
 
 +
<cmath>y_B=\frac{3\sqrt{3}}{2}.</cmath> 
 +
 
 +
Since we’re asked to find \(\cos\angle C\), extend \(\overline{AC}\) to \(F\) where \(F=(-\frac{1}{2},0)\), forming right triangle \(\triangle BFC\). Using the Pythagorean theorem for \(\overline{BC}\): 
 +
 
 +
<cmath>\left(\frac{5}{2}\right)^2+\left(\frac{3\sqrt{3}}{2}\right)^2=BC^2,\quad BC=\sqrt{13}.</cmath> 
 +
 
 +
Thus 
 +
 
 +
<cmath>\cos\angle C=\frac{\frac{5}{2}}{\sqrt{13}}=\frac{5\sqrt{13}}{26}.</cmath> 
 +
 
 +
The problem asks for \(m+n+p\), so \(5+13+26=\boxed{44}\).
 +
 
 +
~Voidling
 +
 
 +
==Video Solution by MOP 2024==
 +
https://youtu.be/QNjvpYI1V5g
 +
 
 +
~r00tsOfUnity
 +
 
 +
==Video Solution (Just 3 min!)==
 +
https://youtu.be/Q54sH65AJa4
 +
 
 +
<i>~Education, the Study of Everything</i>
 +
 
 +
==Video Solution(Length & Angle Chasing)==
 +
https://youtu.be/JVDlHCSPF6k
 +
 
 +
~Hayabusa1
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2022|ab=B|num-b=18|num-a=20}}
 
{{AMC12 box|year=2022|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:19, 21 September 2025

Problem

In $\triangle{ABC}$ medians $\overline{AD}$ and $\overline{BE}$ intersect at $G$ and $\triangle{AGE}$ is equilateral. Then $\cos(C)$ can be written as $\frac{m\sqrt p}n$, where $m$ and $n$ are relatively prime positive integers and $p$ is a positive integer not divisible by the square of any prime. What is $m+n+p?$

$\textbf{(A) }44 \qquad \textbf{(B) }48 \qquad \textbf{(C) }52 \qquad \textbf{(D) }56 \qquad \textbf{(E) }60$

Diagram

[asy] import geometry; unitsize(2cm); real arg(pair p) { return atan2(p.y, p.x) * 180/pi; } pair G=(0,0),E=(1,0),A=(1/2,sqrt(3)/2),D=1.5*G-0.5*A,C=2*E-A,B=2*D-C; pair t(pair p) { return rotate(-arg(dir(B--C)))*p; } path t(path p) { return rotate(-arg(dir(B--C)))*p; } void d(path p, pen q = black+linewidth(1.5)) { draw(t(p),q); } void o(pair p, pen q = 5+black) { dot(t(p),q); } void l(string s, pair p, pair d) { label(s, t(p),d); } d(A--B--C--cycle); d(A--D); d(B--E); o(A); o(B); o(C); o(D); o(E); o(G); l("$A$",A,N); l("$B$",B,SW); l("$C$",C,SE); l("$D$",D,S); l("$E$",E,NE); l("$G$",G,NW); [/asy]

Solution 1 (Law of Cosines)

Let $AG=AE=EG=2x$. Since $E$ is the midpoint of $\overline{AC}$, we must have $EC=2x$.

Since the centroid splits the median in a $2:1$ ratio, $GD=x$ and $BG=4x$.

Applying Law of Cosines on $\triangle ADC$ and $\triangle{}AGB$ yields $AB=\sqrt{28}x$ and $CD=BD=\sqrt{13}x$. Finally, applying Law of Cosines on $\triangle ABC$ yields $\cos(C)=\frac{5}{2\sqrt{13}}=\frac{5\sqrt{13}}{26}$. The requested sum is $5+13+26=44$.

Solution 2 (Law of Cosines: One Fewer Step)

Let $AG = 1$. Since $\frac{BG}{GE}=2$ (as $G$ is the centroid), $BE = 3$. Also, $EC = 1$ and $\angle{BEC} = 120^{\circ}$. By the law of cosines (applied on $\triangle BEC$), $BC = \sqrt{13}$.

Applying the law of cosines again on $\triangle BEC$ gives $\cos{\angle{C}} = \frac{1 + 13 - 9}{2\sqrt{13}} = \frac{5\sqrt{13}}{26}$, so the answer is $\fbox{\textbf{(A)}\ 44}$.

~ Bxiao31415

Solution 3 (Law of Cosine)

Let $AG = AE = GE = CE = 1$. Since $G$ is the centroid, $DG = \frac12$, $BG = 2$.

\[\angle BGD = \angle AGE = 60^{\circ}\]

By the Law of Cosine in $\triangle BGD$

\[BD^2 = BG^2 + DG^2 - 2 \cdot BG \cdot DG \cdot \cos \angle BGD\]

\[BD = \sqrt {2^2 + \left( \frac{1}{2} \right)^2 - 2 \cdot 2 \cdot \frac12 \cdot \cos \angle BGD} = \frac{\sqrt{13}}{2}, \quad CD = \frac{\sqrt{13}}{2}\]

By the Law of Cosine in $\triangle ACD$

\[AD^2 = AC^2 + CD^2 - 2 \cdot AC \cdot CD \cdot \cos \angle C\]

\[\cos \angle C = \frac{ AC^2 + CD^2 - AD^2 }{ 2 \cdot AC \cdot CD } = \frac{ 2^2 + \left( \frac{\sqrt{13}}{2} \right)^2 - \left( \frac{3}{2} \right)^2 }{ 2 \cdot 2 \cdot \frac{\sqrt{13}}{2} } = \frac{ 5 \sqrt{13} }{26}\]

\[5 + 13 + 26 = \boxed{\textbf{(A) }44}\]

~isabelchen

Solution 4 (Barycentric Coordinates)

Using reference triangle $\triangle AGE$, we can let \[A=(1,0,0),G=(0,1,0),E=(0,0,1),C=(-1,0,2),D=(-\tfrac{1}{2},\tfrac{3}{2},0),B=(0,3,-2).\] If we move $A,B,C$ each over by $(1,0,-2)$, leaving $\angle C$ unchanged, we have \[A=(2,0,-2),B=(1,3,-4),C=(0,0,0).\] The angle $\theta$ between vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ satisfies \[\cos\theta=\frac{(2)(1)+(0)(3)+(-2)(-4)}{\sqrt{\left[2^{2}+0^{2}+(-2)^{2}\right]\left[1^{2}+3^{2}+(-4)^{2}\right]}}=\frac{10}{\sqrt{8\cdot 26}}=\frac{10}{4\sqrt{13}}=\frac{5\sqrt{13}}{26},\] giving the answer, $5+13+26=\boxed{\textbf{(A)}~44}$.

~r00tsOfUnity

Solution 5 (Coordinate Bash)

Let \(A\) be at \((0,0)\), with \(E\) at \((1,0)\) and \(G\) at \((\tfrac{1}{2},\tfrac{\sqrt{3}}{2})\). Because \(\triangle AGE\) is equilateral, the equation of line \(AD\) is \(y=x\sqrt{3}\), and the equation for \(BE\) is \(y=-x\sqrt{3}+\sqrt{3}\). Because \(BE\) is a median, we know that \(C\) is at \((2,0)\).

Therefore, the equation for line \(BC\) is \(y=m(x-2)\), where \(m\) is the slope of the line. Since \(AD\) is also a median, \(D\) is the midpoint of \(\overline{BC}\).

To find \(m\), solve for the coordinates of \(D\) in terms of \(m\):

\[x\sqrt{3}=mx-2m\] \[x(m-\sqrt{3})=2m\] \[x=\frac{2m}{m-\sqrt{3}}\]

The \(y\)-value of \(D\) is then \(y=x\sqrt{3}\):

\[y=\frac{2m\sqrt{3}}{m-\sqrt{3}}\]

Because \(D\) is the midpoint of \(\overline{BC}\) and \(y_C=0\), the \(y\)-value of \(B\) is double that of \(D\):

\[y_B=\frac{4m\sqrt{3}}{m-\sqrt{3}}\]

Now use the equation for line \(BE\) to express \(B\):

\[mx-2m=-x\sqrt{3}+\sqrt{3}\] \[x(m+\sqrt{3})=2m+\sqrt{3}\] \[x_B=\frac{2m+\sqrt{3}}{m+\sqrt{3}}\]

Plug this back into \(BE\) to find \(y_B\):

\[y_B=\frac{-2m-\sqrt{3}}{m+\sqrt{3}}+\sqrt{3}=\frac{-m\sqrt{3}}{m+\sqrt{3}}\]

Equating the two expressions for \(y_B\):

\[-\frac{m\sqrt{3}}{m+\sqrt{3}}=\frac{4m\sqrt{3}}{m-\sqrt{3}}\] \[-\sqrt{3}m^2+3m=4\sqrt{3}m^2+12m\] \[5\sqrt{3}m^2+9m=0\] \[m(5m\sqrt{3}+9)=0\]

Since \(m\neq0\), we have

\[m=-\frac{3\sqrt{3}}{5}.\]

Now find \(B\):

\[-x\sqrt{3}+\sqrt{3}=-\frac{3\sqrt{3}}{5}x+\frac{6\sqrt{3}}{5}\] \[\frac{-2\sqrt{3}}{5}x=\frac{\sqrt{3}}{5}\] \[x_B=-\frac{1}{2}\]

Plug this into \(BE\) to get

\[y_B=\frac{3\sqrt{3}}{2}.\]

Since we’re asked to find \(\cos\angle C\), extend \(\overline{AC}\) to \(F\) where \(F=(-\frac{1}{2},0)\), forming right triangle \(\triangle BFC\). Using the Pythagorean theorem for \(\overline{BC}\):

\[\left(\frac{5}{2}\right)^2+\left(\frac{3\sqrt{3}}{2}\right)^2=BC^2,\quad BC=\sqrt{13}.\]

Thus

\[\cos\angle C=\frac{\frac{5}{2}}{\sqrt{13}}=\frac{5\sqrt{13}}{26}.\]

The problem asks for \(m+n+p\), so \(5+13+26=\boxed{44}\).

~Voidling

Video Solution by MOP 2024

https://youtu.be/QNjvpYI1V5g

~r00tsOfUnity

Video Solution (Just 3 min!)

https://youtu.be/Q54sH65AJa4

~Education, the Study of Everything

Video Solution(Length & Angle Chasing)

https://youtu.be/JVDlHCSPF6k

~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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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