Difference between revisions of "2019 AMC 8 Problems/Problem 24"

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==Problem==
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==Problem 24==
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In triangle <math>ABC</math>, point <math>D</math> divides side <math>\overline{AC}</math> s that <math>AD</math>:<math>DC=1</math>:<math>2</math>. Let <math>E</math> be the midpoint of <math>\overline{BD}</math> and left <math>F</math> be the point of intersection of line <math>BC</math> and line <math>AE</math>. Given that the area of <math>\triangle ABC</math> is <math>360</math>, what is the area of <math>\triangle EBF</math>?
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<asy>
 
<asy>
 
unitsize(2cm);
 
unitsize(2cm);
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label("$F$",FF,S);
 
label("$F$",FF,S);
 
</asy>
 
</asy>
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<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math>
  
 
==Solution 1==
 
==Solution 1==
==Solution 2==
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{{AMC8 box|year=2019|num-b=23|num-a=25}}
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==See Also==
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{{AMC8 box|year=2019|num-b=19|num-a=21}}
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{{MAA Notice}}

Revision as of 16:37, 20 November 2019


Problem 24

In triangle $ABC$, point $D$ divides side $\overline{AC}$ s that $AD$:$DC=1$:$2$. Let $E$ be the midpoint of $\overline{BD}$ and left $F$ be the point of intersection of line $BC$ and line $AE$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?

[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0);  A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF);  draw(B--DD);dot(A);  label("$A$",A,N); dot(B);  label("$B$", B,SW);dot(C);  label("$C$",C,SE); dot(DD);  label("$D$",DD,NE); dot(EE);  label("$E$",EE,NW); dot(FF);  label("$F$",FF,S); [/asy]


$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

Solution 1

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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 AJHSME/AMC 8 Problems and Solutions

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