2019 AMC 8 Problems/Problem 24

Revision as of 18:42, 20 November 2019 by Heeeeeeeheeeee (talk | contribs) (Solution 1)


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

Draw $X$ so that $XD$ is parallel to $BC$. That makes triangles $BEF$ and $EXD$ congruent since $BE$=$ED$. $FC$=3$XD$ so $BC$=4$BC$. Since $AF$=3$EF$( $XE$=$EF$ and $AX$=1/3$AF$, so $XE$=$EF$=1/3$AF$), the altitude of triangle $BEF$ is equal to 1/3 of the altitude of $ABC$. The area of $ABC$ is 360, so the area of $BEF$=1/3*1/4*360=$\boxed{(B) 30}$

See Also

2019 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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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