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

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

Revision as of 18:43, 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

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=$\boxed{(B) 30}$~heeeeeeeheeeee

See Also

2019 AMC 8 (Problems • Answer Key • Resources)
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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