Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2019 AMC 8 Problems/Problem 24

Revision as of 23:42, 20 November 2019 by Frestho (talk | contribs) (→‎Solution 4 (Similar Triangles))


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

Solution 2 (Mass Points)

[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.28,2.39),dotstyle); label("$A$", (0.36,2.59), NE * labelscalefactor); dot((-2.8,-1.17),dotstyle); label("$B$", (-2.72,-0.97), NE * labelscalefactor); dot((3.78,-1.05),dotstyle); label("$C$", (3.86,-0.85), NE * labelscalefactor); dot((1.2887445398528459,1.3985482236874887),dotstyle); label("$D$", (1.36,1.59), NE * labelscalefactor); dot((-0.7199623188673492,-1.1320661821070033),dotstyle); label("$F$", (-0.64,-0.93), NE * labelscalefactor); dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); label("$E$", (-0.2,0.57), NE * labelscalefactor); label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.

First, we assign a mass of $2$ to point $A$. We figure out that $C$ has a mass of $1$ since $2\times1 = 1\times2$. Then, by adding $1+2 = 3$, we get that point $D$ has a mass of 3. By equality, point $B$ has a mass of 3 also.

Now, we add $3+3 = 6$ for point $E$ and $3+1 = 4$ for point $F$.

Now, $BF$ is a common base for triangles $ABF$ and $EBF$, so we figure out that the ratios of the areas is the ratios of the heights which is $\frac{AE}{EF} = 2:1$. So, $EBF$'s area is one third the area of $ABF$, and we know the area of $ABF$ is $\frac{1}{4}$ the area of $ABC$ since they have the same heights but different bases.

So we get the area of $EBF$ as $\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}(30)$ -Brudder Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of $EBF$ over the product of the mass points of $ABC$ which is $\frac{2\times3\times1}{3\times6\times4}\times360$ which also yields $\boxed{B}$ -Brudder

Solution 3

$\frac{BF}{FC}$ is equal to $\frac{The area of triangle ABE}{The area of triangle ACE}$. The area of triangle $ABE$ is equal to $60$ because it is equal to on half of the area of triangle $ABD$, which is equal to one third of the area of triangle $ABC$, which is $360$. The area of triangle $ACE$ is the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\frac{BF}{FC}$ is equal to $\frac{60}{180}$=$\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\boxed{(B) 30}$. ~~SmileKat32

Solution 4 (Similar Triangles)

Extend $\overline{BD}$ to $G$ such that $\overline{AG} \parallel \overline{BC}$ as shown: [asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SE); label("$F$", F, S); label("$G$", G, ENE); [/asy] Then $\triangle ADG \sim \triangle CDB$ and $\triangle AEG \sim \triangle FEB$. Since $AD = 2CD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$.

Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}$.

[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SE); label("$F$", F, S); label("$G$", G, ENE); label("$60$", (A+E+D)/3); label("$60$", (A+E+B)/3); label("$60$", (A+G+D)/3); label("$30$", (B+E+F)/3); [/asy] (Credit to MP8148 for the idea)

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_8_Problems/Problem_24&oldid=111882"