2019 AMC 8 Problems/Problem 24

Problem 24

In triangle $\triangle ABC$ , point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$ . Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $\overline{BC}$ and line $\overline{AE}$ . 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 0 (middle-school knowledge)

We use the line-segment ratios to infer area ratios and height ratios.

Areas:

$AD:DC = 1:2 \implies AD:AC = 1:3 \implies [ABD] =\frac{[ABC]}{3} = 120$ .

$BE:BD = 1:2 \text{ (midpoint)} \implies [ABE] = \frac{[ABD]}{2} = \frac{120}{2} = 60$ .

Heights:

Let $h_A$ = height (of altitude) from $\overline{BC}$ to $A$ .

$AD:DC = 1:2 \implies CD:CA = 2:3 \implies \text{height } h_D$ from $\overline{BC}$ to $D$ is $\frac{2}{3}h_A$ .

$BE:BD = 1:2 \text{ (midpoint)} \implies \text{height } h_E$ from $\overline{BC}$ to $D$ is $\frac{1}{2} h_E = \frac{1}{2}(\frac{2}{3} h_A) = \frac{1}{3} h_A$ .

Conclusion:

$\frac{[EBF]} {[ABF]} = \frac{[EBF]} {[EBF] + [ABE]} = \frac{[EBF]} {[EBF]+60}$ , and also $\frac{[EBF]} {[ABF]} = \frac{h_E}{h_A} = \frac{1}{3}$ .

So, $\frac{[EBF]} {[EBF] + 60} = \frac{1}{3}$ , and thus, $[EBF] = \boxed{30}$ .

Solution 1

Draw $X$ on $\overline{AF}$ such that $\overline{XD}$ is parallel to $\overline{BC}$ .

Triangles $BEF$ and $EXD$ are similar, and since $BE = ED$ , they are also congruent, and so $XE=EF$ and $XD=BF$ .

$AD:AC = 3$ implies $\frac{AX}{AF} = 3 = \frac{FC}{XD} = \frac{FC}{BF}$ , so $BC=BF + 3BF = 4BF$ , $BF=\frac{BC}{4}$ .

Since $XE=EF$ , $AX = XE = EF$ , and since $AX + XE + EF = AF$ , all of these are equal to $\frac{AF}{3}$ , and so the altitude of triangle $BEF$ is equal to $\frac{1}{3}$ of the altitude of $ABC$ .

The area of $ABC$ is $360$ , so the area of $\triangle EBF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}$ .

Solution 2

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.28,2.39),dotstyle); label("$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][Mass_points]] at us.

The triangle we will consider is $\triangle ABC$ (obviously), and we will let $E$ be the center of mass, so that $D$ balances $A$ and $C$ (this is true since $E$ balances $B$ and $D$ , but $E$ also balances $A$ and $B$ and $C$ so $D$ balances $A$ and $C$ ), and $F$ balances $B$ and $C$ .

We know that $AD:CD=1:2$ and $D$ balances $A$ and $C$ so we assign $2$ to $A$ and $1$ to $C$ . Then, since $D$ balances $A$ and $C$ , we get $D = A + C = 2 + 1 = 3$ (by mass points addition).

Next, since $E$ balances $B$ and $D$ in a ratio of $BE:DE=1:1$ , we know that $B=D=3$ . Similarly, by mass points addition, $E=B+D=3+3=6$ .

Finally, $F$ balances $B$ and $C$ so $F=B+C=3+1=4$ . We can confirm we have done everything right by noting that $E$ balances $A$ and $F$ , so $E$ should equal $A+F$ , which it does.

Now that our points have weights, we can solve the problem. $BF:FC=1:3$ so $BF:BC=1:4$ so $[ABF]=\frac{1}{4}[ABC]=90$ . Also, $EF:EA=2:4=1:2$ so $EF:AF=1:3$ so $[EBF]=\frac{1}{3}[ABF]=30$ .

So we get the area of $EBF$ as $\boxed{\textbf{(B) }30}$ .

-Firebolt360 and 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 $\boxed{\textbf{(B) }30}$ .

-Brudder

Solution 3

$\frac{BF}{FC}$ is equal to $\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}$ . 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 $\boxed{\textbf{(B) }30}$ .

~~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 $CD = 2AD$ , 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)

Solution 5 (Area Ratios)

$[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SSE); label("$F$", F, S); label("$60$", (A+E+D)/3); label("$60$", (A+E+B)/3); label("$120$", (D+E+C)/3); label("$x$", (B+E+F)/3); label("$120-x$", (F+E+C)/3); [/asy]$ As before, we figure out the areas labeled in the diagram. Then, we note that $\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}$ Even simpler: $\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120}{240}$ Solving gives $x = \boxed{\textbf{(B) }30}$ . (Credit to scrabbler94 for the idea)

Solution 6 (Coordinate Bashing)

Let $ADB$ be a right triangle, and $BD=CD$

Let $A=(-2\sqrt{30}, 0)$

$B=(0, 4\sqrt{30})$

$C=(4\sqrt{30}, 0)$

$D=(0, 0)$

$E=(0, 2\sqrt{30})$

$F=(\sqrt{30}, 3\sqrt{30})$

The line $\overleftrightarrow{AE}$ can be described with the equation $y=x-2\sqrt{30}$

The line $\overleftrightarrow{BC}$ can be described with $x+y=4\sqrt{30}$

Solving, we get $x=3\sqrt{30}$ and $y=\sqrt{30}$

Now we can find $EF=BF=2\sqrt{15}$

$[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare$

-Trex4days

Solution 7

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */ /* draw figures */ draw(circle((0,0), 5), linewidth(2)); draw((-4,-3)--(4,3), linewidth(2)); draw((-4,-3)--(0,5), linewidth(2)); draw((0,5)--(4,3), linewidth(2)); draw((12,-1)--(-4,-3), linewidth(2)); draw((0,5)--(0,-5), linewidth(2)); draw((-4,-3)--(0,-5), linewidth(2)); draw((4,3)--(0,2.48), linewidth(2)); draw((4,3)--(12,-1), linewidth(2)); draw((-4,-3)--(4,3), linewidth(2)); /* dots and labels */ dot((0,0),dotstyle); label("E", (0.27,-0.24), NE * labelscalefactor); dot((-5,0),dotstyle); dot((-4,-3),dotstyle); label("B", (-4.45,-3.38), NE * labelscalefactor); dot((4,3),dotstyle); label("$D$", (4.15,3.2), NE * labelscalefactor); dot((0,5),dotstyle); label("A", (-0.09,5.26), NE * labelscalefactor); dot((12,-1),dotstyle); label("C", (12.23,-1.24), NE * labelscalefactor); dot((0,-5),dotstyle); label("$G$", (0.19,-4.82), NE * labelscalefactor); dot((0,2.48),dotstyle); label("I", (-0.33,2.2), NE * labelscalefactor); dot((0,0),dotstyle); label("E", (0.27,-0.24), NE * labelscalefactor); dot((0,-2.5),dotstyle); label("F", (0.23,-2.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $A[\Delta XYZ]$ = $\text{Area of Triangle XYZ}$

$A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240$

$A[\Delta ABE] = A[\Delta AED] = 60$ (the median divides the area of the triangle into two equal parts)

Construction: Draw a circumcircle around $\Delta ABD$ with $BD$ as is diameter. Extend $AF$ to $G$ such that it meets the circle at $G$ . Draw line $BG$ .

$A[\Delta ABD] = A[\Delta ABG] = 120$ (Since $\square ABGD$ is cyclic)

But $A[\Delta ABE]$ is common in both with an area of 60. So, $A[\Delta AED] = A[\Delta BEG]$ .

Therefore $A[\Delta AED] \cong A[\Delta BEG]$ (SAS Congruency Theorem).

In $\Delta AED$ , let $DI$ be the median of $\Delta AED$ ,

which means $A[\Delta AID] = 30 = A[\Delta EID]$ .

Rotate $\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$ . $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\frac{AE}{EF} = 2:1$ .

$AE$ is a radius and $EF$ is half of it implies $EF$ = $\frac{radius}{2}$ ,

which means $A[\Delta BEF] \cong A[\Delta DEI]$ .

Thus, $A[\Delta BEF] = \boxed{\textbf{(B) }30}$ .

~phoenixfire & flamewavelight

Solution 8

$[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("$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$",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label("$M$",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\overline{AD}$ and $\overline{CD}$ , we find the area of $\triangle ADB$ is $120$ and the area of $\triangle BDC$ is $240$ . Also using the fact that $E$ is the midpoint of $\overline{BD}$ , we know $\triangle ADE = \triangle ABE = 60$ . Let $M$ be a point such $\overline{EM}$ is parellel to $\overline{CD}$ . We immediatley know that $\triangle BEM \sim BDC$ by $2$ . Using that we can conclude $EM$ has ratio $1$ . Using $\triangle EFM \sim \triangle AFC$ , we get $EF:AE = 1:2$ . Therefore using the fact that $\triangle EBF$ is in $\triangle ABF$ , the area has ratio $\triangle BEF : \triangle ABE=1:2$ and we know $\triangle ABE$ has area $60$ so $\triangle BEF$ is $\boxed{\textbf{(B) }30}$ .

- fath2012

Solution 9 (Menelaus's Theorem)

$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("$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$",FF,S); [/asy]$ By Menelaus's Theorem on triangle $BCD$ , we have $\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.$ Therefore, $[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.$

Solution 10 (Graph Paper)

$[asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2); C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed); draw(A--B--C--cycle); draw(A--F); draw(B--D); dot(A); label("$A$",A,NE); dot(B); label("$B$",B,dir(180)); dot(C); label("$C$",C,SE); dot(D); label("$D$",D,dir(0)); dot(E); label("$E$",E,SE); dot(F); label("$F$",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.

As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.

As point $D$ splits line segment $\overline{AC}$ in a $1:2$ ratio, we draw $\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$ . By definition, Point $E$ splits line segment $\overline{BD}$ in a $1:1$ ratio, so we draw $\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$ , $1$ unit away from both.

We then draw line segments $\overline{AB}$ and $\overline{BC}$ . We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$ , we can easily see that triangle $EBF$ forms a right triangle occupying $\frac{1}{4}$ of a square unit of space.

The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus $\frac{1}{4}\div3=\frac{1}{12}$ , and since the area of triangle $ABC$ is $360$ , this means that the area of triangle $EBF$ is $\frac{1}{12}\times360=\boxed{\textbf{(B) }30}$ .

~emerald_block

Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones.

~i_equal_tan_90

Solution 11

$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label("$A$",A,N); label("$B$", B,SW); label("$C$",C,SE); label("$D$",DD,NE); label("$E$",EE,NW); label("$F$",FF,S); label("$G$",G,S); [/asy]$ We know that $AD = \dfrac{1}{3} AC$ , so $[ABD] = \dfrac{1}{3} [ABC] = 120$ . Using the same method, since $BE = \dfrac{1}{2} BD$ , $[ABE] = \dfrac{1}{2} [ABD] = 60$ . Next, we draw $G$ on $\overline{BC}$ such that $\overline{DG}$ is parallel to $\overline{AF}$ and create segment $DG$ . We then observe that $\triangle AFC \sim \triangle DGC$ , and since $AD:DC = 1:2$ , $FG:GC$ is also equal to $1:2$ . Similarly (no pun intended), $\triangle DBG \sim \triangle EBF$ , and since $BE:ED = 1:1$ , $BF:FG$ is also equal to $1:1$ . Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$ , or equivalently, $BF = \dfrac{1}{4} BC$ . Thus, $[BFA] = \dfrac{1}{4} [BCA] = 90$ . We already know that $[ABE] = 60$ , so the area of $\triangle EBF$ is $[BFA] - [ABE] = \boxed{\textbf{(B) }30}$ .

~i_equal_tan_90

Solution 12 (Fastest Solution if you have no time)

The picture is misleading. Assume that the triangle ABC is right.

Then, find two factors of $720$ that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near $720$ to use difference of squares, we find $24$ and $30$ as our numbers. Then, the coordinates of D are $(10,16)$ (note, A=0,0). E is then $(5,8)$ . Then the equation of the line AE is $-16x/5+24=y$ . Plugging in $y=0$ , we have $x=\dfrac{15}{2}$ . Now notice that we have both the height and the base of EBF.

Solving for the area, we have $(8)(15/2)(1/2)=30$ .

Solution 13

$AD : DC = 1:2$ , so $ADB$ has area $120$ and $CDB$ has area $240$ . $BE = ED$ so the area of $ABE$ is equal to the area of $ADE = 60$ . Draw $\overline{DG}$ parallel to $\overline{AF}$ .
Set area of BEF = $x$ . BEF is similar to BDG in ratio of 1:2
so area of BDG = $4x$ , area of EFDG= $3x$ , and area of CDG $=240-4x$ .
CDG is similar to CAF in ratio of 2:3 so area CDG = $4/9$ area CAF, and area AFDG= $5/4$ area CDG.
Thus, $60+3x=5/4(240-4x)$ and $x=30$ .

~EFrame

Solution 14 - Geometry & Algebra

$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label("$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$",FF,S); [/asy]$

We draw line $FD$ so that we can define a variable $x$ for the area of $\triangle BEF = \triangle DEF$ . Knowing that $\triangle ABE$ and $\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$ .

Since we have a rule where 2 triangles, ( $\triangle A$ which has base $a$ and vertex $c$ ), and ( $\triangle B$ which has Base $b$ and vertex $c$ )who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}$ . Substituting $x$ into the equation we get:

$\frac{x+60}{300-x} = \frac{2x}{240-2x}$ $(2x)(300-x) = (60+x)(240-x)$ $600-2x^2 = 14400 - 120x + 240x - 2x^2$ $480x = 14400$ and we now have that $\triangle BEF=30$ .

~ $\bold{\color{blue}{onionheadjr}}$

Solution 15 (Straightfoward & Simple Solution)

$[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SSE); label("$F$", F, S); label("$60$", (A+E+D)/3); label("$60$", (A+E+B)/3); label("$120$", (D+E+C)/3); label("$x$", (B+E+F)/3); label("$120-x$", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\triangle ABD=\frac{1}{3} \cdot 360 = 120.$

Similarly, $\triangle DBC = \frac{2}{3} \cdot 360 = 240.$

Now, since $E$ is a midpoint of $BD$ , $\triangle ABE = \triangle AED = 120 \div 2 = 60.$

We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\triangle BEC$ and $\triangle DEC$ share 2 sides.

We know that $\triangle BEC=\triangle DEC=240 \div 2 = 120$ since $E$ is a midpoint of $BD.$

Let's label $\triangle BEF$ $x$ . We know that $\triangle EFC$ is $120-x$ since $\triangle BEC = 120.$

Note that with this information now, we can deduct more things that are needed to finish the solution.

Note that $\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$

We want to find $x.$

This is a simple equation, and solving we get $x=\boxed{\textbf{(B)}30}.$

~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.

Solution 16

$[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("$A$", A, N); label("$B$", B, WSW); label("$C$", C, ESE); label("$D$", D, dir(0)*1.5); label("$E$", E, SSE); label("$F$", F, S); label("$60$", (A+E+D)/3); label("$60$", (A+E+B)/3); label("$120$", (D+E+C)/3); [/asy]$

Because $AD:DC=1:2$ and $E$ is the midpoint of $BD$ , we know that the areas of $ABE$ and $AED$ are $60$ and the areas of $DEC$ and $EBC$ are $120$ . $\frac{[EBF]}{[EFC]} = \frac{[ABF]}{[AFC]} = \frac{ [ABE]}{[AEC]} = \frac{60}{180}$ $[EBF] = \frac{120}{4} = \boxed{\textbf{(B) }30}$

Note

This question is extremely similar to 1971 AHSME Problem 26.

Video Solutions

https://www.youtube.com/watch?v=AY4mByrL8v0

Associated video

https://www.youtube.com/watch?v=DMNbExrK2oo

https://youtu.be/Ns34Jiq9ofc

—DSA_Catachu

https://www.youtube.com/watch?v=nm-Vj_fsXt4

- Happytwin (Another video solution)

https://www.youtube.com/watch?v=nyevg9w-CCI&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=6

~ MathEx

https://www.youtube.com/watch?v=m04K0Q2SNXY&t=1s

https://youtu.be/vZjPUW_ZupA

~savannahsolver

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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