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Difference between revisions of "1983 AHSME Problems/Problem 28"

(Created page with "Clearly since <math>[DBEF] = [ABE]</math> it follows that <math>[ADF] = [AFE]</math>. This implies that <math>AC \parallel DE</math> and so <math>\frac{AD}{DB} = \frac{CE}{EB}...")
 
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Clearly since <math>[DBEF] = [ABE]</math> it follows that <math>[ADF] = [AFE]</math>. This implies that <math>AC \parallel DE</math> and so <math>\frac{AD}{DB} = \frac{CE}{EB} = \frac{2}{3}</math>. Thus <math>[AEB] = \frac{3}{2+3} \cdot 10 = \boxed{6}</math>
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== Problem 28 ==
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Triangle <math>\triangle ABC</math> in the figure has area <math>10</math>. Points <math>D, E</math> and <math>F</math>, all distinct from <math>A, B</math> and <math>C</math>,
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are on sides <math>AB, BC</math> and <math>CA</math> respectively, and <math>AD = 2, DB = 3</math>. If triangle <math>\triangle ABE</math> and quadrilateral <math>DBEF</math>
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have equal areas, then that area is
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<asy>
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defaultpen(linewidth(0.7)+fontsize(10));
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pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B);
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draw(A--B--C--A--E--F--D);
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pair point=incenter(A,B,C);
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label("$A$", A, dir(point--A));
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label("$B$", B, dir(point--B));
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label("$C$", C, dir(point--C));
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label("$D$", D, dir(point--D));
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label("$E$", E, dir(point--E));
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label("$F$", F, dir(point--F));
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label("$2$", (2,0), S);
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label("$3$", (7,0), S);</asy>
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<math>\textbf{(A)}\ 4\qquad
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\textbf{(B)}\ 5\qquad
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\textbf{(C)}\ 6\qquad
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\textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad
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\textbf{(E)}\ \text{not uniquely determined} </math>   
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== Solution ==
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Let <math>G</math> be the intersection point of <math>AE</math> and <math>DF</math>. Since <math>[DBEF] = [ABE]</math>, we have <math>[DBEG] + [EFG] = [DBEG] + [ADG] </math>, i.e. <math>[EFG] = [ADG]</math>. It therefore follows that <math>[ADG] + [AGF] = [EFG] + [AGF]</math>, so <math>[ADF] = [AFE]</math>. Now, taking <math>AF</math> as the base of both <math>\triangle ADF</math> and <math>\triangle AFE</math>, and using the fact that triangles with the same base and same perpendicular height have the same area, we deduce that the perpendicular distance from <math>D</math> to <math>AF</math> is the same as the perpendicular distance from <math>E</math> to <math>AF</math>. This in turn implies that <math>AF \parallel DE</math>, and so as <math>A</math>, <math>F</math>, and <math>C</math> are collinear, <math>AC \parallel DE</math>. Thus <math>\triangle DBE \sim \triangle ABC</math>, so <math>\frac{BE}{BC} = \frac{BD}{BA} = \frac{3}{5}</math>. Since <math>\triangle ABE</math> and <math>\triangle ABC</math> have the same perpendicular height (taking <math>AB</math> as the base), <math>[ABE] = \frac{3}{5} \cdot [ABC]=\frac{3}{5}\cdot 10 = 6</math>, and hence the answer is <math>\boxed{\textbf{(C)}}</math>.
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==See Also==
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{{AHSME box|year=1983|num-b=27|num-a=29}}
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{{MAA Notice}}

Latest revision as of 00:11, 20 February 2019

Problem 28

Triangle $\triangle ABC$ in the figure has area $10$. Points $D, E$ and $F$, all distinct from $A, B$ and $C$, are on sides $AB, BC$ and $CA$ respectively, and $AD = 2, DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$ have equal areas, then that area is

[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B); draw(A--B--C--A--E--F--D); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$2$", (2,0), S); label("$3$", (7,0), S);[/asy]

$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad \textbf{(E)}\ \text{not uniquely determined}$

Solution

Let $G$ be the intersection point of $AE$ and $DF$. Since $[DBEF] = [ABE]$, we have $[DBEG] + [EFG] = [DBEG] + [ADG]$, i.e. $[EFG] = [ADG]$. It therefore follows that $[ADG] + [AGF] = [EFG] + [AGF]$, so $[ADF] = [AFE]$. Now, taking $AF$ as the base of both $\triangle ADF$ and $\triangle AFE$, and using the fact that triangles with the same base and same perpendicular height have the same area, we deduce that the perpendicular distance from $D$ to $AF$ is the same as the perpendicular distance from $E$ to $AF$. This in turn implies that $AF \parallel DE$, and so as $A$, $F$, and $C$ are collinear, $AC \parallel DE$. Thus $\triangle DBE \sim \triangle ABC$, so $\frac{BE}{BC} = \frac{BD}{BA} = \frac{3}{5}$. Since $\triangle ABE$ and $\triangle ABC$ have the same perpendicular height (taking $AB$ as the base), $[ABE] = \frac{3}{5} \cdot [ABC]=\frac{3}{5}\cdot 10 = 6$, and hence the answer is $\boxed{\textbf{(C)}}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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