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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}...")
 
(Added question, made the solution a bit clearer.)
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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>,
 +
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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 +
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{BE}{BC} = \frac{BD}{DA} = \frac{3}{5}</math>. Since <math>\triangle ABE</math> and <math>\triangle ABC</math> have the same height, <math>[ABE] = \frac{3}{5} \cdot [ABC]=\frac{3}{5}\cdot 10 = 6</math>, hence our answer is <math>\fbox{C}</math>

Revision as of 02:07, 9 July 2018

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

Clearly since $[DBEF] = [ABE]$ it follows that $[ADF] = [AFE]$. This implies that $AC \parallel DE$ and so $\frac{BE}{BC} = \frac{BD}{DA} = \frac{3}{5}$. Since $\triangle ABE$ and $\triangle ABC$ have the same height, $[ABE] = \frac{3}{5} \cdot [ABC]=\frac{3}{5}\cdot 10 = 6$, hence our answer is $\fbox{C}$

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