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Difference between revisions of "2008 AMC 8 Problems/Problem 23"

(→‎Solution 2~Mr.BigBrain_AoPS)
(→‎Solution 2~Mr.BigBrain_AoPS)
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==Solution 2~Mr.BigBrain_AoPS==
 
==Solution 2~Mr.BigBrain_AoPS==
 
Say that <math>\overline{FE}</math> has length <math>x</math>, and that from there we can infer that <math>\overline{AF} = 2x</math>. We also know that <math>\overline{ED} = x</math>, and that <math>\overline{DC} = 2x</math>. The area of triangle <math>BFD</math> is the square's area subtracted from the area of the excess triangles, which is simply these equations:
 
Say that <math>\overline{FE}</math> has length <math>x</math>, and that from there we can infer that <math>\overline{AF} = 2x</math>. We also know that <math>\overline{ED} = x</math>, and that <math>\overline{DC} = 2x</math>. The area of triangle <math>BFD</math> is the square's area subtracted from the area of the excess triangles, which is simply these equations:
−
\begin{align*}
+
<cmath>9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2) </cmath>
−
9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2) \\
+
<cmath>9x^2 - 6.5x^2</cmath>
−
9x^2 - 6.5x^2\\
+
<cmath>2.5x^2</cmath>
−
2.5x^2
 
−
\end{align*}
 
 
Thus, the area of the triangle is <math>2.5x^2</math>. We can now put the ratio of triangle <math>BFD</math>'s area to the area of the square <math>ABCE</math> as a fraction. We have:
 
Thus, the area of the triangle is <math>2.5x^2</math>. We can now put the ratio of triangle <math>BFD</math>'s area to the area of the square <math>ABCE</math> as a fraction. We have:
−
\begin{align*}
+
<cmath>\dfrac{2.5x^2}{9x^2}</cmath>
−
\dfrac{2.5x^2}{9x^2} \\
+
<cmath>\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}} </cmath>
−
\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}} \\
+
<cmath>\dfrac{2.5}{9}</cmath>
−
\dfrac{2.5}{9} \\
+
<cmath>\dfrac{5}{18} </cmath>
−
\dfrac{5}{18} \end{align*}
 
 
Thus, our answer is <math>\boxed{C}</math>, <math>\boxed{\dfrac{5}{18}}</math>.
 
Thus, our answer is <math>\boxed{C}</math>, <math>\boxed{\dfrac{5}{18}}</math>.
  

Revision as of 15:44, 20 November 2022

Problem

In square $ABCE$, $AF=2FE$ and $CD=2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$? [asy] size((100)); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(9,9)--(0,3)--cycle); dot((3,0)); dot((0,3)); dot((9,9)); dot((0,0)); dot((9,0)); dot((0,9)); label("$A$", (0,9), NW); label("$B$", (9,9), NE); label("$C$", (9,0), SE); label("$D$", (3,0), S); label("$E$", (0,0), SW); label("$F$", (0,3), W); [/asy]

$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20}$

Solution

The area of $\triangle BFD$ is the area of square $ABCE$ subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be $6$.

[asy] size((100)); pair A=(0,9), B=(9,9), C=(9,0), D=(3,0), E=(0,0), F=(0,3); pair[] ps={A,B,C,D,E,F}; dot(ps); draw(A--B--C--E--cycle); draw(B--F--D--cycle); label("$A$",A, NW); label("$B$",B, NE); label("$C$",C, SE); label("$D$",D, S); label("$E$",E, SW); label("$F$",F, W); label("$6$",A--B,N); label("$6$",(10,4.5),E); label("$4$",D--C,S); label("$2$",E--D,S); label("$2$",E--F,W); label("$4$",F--A,W); [/asy]

The ratio of the area of $\triangle BFD$ to the area of $ABCE$ is

\[\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}\]

Solution 2~Mr.BigBrain_AoPS

Say that $\overline{FE}$ has length $x$, and that from there we can infer that $\overline{AF} = 2x$. We also know that $\overline{ED} = x$, and that $\overline{DC} = 2x$. The area of triangle $BFD$ is the square's area subtracted from the area of the excess triangles, which is simply these equations: \[9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2)\] \[9x^2 - 6.5x^2\] \[2.5x^2\] Thus, the area of the triangle is $2.5x^2$. We can now put the ratio of triangle $BFD$'s area to the area of the square $ABCE$ as a fraction. We have: \[\dfrac{2.5x^2}{9x^2}\] \[\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}}\] \[\dfrac{2.5}{9}\] \[\dfrac{5}{18}\] Thus, our answer is $\boxed{C}$, $\boxed{\dfrac{5}{18}}$.

See Also

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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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