Difference between revisions of "2008 AMC 8 Problems/Problem 23"
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<cmath>\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}</cmath> | <cmath>\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}</cmath> | ||
| − | ==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*} | \begin{align*} | ||
Revision as of 15:41, 20 November 2022
Problem
In square 
, 
and 
. What is the ratio of the area of 
to the area of square ?
![[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]](http://latex.artofproblemsolving.com/6/c/d/6cd0313c1b06e23b6ae03cfe53c292b0ac2371db.png)
Solution
The area of is the area of square subtracted by the the area of the three triangles around it. Arbitrarily assign the side length of the square to be 
.
![[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]](http://latex.artofproblemsolving.com/e/7/0/e70951e5f1b40046a0a06bfca96b4332908164d9.png)
The ratio of the area of to the area of is
![\[\frac{36-12-12-2}{36} = \frac{10}{36} = \boxed{\textbf{(C)}\ \frac{5}{18}}\]](http://latex.artofproblemsolving.com/9/c/8/9c8ba1a5d01139d2d5b5a9f885680737cb9d92f5.png)
Solution 2~Mr.BigBrain_AoPS
Say that 
has length 
, and that from there we can infer that 
. We also know that 
, and that 
. The area of triangle 
is the square's area subtracted from the area of the excess triangles, which is simply these equations:
\begin{align*}
9x^2 - (3x^2 + \dfrac{x}{2}^2 + 3x^2) \\
9x^2 - 6.5x^2\\
2.5x^2
\end{align*}
Thus, the area of the triangle is 
. We can now put the ratio of triangle 's area to the area of the square as a fraction. We have:
\begin{align*}
\dfrac{2.5x^2}{9x^2} \\
\dfrac{2.5\cancel{x^2}}{9\cancel{x^2}} \\
\dfrac{2.5}{9} \\
\dfrac{5}{18} \end{align*}
Thus, our answer is 
, 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
