Difference between revisions of "2003 AMC 10B Problems/Problem 20"
m (→Solution 2) |
m (→Solution 2) |
||
| Line 51: | Line 51: | ||
<cmath>x=\frac{5}{3}</cmath> | <cmath>x=\frac{5}{3}</cmath> | ||
| − | We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math> | + | We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math> |
==See Also== | ==See Also== | ||
Revision as of 12:22, 3 January 2016
- The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.
Contents
Problem
In rectangle 
and 
. Points 
and 
are on 
so that 
and 
. Lines 
and 
intersect at 
. Find the area of 
.
![[asy] unitsize(8mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(5,3), D=(0,3); pair F=(1,3), G=(3,3); pair E=(5/3,5); draw(A--B--C--D--cycle); draw(A--E); draw(B--E); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$1$",midpoint(D--F),N); label("$2$",midpoint(G--C),N); label("$5$",midpoint(A--B),S); label("$3$",midpoint(A--D),W); [/asy]](http://latex.artofproblemsolving.com/1/a/c/1ac3fc6b18ce26a1ce9b9a1e0e9c2c73bed94a30.png)
Solution 1
because 
The ratio of 
to 
is 
since 
and 
from subtraction. If we let 
be the height of 
![\[\frac{2}{5} = \frac{h-3}{h}\]](http://latex.artofproblemsolving.com/1/4/0/140780e0cdc7959d758ae0976f1715fcc6ffbfa0.png)
![\[2h = 5h-15\]](http://latex.artofproblemsolving.com/d/1/8/d18ea1812352a00725cd0a4938c2adbc85eb9332.png)
![\[3h = 15\]](http://latex.artofproblemsolving.com/a/0/9/a094622098ec270ad387c750fdeb057742bb8196.png)
![\[h = 5\]](http://latex.artofproblemsolving.com/a/d/f/adf85c076b2be16b9a1042f4742a0572d100110d.png)
The height is 
so the area of is 
.
Solution 2
We can look at this diagram as if it were a coordinate plane with point 
being 
. This means that the equation of the line 
is 
and the equation of the line 
is 
. From this we can set of the follow equation to find the 
coordinate of point :
![\[3x=\frac{-3}{2}x+\frac{15}{2}\]](http://latex.artofproblemsolving.com/4/a/c/4ac3116a22f1f37425cc2781f4d32a183bbb294b.png)
![\[6x=-3x+15\]](http://latex.artofproblemsolving.com/d/c/a/dca5153dba12a953a9ce8c260b780f2ee3374b56.png)
![\[9x=15\]](http://latex.artofproblemsolving.com/c/9/9/c9904b9484433a55aead3c805324fdec0f7aed94.png)
![\[x=\frac{5}{3}\]](http://latex.artofproblemsolving.com/5/f/b/5fb92a0b1dfdfd971d296b733854fd122d30dcc3.png)
We can plug this into one of our original equations to find that the 
coordinate is , meaning the area of is
See Also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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 AMC 12 Problems and Solutions | |
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
