Difference between revisions of "2007 AMC 12B Problems/Problem 20"

(New page: ==Problem== The parallelogram bounded by the lines <math>y=ax+c</math>, <math>y=ax+d</math>, <math>y=bx+c</math>, and <math>y=bx+d</math> has area <math>18</math>. The parallelogram bounde...)
 
m (See also)
Line 8: Line 8:
  
 
==See also==
 
==See also==
 +
{{AMC12 box|year=2007|ab=B|num-b=21|num-a=23}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 09:57, 15 October 2008

Problem

The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?

$\mathrm {(A)} 13\qquad \mathrm {(B)} 14\qquad \mathrm {(C)} 15\qquad \mathrm {(D)} 16\qquad \mathrm {(E)} 17$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

2007 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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