Difference between revisions of "2020 AMC 8 Problems/Problem 25"
(→Solution 3) |
m |
||
| Line 31: | Line 31: | ||
Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{(\text{A}) 651}.</math> | Since each pair of boxes has a sum of <math>3322</math> or <math>2020</math> and a difference of <math>S_2</math>, we see that the answer is <math>\dfrac{3322 - 2020}{2} = \dfrac{1302}{2} = \boxed{(\text{A}) 651}.</math> | ||
| − | -A_MatheMagician | + | -A_MatheMagician. Note this is just a more quicker way to do it to get \boxed{(\text{A}) 651}.$ |
| − | + | https://artofproblemsolving.com/community/my-aops | |
==See also== | ==See also== | ||
{{AMC8 box|year=2020|num-b=24|after=Last Problem}} | {{AMC8 box|year=2020|num-b=24|after=Last Problem}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:12, 19 November 2020
Rectangles 
and 
and squares 
and 
shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of 
in units?
![[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]](http://latex.artofproblemsolving.com/a/5/5/a55c518a1179d67520c6a03827b4ba25ca1413f2.png)
Contents
Solution 1
For each square 
, let the sidelength of this square be denoted by 
.
As the diagram shows, 
We subtract the second equation from the first, getting 
, and thus 
, so the answer is 
~icematrix, edits by starrynight7210
Solution 2
WLOG, assume that 
and 
. Let the sum of the lengths of 
and be 
and let the length of be 
. We have the system ![\[x+y =3322\]](http://latex.artofproblemsolving.com/c/e/8/ce8d1df0cd0bee0e757e4760e9bcf9a11c465a62.png)
![\[x-y=2020\]](http://latex.artofproblemsolving.com/d/a/a/daa50e1052cf0a74ed930564aee65db09b5f4fb3.png)
which we solve to find that 
.
-franzliszt
Solution 3
Since each pair of boxes has a sum of 
or 
and a difference of , we see that the answer is 
-A_MatheMagician. Note this is just a more quicker way to do it to get \boxed{(\text{A}) 651}.$ https://artofproblemsolving.com/community/my-aops
See also
| 2020 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 24 |
Followed by Last Problem | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
