Difference between revisions of "2007 AMC 8 Problems/Problem 23"
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== Video Solution == | == Video Solution == | ||
https://youtu.be/KOZBOvI9WTs -Happytwin | https://youtu.be/KOZBOvI9WTs -Happytwin | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/abSgjn4Qs34?t=748 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== See Also == | == See Also == | ||
{{AMC8 box|year=2007|num-b=22|num-a=24}} | {{AMC8 box|year=2007|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:50, 3 January 2023
Contents
Problem
What is the area of the shaded pinwheel shown in the 
grid?
![[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]](http://latex.artofproblemsolving.com/b/b/1/bb1fcead3fb785c17349c43a31ed1e90bfc319e0.png)
Solution
The area of the square around the pinwheel is 25. The area of the pinwheel is equal to 
Each of the four triangles have a base of 3 units and a height of 2.5 units, and so their combined area is 15 units squared. Then the unshaded space consists of the four triangles with total area of 15, and there are four white corner squares. Therefore the area of the pinwheel is 
which is 
Solution 2
Notice that the pinwheel is symmetrical, thus we only have to find the area of one of the sections and multiply that by four. Using Pick’s Theorem, b/2+i-1, we can count 3 lattice points on the border, so b/2 is 1.5. The number of interior points is 1. So the area of one section of the pinwheel is 1.5. Multiplying that by 4 yields
Video Solution
https://youtu.be/KOZBOvI9WTs -Happytwin
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=748
~ pi_is_3.14
See Also
| 2007 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. 
