Difference between revisions of "2007 AMC 8 Problems/Problem 23"
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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 <math>\boxed{\textbf{(B) 6}}</math> | 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 <math>\boxed{\textbf{(B) 6}}</math> | ||
| + | |||
| + | == Solution 3 (area of a kite)== | ||
| + | |||
| + | The area of any kite (concave OR convex) with diagonals <math>p</math>, <math>q</math> is <math>\frac{1}{2}pq</math>. Let <math>p</math> be the smaller diagonal and <math>q</math> be the longer diagonal. Then by Pythagorean Theorem <math>p=\sqrt{2}</math>. Similarly, <math>q</math> is <math>\sqrt{2}</math> less than half of the diagonal of the <math>5 \times 5</math> grid, or <math>q=\frac{5\sqrt{2}}{2}-\sqrt{2}=\frac{3\sqrt{2}}{2}</math>. Therefore the area of the four kites is just: | ||
| + | |||
| + | <cmath>A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{\textbf{(B) 6}}</cmath> | ||
| + | |||
| + | ~ proloto | ||
== Video Solution == | == Video Solution == | ||
Revision as of 13:31, 13 July 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
Solution 3 (area of a kite)
The area of any kite (concave OR convex) with diagonals 
, 
is 
. Let be the smaller diagonal and be the longer diagonal. Then by Pythagorean Theorem 
. Similarly, is 
less than half of the diagonal of the grid, or 
. Therefore the area of the four kites is just:
![\[A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{\textbf{(B) 6}}\]](http://latex.artofproblemsolving.com/7/c/0/7c0d01b9c96a039004cc322f7023b155de80a051.png)
~ proloto
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. 
