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Difference between revisions of "2007 AMC 8 Problems/Problem 23"

(Added another direct solution involving the area of a kite)
(Previous solution using Pick's Theorem was bogus and "coincidentally" gave the correct answer. Adding a solution properly uses Pick's Theorem.)
Line 16: Line 16:
 
The area of the square around the pinwheel is 25. The area of the pinwheel is equal to <math>\text{the square } - \text{ the white space.}</math> 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 <math>25-(15+4)</math> which is <math>\boxed{\textbf{(B) 6}}</math>
 
The area of the square around the pinwheel is 25. The area of the pinwheel is equal to <math>\text{the square } - \text{ the white space.}</math> 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 <math>25-(15+4)</math> which is <math>\boxed{\textbf{(B) 6}}</math>
  
−
== Solution 2 ==
+
== Solution 2 (Pick's Theorem) ==
  
−
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>
+
We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.
 +
 
 +
We can remedy this be pretending the figure is twice as big:
 +
 
 +
<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) {
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draw((i,0)--(i,5));
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draw((0,i)--(5,i));
 +
}
 +
for(i=0; i<5; i=i+1) {
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draw((i+.5,0)--(i+.5,5), rgb(0.6,0.6,0.6));
 +
draw((0,i+.5)--(5,i+.5), rgb(0.6,0.6,0.6));
 +
}
 +
</asy>
 +
 
 +
Now we can safely use Pick's Theorem:
 +
 
 +
<cmath>A=\frac{b}{2}+i-1=\frac{6}{2}+4-1=6</cmath>
 +
 
 +
However since we scaled the figure's dimensions by <math>2</math>, we scaled its area by <math>4</math> (since the area of similar shapes scales quadratically with the scaling factor). Therefore the area of each kite is <math>\frac{6}{4}</math> and the area of all four kites combined is <math>\boxed{\textbf{(B) 6}}</math>.
 +
 
 +
~ proloto
  
 
== Solution 3 (area of a kite)==
 
== Solution 3 (area of a kite)==

Revision as of 08:15, 28 July 2023

Problem

What is the area of the shaded pinwheel shown in the $5 \times 5$ 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]

$\textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12$

Solution

The area of the square around the pinwheel is 25. The area of the pinwheel is equal to $\text{the square } - \text{ the white space.}$ 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 $25-(15+4)$ which is $\boxed{\textbf{(B) 6}}$

Solution 2 (Pick's Theorem)

We'd like to use Pick's Theorem on one of the kites, except it doesn't immediately apply since there is a single vertex (in the middle of the diagram) of each kite that does not lie on a lattice point.

We can remedy this be pretending the figure is twice as big:

[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)); } for(i=0; i<5; i=i+1) { draw((i+.5,0)--(i+.5,5), rgb(0.6,0.6,0.6)); draw((0,i+.5)--(5,i+.5), rgb(0.6,0.6,0.6)); } [/asy]

Now we can safely use Pick's Theorem:

\[A=\frac{b}{2}+i-1=\frac{6}{2}+4-1=6\]

However since we scaled the figure's dimensions by $2$, we scaled its area by $4$ (since the area of similar shapes scales quadratically with the scaling factor). Therefore the area of each kite is $\frac{6}{4}$ and the area of all four kites combined is $\boxed{\textbf{(B) 6}}$.

~ proloto

Solution 3 (area of a kite)

The area of any kite (concave OR convex) with diagonals $p$, $q$ is $\frac{1}{2}pq$. Let $p$ be the smaller diagonal and $q$ be the longer diagonal. Then by Pythagorean Theorem $p=\sqrt{2}$. Similarly, $q$ is $\sqrt{2}$ less than half of the diagonal of the $5 \times 5$ grid, or $q=\frac{5\sqrt{2}}{2}-\sqrt{2}=\frac{3\sqrt{2}}{2}$. Therefore the area of the four kites is just:

\[A=4\cdot\frac{1}{2}pq=4\cdot\frac{1}{2}\cdot\sqrt{2}\cdot\frac{3\sqrt{2}}{2}=\boxed{\textbf{(B) 6}}\]

~ 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

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