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

(Previous solution using Pick's Theorem was bogus and "coincidentally" gave the correct answer. Adding a solution properly uses Pick's Theorem.)
 
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<math> \textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12 </math>
 
<math> \textbf{(A)}\: 4\qquad\textbf{(B)}\: 6\qquad\textbf{(C)}\: 8\qquad\textbf{(D)}\: 10\qquad\textbf{(E)}\: 12 </math>
  
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== Solution ==
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== Solution 1 ==
 
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>
  
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== Solution 2 (Pick's Theorem) ==
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== Solution 2 ==
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The pinwheel is composed of <math>8</math> congruent obtuse triangles whose base measures length <math>1</math> and height measures length <math>1.5</math>. Using the area formula for triangles, the pinwheel has an area of
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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.
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<cmath>8(\frac12\cdot1\cdot1.5)=8(0.75)=\boxed{\textbf{(B) 6}}.</cmath>
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We can remedy this be pretending the figure is twice as big:
 
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<asy>
 
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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);
 
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int i;
 
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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));
 
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}
 
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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));
 
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draw((0,i+.5)--(5,i+.5), rgb(0.6,0.6,0.6));
 
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}
 
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</asy>
 
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Now we can safely use Pick's Theorem:
 
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<cmath>A=\frac{b}{2}+i-1=\frac{6}{2}+4-1=6</cmath>
 
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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>.
 
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~ proloto
 
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== Solution 3 (area of a kite)==
 
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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:
 
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<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>
 
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~ proloto
 
  
 
== Video Solution ==
 
== Video Solution ==
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https://youtu.be/KOZBOvI9WTs -Happytwin
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https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin
  
 
==Video Solution by OmegaLearn==
 
==Video Solution by OmegaLearn==
Line 58: Line 27:
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/CGpR3VKeVV0
  
 
== 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}}

Latest revision as of 13:17, 29 October 2024

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 1

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

The pinwheel is composed of $8$ congruent obtuse triangles whose base measures length $1$ and height measures length $1.5$. Using the area formula for triangles, the pinwheel has an area of \[8(\frac12\cdot1\cdot1.5)=8(0.75)=\boxed{\textbf{(B) 6}}.\]

Video Solution

https://www.youtube.com/watch?v=dQw4w9WgXcQ -Happytwin

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=748

~ pi_is_3.14

Video Solution by WhyMath

https://youtu.be/CGpR3VKeVV0

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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