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Difference between revisions of "2002 AMC 8 Problems/Problem 15"

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==Solution 1==
 
==Solution 1==
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Each polygon can be partitioned into unit squares and right triangles with sidelength <math>1</math>. Count the number of boxes enclosed by each polygon, with the unit square being <math>1</math>, and the triangle being being <math>.5</math>. A has 5, B has 5, C has 5, D has 4.5, and E has 5.5. Therefore, the polygon with the largest area is <math>\boxed{\textbf{(E)}\ \text{E}}</math>.
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Each polygon can be partitioned into unit squares and right triangles with sidelength <math>1</math>. Count the number of boxes enclosed by each polygon, with the unit square being <math>1</math>, and the triangle being <math>.5</math>. A has 5, B has 5, C has 5, D has 4.5, and E has 5.5. Therefore, the polygon with the largest area is <math>\boxed{\textbf{(E)}\ \text{E}}</math>.
  
 
==Solution 2 (Under 15 seconds) ==
 
==Solution 2 (Under 15 seconds) ==

Latest revision as of 13:57, 24 December 2024

Problem

Which of the following polygons has the largest area?

[asy] size(330); int i,j,k; for(i=0;i<5; i=i+1) { for(j=0;j<5;j=j+1) { for(k=0;k<5;k=k+1) { dot((6i+j, k)); }}} draw((0,0)--(4,0)--(3,1)--(3,3)--(2,3)--(2,1)--(1,1)--cycle); draw(shift(6,0)*((0,0)--(4,0)--(4,1)--(3,1)--(3,2)--(2,1)--(1,1)--(0,2)--cycle)); draw(shift(12,0)*((0,1)--(1,0)--(3,2)--(3,3)--(1,1)--(1,3)--(0,4)--cycle)); draw(shift(18,0)*((0,1)--(2,1)--(3,0)--(3,3)--(2,2)--(1,3)--(1,2)--(0,2)--cycle)); draw(shift(24,0)*((1,0)--(2,1)--(2,3)--(3,2)--(3,4)--(0,4)--(1,3)--cycle)); label("$A$", (0*6+2, 0), S); label("$B$", (1*6+2, 0), S); label("$C$", (2*6+2, 0), S); label("$D$", (3*6+2, 0), S); label("$E$", (4*6+2, 0), S);[/asy]


$\textbf{(A)}\text{A}\qquad\textbf{(B)}\ \text{B}\qquad\textbf{(C)}\ \text{C}\qquad\textbf{(D)}\ \text{D}\qquad\textbf{(E)}\ \text{E}$

Solution 1

Each polygon can be partitioned into unit squares and right triangles with sidelength $1$. Count the number of boxes enclosed by each polygon, with the unit square being $1$, and the triangle being $.5$. A has 5, B has 5, C has 5, D has 4.5, and E has 5.5. Therefore, the polygon with the largest area is $\boxed{\textbf{(E)}\ \text{E}}$.

Solution 2 (Under 15 seconds)

Pick's Theorem:

$A = I + \frac{1}{2}B - 1$

where $I$ is the number of lattice points in the interior and $B$ being the number of lattice points on the boundary.

Ok, so we want to find the polygon with the largest area, first notice that not a single polygon has a lattice point in the interior. Now look at the formula:

$A = \frac{1}{2}B - 1$

So now just look at the polygon with the most lattice points on the boundary, and that is figure $\text{E}$, with 13 lattice points on the boundary. So the polygon with the largest area is $\boxed{\textbf{(E)}\ \text{E}}$.

~blankbox

See Also

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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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