Difference between revisions of "2002 AMC 8 Problems/Problem 15"
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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>. | 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>. | ||
| − | ==Solution 2(Under 15 seconds) == | + | ==Solution 2 (Under 15 seconds) == |
Pick's Theorem: | Pick's Theorem: | ||
Revision as of 23:24, 7 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]](http://latex.artofproblemsolving.com/6/b/8/6b83ac33b3e3e61ab2ca2c1c80dcbd8766f52066.png)
Solution 1
Each polygon can be partitioned into unit squares and right triangles with sidelength 
. Count the number of boxes enclosed by each polygon, with the unit square being , and the triangle being being 
. 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 
.
Solution 2 (Under 15 seconds)
Pick's Theorem:
where 
is the number of lattice points in the interior and 
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:
So now just look at the polygon with the most lattice points on the boundary, and that is figure 
, with 13 lattice points on the boundary.
So the polygon with the largest area is .
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
