Difference between revisions of "1998 AJHSME Problems/Problem 6"

Latest revision as of 16:52, 26 October 2020

Problem

Dots are spaced one unit apart, horizontally and vertically. The number of square units enclosed by the polygon is

$[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]$

$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solutions

Solution 1

We could count the area contributed by each square on the $3 \times 3$ grid:

Top-left: $0$

Top: Triangle with area $\frac{1}{2}$

Top-right: $0$

Left: Square with area $1$

Center: Square with area $1$

Right: Square with area $1$

Bottom-left: Square with area $1$

Bottom: Triangle with area $\frac{1}{2}$

Bottom-right: Square with area $1$

Adding all of these together, we get $6$ which is the same as $\boxed{B}$

Solution 2

By Pick's Theorem, we get the formula, $A=I+\frac{b}{2}-1$ where $I$ is the number of lattice points in the interior and $b$ being the number of lattice points on the boundary. In this problem, we can see that $I=1$ and $B=12$ . Substituting gives us $A=1+\frac{12}{2}-1=6$ Thus, the answer is $\boxed{\text{(B) 6}}$

Solution 3

Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$ . This creates a $2*3$ rectangle, with a area of $6$ . The answer is $\boxed{\text{(B) 6}}$ ~sakshamsethi

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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.

@@ Line 18: / Line 18: @@
 == Solutions ==
 ===Solution 1===
-By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.
-This creates a <math>2\times3</math> box, which has area <math>2\times3=\boxed{6}</math>
-===Solution 2===
 We could count the area contributed by each square on the <math>3 \times 3</math> grid:
@@ Line 45: / Line 38: @@
 Bottom-right: Square with area <math>1</math>
-Adding all of these together, we get <math>\boxed{6}</math> or <math>\boxed{B}</math>
+Adding all of these together, we get <math>6</math> which is the same as <math>\boxed{B}</math>
+===Solution 2===
+By Pick's Theorem, we get the formula, <math>A=I+\frac{b}{2}-1</math> where <math>I</math> is the number of lattice points in the interior and <math>b</math> being the number of lattice points on the boundary. In this problem, we can see that <math>I=1</math> and <math>B=12</math>. Substituting gives us <math>A=1+\frac{12}{2}-1=6</math> Thus, the answer is <math>\boxed{\text{(B) 6}}</math>
 ===Solution 3===
-By [[Pick's Theorem]], we get the formula, <math>A=I+\frac{b}{2}-1</math> where <math>I</math> is the number of lattice points in the interior and <math>b</math> being the number of lattice points on the boundary. In this problem, we can see that <math>I=1</math> and <math>B=12</math>. Substituting gives us <math>A=1+\frac{12}{2}-1=6</math> Thus, the answer is <math>\boxed{\text{(B) 6}}</math>
+Notice that the extra triangle on the top with area <math>1</math> can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area <math>1</math>. This creates a <math>2*3</math> rectangle, with a area of <math>6</math>. The answer is <math>\boxed{\text{(B) 6}}</math>
+~sakshamsethi
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