Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created page, added problem, 2 solutions, navigational box)
 
m (→‎Solution 1)
 
(15 intermediate revisions by 6 users not shown)
Line 1: Line 1:
−
==Problem 6==
+
==Problem==
  
 
Dots are spaced one unit apart, horizontally and vertically.  The number of square units enclosed by the polygon is
 
Dots are spaced one unit apart, horizontally and vertically.  The number of square units enclosed by the polygon is
Line 16: Line 16:
 
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
 
<math>\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
  
−
==Solution 1==
+
== Solutions ==
 +
===Solution 1===
 +
We could count the area contributed by each square on the <math>3 \times 3</math> grid:
  
−
By inspection, you can notice that the triangle on the top row matches the hole in the bottom row.
+
Top-left: <math>0</math>
  
−
This creates a <math>2\times3</math> box, which has area <math>2\times3=\boxed{6}</math>
+
Top: Triangle with area <math>\frac{1}{2}</math>
  
−
==Solution 2==
+
Top-right: <math>0</math>
  
−
We could take the area of each square.
+
Left: Square with area <math>1</math>
  
−
Top-left: <math>0</math>
 
−
Top: Triangle with area <math>\frac{1}{2}</math>
 
−
Top-right: <math>0</math>
 
−
Left: Square with area <math>1</math>
 
 
Center: Square with area <math>1</math>
 
Center: Square with area <math>1</math>
 +
 
Right: Square with area <math>1</math>
 
Right: Square with area <math>1</math>
 +
 
Bottom-left: Square with area <math>1</math>
 
Bottom-left: Square with area <math>1</math>
 +
 
Bottom: Triangle with area <math>\frac{1}{2}</math>
 
Bottom: Triangle with area <math>\frac{1}{2}</math>
 +
 
Bottom-right: Square with area <math>1</math>
 
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===
  
 +
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
  
−
== See also ==
+
== See Also ==
−
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+
{{AJHSME box|year=1998|num-b=5|num-a=7}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

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

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AJHSME_Problems/Problem_6&oldid=135885"