2022 AMC 8 Problems/Problem 1
Contents
Problem
The Math Team designed a hamburger shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
usepackage("mathptmx");
defaultpen(linewidth(0.5));
size(5cm);
defaultpen(fontsize(14pt));
label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7));
label("$\textbf{Team}$", (2.1,3)--(3.9,3));
filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5);
draw((0,0)--(6,0), gray);
draw((0,1)--(6,1), gray);
draw((0,2)--(6,2), gray);
draw((0,3)--(6,3), gray);
draw((0,4)--(6,4), gray);
draw((0,5)--(6,5), gray);
draw((0,6)--(6,6), gray);
draw((0,0)--(0,6), gray);
draw((1,0)--(1,6), gray);
draw((2,0)--(2,6), gray);
draw((3,0)--(3,6), gray);
draw((4,0)--(4,6), gray);
draw((5,0)--(5,6), gray);
draw((6,0)--(6,6), gray);
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Solution 1
Draw the following four lines as shown:
![[asy] usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); draw((3,4)--(4,3), red); draw((4,3)--(3,2), red); draw((3,2)--(2,3), red); draw((2,3)--(3,4), red); [/asy]](http://latex.artofproblemsolving.com/7/1/0/71090e9bb25f5c8e950f50e21b102edfdb73e10d.png)
We see these lines split the figure into five squares with side length 
. Thus, the area is 
.
~pog ~wamofan
Solution 2
We can apply Pick's Theorem: There are 
lattice points in the interior and 
lattice points on the boundary of the figure. As a result, the area is 
.
~MathFun1000
Solution 3
Notice that the area of the figure is equal to the area of the 
square subtracted by the triangles that are half the area of each square, which is 
. The total area of the triangles not in the figure is 
, so the answer is 
.
~hh99754539
Solution 4
Draw the following four lines as shown:
![[asy] usepackage("mathptmx"); defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); draw((2,4)--(4,4), red); draw((4,4)--(4,2), red); draw((4,2)--(2,2), red); draw((2,2)--(2,4), red); [/asy]](http://latex.artofproblemsolving.com/8/1/f/81f0aead6447f752fc17ac705810fc2bb14e9fe0.png)
The area of the big square is 
, and the area of each triangle is 
. There are of these triangles, so the total area of all the triangles is 
. Therefore, the area of the entire figure is 
.
~RocketScientist
Solution 5 (Shoelace Theorem)
The coordinates are 
Use the Shoelace Theorem to get 
.
Solution 6 (Quick)
If the triangles are rearranged such that the gaps are filled, there would be a by 
rectangle, and two by squares are present. Thus, the answer is .
~peelybonehead i personally agree
Video Solution
https://www.youtube.com/watch?v=Ij9pAy6tQSg ~Interstigation
Video Solution 2
~savannahsolver
Video Solution
~STEMbreezy
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
