2022 AMC 8 Problems/Problem 1: Difference between revisions
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We see these lines split the figure into five squares with side length <math>\sqrt2</math>. Thus, the area is <math>5\cdot\left(\sqrt2\right)^2=5\cdot 2 = \boxed{\textbf{(A) } 10}</math>. | We see these lines split the figure into five squares with side length <math>\sqrt2</math>. Thus, the area is <math>5\cdot\left(\sqrt2\right)^2=5\cdot 2 = \boxed{\textbf{(A) } 10}</math>. | ||
~pog ~ | ~pog ~wamofanssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss | ||
==Solution 2== | ==Solution 2== | ||
Revision as of 18:10, 21 March 2022
Problem
The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the area of the logo in square inches?
![[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); [/asy]](http://latex-new.aopstest.com/a/a/8/aa8ae8c181def4cc4946be20a97f03d34db3ea20.png)
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 ~wamofanssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss
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
Video Solution
~Interstigation
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
