2022 AMC 8 Problems/Problem 1

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?

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);
 (Error making remote request. Unexpected URL sent back)

$\textbf{(A) } 10 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

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}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); draw((3,4)--(4,3), red); draw((4,3)--(3,2), red); draw((3,2)--(2,3), red); draw((2,3)--(3,4), red); [/asy]$

We see these lines split the figure into five squares with side length $\sqrt2$ . Thus, the area is $5\cdot\left(\sqrt2\right)^2=5\cdot 2 = \boxed{\textbf{(A) } 10}$ .

~pog ~wamofan

Solution 2

We can apply Pick's Theorem: There are $5$ lattice points in the interior and $12$ lattice points on the boundary of the figure. As a result, the area is $5+\frac{12}{2}-1=\boxed{\textbf{(A) } 10}$ .

~MathFun1000

Solution 3

Notice that the area of the figure is equal to the area of the $4 \times 4$ square subtracted by the $12$ triangles that are half the area of each square, which is $1$ . The total area of the triangles not in the figure is $12 \cdot \frac{1}{2} = 6$ , so the answer is $16-6 = \boxed{\textbf{(A) } 10}$ .

~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}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); draw((2,4)--(4,4), red); draw((4,4)--(4,2), red); draw((4,2)--(2,2), red); draw((2,2)--(2,4), red); [/asy]$

The area of the big square is $4$ , and the area of each triangle is $0.5$ . There are $12$ of these triangles, so the total area of all the triangles is $0.5\cdot12=6$ . Therefore, the area of the entire figure is $4+6=\boxed{\textbf{(A) } 10}$ . ~RocketScientist

Bad Solution

The coordinates are $(1,2), (2,1), (3,2), (4,1), (5,2), (4,3), (5,4), (4,5), (3,4), (2,5), (1,4), (2,3)$ Use the Shoelace Theorem to get $\boxed{\textbf{(A)} ~10}$

Quick solution

If the triangles are rearanged such that the gaps are filled, a $4$ by $2$ rectangle and $2$ $1$ by $1$ squares are present. Thus, the answer is $\boxed{(A) 10}$ ~peelybonehead

Video Solution

https://youtu.be/Ij9pAy6tQSg

~Interstigation

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

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