Difference between revisions of "2022 AMC 8 Problems/Problem 1"

m (Solution 2: Emphasized LATTICE points.)
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==Solution 2==
 
==Solution 2==
  
We can use Pick's Theorem. There is a total of <math>5</math> lattice points in the interior and <math>12</math> lattice points on the boundary of the figure. As a result, the area is <math>5+\frac{12}{2}-1=\boxed{\textbf{(A) } 10}</math>.
+
We can apply Pick's Theorem: There are <math>5</math> lattice points in the interior and <math>12</math> lattice points on the boundary of the figure. As a result, the area is <math>5+\frac{12}{2}-1=\boxed{\textbf{(A) } 10}</math>.
  
 
~MathFun1000
 
~MathFun1000

Revision as of 21:33, 28 January 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?

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

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

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
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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