2022 AMC 8 Problems/Problem 1
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[hide]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)
Solution 1
Draw the following four lines as shown:
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
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), and (2,3) Use the shoelace forula to get
Quick solution
If the triangles are rearanged such that the gaps are filled, a by rectangle and by squares are present. Thus, the answer is ~peelybonehead
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.