Difference between revisions of "2022 AMC 8 Problems/Problem 1"
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<i>pog</i> | <i>pog</i> | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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 <math>\sqrt2</math>. Thus, the area is <math>5(\sqrt2)^2=5\cdot 2 = \boxed{\textbf{(A) } 10}</math> | ||
+ | |||
+ | ~wamofan | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|before=First Problem|num-a=2}} | {{AMC8 box|year=2022|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:42, 28 January 2022
Contents
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
Note that the given symbol can be split into diamonds. Each of the diamonds contain triangles, so each of the diamonds has an area of . Hence, our answer is .
pog
Solution 2
Draw the following four lines as shown:
We see these lines split the figure into five squares with side length . Thus, the area is
~wamofan
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.