Difference between revisions of "2024 AMC 12A Problems/Problem 7"
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== Solution 4 == | == Solution 4 == | ||
− | Using | + | Using the Pythagorean theorem, we can see that the length of the hypotenuse is <math>2</math>. There are 2024 equally-spaced points on <math>AC</math>, so there are 2025 line segments along that hypotenuse. <math>\frac{2}{2025}</math> is the length of each line segment. We get <math>\frac{2}{2025}+\frac{4}{2025}+...+\frac{4048}{2025} = \frac{2}{2025} \times \frac{2024*2025}{2}=\fbox{(D) 2024}</math> |
Someone please clean this up lol | Someone please clean this up lol | ||
~helpmebro | ~helpmebro |
Revision as of 21:04, 9 November 2024
Contents
Problem
In , and . Points lie on hypotenuse so that . What is the length of the vector sum
Solution 1 (technical vector bash)
Let us find an expression for the - and -components of . Note that , so . All of the vectors and so on up to are equal; moreover, they equal .
We now note that ( copies of added together). Furthermore, note that
We want 's length, which can be determined from the - and -components. Note that the two values should actually be the same - in this problem, everything is symmetric with respect to the line , so the magnitudes of the - and -components should be identical. The -component is easier to calculate.
One can similarly evaulate the -component and obtain an identical answer; thus, our desired length is .
~Technodoggo
Solution 2
Notice that the average vector sum is 1. Multiplying the 2024 by 1, our answer is
~MC
Solution 3 (Pair Sum)
Let point reflect over
We can see that for all , As a result, ~lptoggled image and edited by ~luckuso
Solution 4
Using the Pythagorean theorem, we can see that the length of the hypotenuse is . There are 2024 equally-spaced points on , so there are 2025 line segments along that hypotenuse. is the length of each line segment. We get Someone please clean this up lol ~helpmebro
Solution 5 (Complex Number)
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.