2024 AMC 12A Problems/Problem 7
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
Let be the origin, and set the and axes so that the axis bisects , and the axis is parallel to Notice that the endpoints of each vector all lie on , so each vector is of the form . Furthermore, observe that for each , we have , by properties of reflections about the -axis: therefore Since there are pairs, the resultant vector is , the magnitude of which is
--Benedict T (countmath1)
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 |
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