2024 AMC 12A Problems/Problem 7

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 Pythagoream theorom, we see the length of the hypotenuse is . There are 2024 points on , that are equal spaced, 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 (Similar to Solution 4)

Let \mathbf{A} = (0, \sqrt{2}), \mathbf{B} = (0, 0), \mathbf{C} = (\sqrt{2}, 0). \\ \text{We know that } \mathbf{A} \mathbf{P}_1 = \mathbf{P}_1 \mathbf{P}_2 = \dots = \mathbf{P}_{2024} \mathbf{C}, \\ \text{and that the length of each segment is } \frac{1}{2025}. \\ \text{The point } \mathbf{P}_1 \text{ is } \frac{2024}{2025} \text{ of the way from } \mathbf{C} \text{ to } \mathbf{A}, \text{ so we can write:} \\ \mathbf{P}_1 = \frac{2024}{2025} \mathbf{A} + \frac{1}{2025} \mathbf{C}. \\ \text{Similarly, for } \mathbf{P}_2, \text{ we write: } \mathbf{P}_2 = \frac{2023}{2025} \mathbf{A} + \frac{2}{2025} \mathbf{C}, \\ \text{and so on.} \\ \text{The sum of the vectors } \mathbf{B} \mathbf{P}_1 + \mathbf{B} \mathbf{P}_2 + \dots + \mathbf{B} \mathbf{P}_{2024} \text{ is:} \\ \sum_{i=1}^{2024} \mathbf{B} \mathbf{P}_i = \left( \sum_{i=1}^{2024} \frac{2025-i}{2025} \right) \mathbf{A} + \left( \sum_{i=1}^{2024} \frac{i}{2025} \right) \mathbf{C}. \\ \text{We compute the sums:} \\ \sum_{i=1}^{2024} (2025 - i) = \frac{2024 \times 2025}{2}, \quad \sum_{i=1}^{2024} i = \frac{2024 \times 2025}{2}. \\ \text{Thus, the sum is:} \\ \sum_{i=1}^{2024} \mathbf{B} \mathbf{P}_i = \frac{2024 \times 2025}{2025 \times 2} \mathbf{A} + \frac{2024 \times 2025}{2025 \times 2} \mathbf{C} = 1012 (\mathbf{A} + \mathbf{C}). \\ \mathbf{A} + \mathbf{C} = (\sqrt{2}, \sqrt{2}), \\ \text{and the magnitude of } \mathbf{A} + \mathbf{C} \text{ is } \sqrt{2^2 + 2^2} = \sqrt{4} = 2. \\ \text{Thus, the magnitude of the sum is:} \\ \| 1012 (\mathbf{A} + \mathbf{C}) \| = 1012 \times 2 = 2024. \\ \boxed{2024}.

See also

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