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

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 (Physics-Inspired)

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 6 (Complex Number)

let B be the origin, put C at c= 1+i

complex number

= c +

= c +

...

= c +

we want to find sum of complex number + + ... + = 2024 *c + (1+2+...+2024) = 2024c +

now we can plug in c= 1+i, = = -

2024c + = 2024 ( 1+i) - 2024 = 2024i

so the length is

~luckuso

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.