Difference between revisions of "2024 AMC 12A Problems/Problem 7"

Revision as of 02:02, 13 January 2025

Problem

In $\Delta ABC$ , $\angle ABC = 90^\circ$ and $BA = BC = \sqrt{2}$ . Points $P_1, P_2, \dots, P_{2024}$ lie on hypotenuse $\overline{AC}$ so that $AP_1= P_1P_2 = P_2P_3 = \dots = P_{2023}P_{2024} = P_{2024}C$ . What is the length of the vector sum $\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}?$

$\textbf{(A) }1011 \qquad \textbf{(B) }1012 \qquad \textbf{(C) }2023 \qquad \textbf{(D) }2024 \qquad \textbf{(E) }2025 \qquad$

Solution 1 (technical vector bash)

Let us find an expression for the $x$ - and $y$ -components of $\overrightarrow{BP_i}$ . Note that $AP_1+P_1P_2+\dots+P_{2023}P_{2024}+P_{2024}C=AC=2$ , so $AP_1=P_1P_2=\dots=P_{2023}P_{2024}=P_{2024}C=\dfrac2{2025}$ . All of the vectors $\overrightarrow{AP_1},\overrightarrow{P_1P_2},$ and so on up to $\overrightarrow{P_{2024}C}$ are equal; moreover, they equal $\textbf v=\left\langle\dfrac1{\sqrt2}\cdot\dfrac2{2025},-\dfrac1{\sqrt2}\cdot\dfrac2{2025}\right\rangle=\left\langle\dfrac{\sqrt2}{2025},-\dfrac{\sqrt2}{2025}\right\rangle$ .

We now note that $\overrightarrow{AP_i}=i\textbf v=\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle$ ( $i$ copies of $\textbf v$ added together). Furthermore, note that $\overrightarrow{BP_i}=\overrightarrow{BA}+\overrightarrow{AP_i}=\left\langle0,\sqrt2\right\rangle+\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle=\left\langle\dfrac{i\sqrt2}{2025},\sqrt2-\dfrac{i\sqrt2}{2025}\right\rangle.$

We want $\sum_{i=1}^{2024}\overrightarrow{BP_i}$ 's length, which can be determined from the $x$ - and $y$ -components. Note that the two values should actually be the same - in this problem, everything is symmetric with respect to the line $x=y$ , so the magnitudes of the $x$ - and $y$ -components should be identical. The $x$ -component is easier to calculate.

$\sum_{i=1}^{2024}\left(\overrightarrow{BP_i}\right)_x=\sum_{i=1}^{2024}\dfrac{i\sqrt2}{2025}=\dfrac{\sqrt2}{2025}\sum_{i=1}^{2024}i=\dfrac{\sqrt2}{2025}\cdot\dfrac{2024\cdot2025}2=1012\sqrt2.$

One can similarly evaulate the $y$ -component and obtain an identical answer; thus, our desired length is $\sqrt{\left(1012\sqrt2\right)^2+\left(1012\sqrt2\right)^2}=\sqrt{4\cdot1012^2}=\boxed{\textbf{(D) }2024}$ .

~Technodoggo

Solution 2

Notice that the average vector sum is 1. Multiplying the 2024 by 1, our answer is $\boxed{D}$

~MC

Solution 3 (Pair Sum)

Let point $B$ reflect over $AC \longrightarrow B'$

We can see that for all $n$ , $\overrightarrow{BP_n}+\overrightarrow{BP_{2025-n}}=\overrightarrow{BB'}=2$ As a result, $\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}$ ~lptoggled image

edited by luckuso

Solution 4

Using the Pythagorean theorem, we can see that the length of the hypotenuse is $2$ . There are 2024 equally-spaced points on $AC$ , so there are 2025 line segments along that hypotenuse. $\frac{2}{2025}$ is the length of each line segment. We get $\frac{2}{2025}+\frac{4}{2025}+...+\frac{4048}{2025} = \frac{2}{2025} \times \frac{2024*2025}{2}=\fbox{(D) 2024}$ Someone please clean this up lol ~helpmebro

Solution 5 (Physics-Inspired)

Let $B$ be the origin, and set the $x$ and $y$ axes so that the $x$ axis bisects $\angle ABC$ , and the $y$ axis is parallel to $\overline{AC}.$ Notice that the endpoints of each vector all lie on $i=1$ , so each vector is of the form $1i + xj$ . Furthermore, observe that for each $v_k=1i + xj$ , we have $v_{2024-k} = 1i - xj$ , by properties of reflections about the $x$ -axis: therefore $v_k + v_{2024-k} = 2i.$ Since there are $1012$ pairs, the resultant vector is $1012\cdot 2i$ , the magnitude of which is $\boxed{\textbf{(D)\ 2024}}.$

--Benedict T (countmath1)

Solution 6 (Complex Number)

Let B be the origin, place C at $C= 1+i$

$\overrightarrow{CP_{1}} = re^{i\theta}$

$\overrightarrow{CP_{n}} = nre^{i\theta}$

Now we'll find $re^{i\theta}$

$\frac{|\overrightarrow{AC}|}{Number\;of\;Equal\;Segments}$

= $\frac{2}{2025} e^{i\pi}$

= $- \frac{2}{2025}$

$P_{1}$ to $P_{2024}$ can be written as such:

$P_{1} = C + \overrightarrow{CP_{1}}$

$P_{2} = C + \overrightarrow{CP_{2}}$

...

$P_{2024} = C + \overrightarrow{CP_{2024}}$

We want to find the sum of the complex numbers:

$P_{1} + P_{2} + ... + P_{2024}$

$= 2024c + re^{i\theta}(1+2+...+2024)$

$= 2024c + \frac{2024\cdot2025}{2} \cdot re^{i\theta}$

Now we can plug in our value for $C$ and $re^{i\theta}$

$2024c + \frac{2024\cdot2025}{2} \cdot re^{i\theta}$

$= 2024 (1+i) - 2024$

$= 2024i$

So the length is $\fbox{(D) 2024}$

~luckuso

Solution 7 (Extreme Case)

Notice that we can put points $P_i$ with odd numbers i on $C$ and those with even numbers i on $A$ . So the sum of vectors $\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}$ is just $1012 \times (\overrightarrow{BC} + \overrightarrow{BA})$ and the length of vector sum is $\boxed{D}$

~Emordnilap

Video Solution 1 (⚡️3 min solve⚡️)

https://youtu.be/abD8gdEI48A

~Education, the Study of Everything

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.

@@ Line 103: / Line 103: @@
 So the length is <math>\fbox{(D) 2024}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
 == Solution 7 (Extreme Case) ==
-Notice that we can put points <math>P_i</math> with odd numbers i on <math>C</math> and those with even numbers i on <math>A</math>. So the sum of vectors <math>\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}</math> is just <math>1012 * (\overrightarrow{BC}+\overrightarrow{BA})</math> and the length of vector sum is <math>\boxed{D}</math>
+Notice that we can put points <math>P_i</math> with odd numbers i on <math>C</math> and those with even numbers i on <math>A</math>. So the sum of vectors <math>\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}</math> is just <math>1012 \times (\overrightarrow{BC} + \overrightarrow{BA})</math> and the length of vector sum is <math>\boxed{D}</math>
 ~Emordnilap
-~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]

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