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

Revision as of 20:54, 8 November 2024

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

Let point $A$ reflect over $BC \longrightarrow A'$ We can see that for all $n$ , $\overrightarrow{BP_n}+\overrightarrow{BP_{2025-n}}=\overrightarrow{AA'}=2$ As a result, $\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}$ ~lptoggled

@@ Line 35: / Line 35: @@
 <cmath>\overrightarrow{BP_n}+\overrightarrow{BP_{2025-n}}=\overrightarrow{AA'}=2</cmath>
 As a result, <cmath>\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}</cmath>
+~lptoggled
 ==See also==
 {{AMC12 box|year=2024|ab=A|num-b=6|num-a=8}}
 {{MAA Notice}}

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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