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

(→‎Solution 1 (technical vector bash))
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~MC
 
~MC
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== Solution 3 ==
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Let point <math>A</math> reflect over <math>BC \longrightarrow A'</math>
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We can see that for all <math>n</math>,
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<cmath>\overrightarrow{BP_n}+\overrightarrow{BP_{2025-n}}=\overrightarrow{AA'}=2</cmath>
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As a result, <cmath>\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}</cmath>
 
==See also==
 
==See also==
 
{{AMC12 box|year=2024|ab=A|num-b=6|num-a=8}}
 
{{AMC12 box|year=2024|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:16, 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, \[\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}\]

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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