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

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~Technodoggo
 
~Technodoggo
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== Solution 2 ==
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Notice that the average vector sum is 1. Multiplying the 2024 by 1, our answer is <math>\boxed{D}</math>
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~MC
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== Solution 3 (Pair Sum)==
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[[Image:2024_amc12A_p7.png|thumb|center|600px|]]
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Let point <math>B</math> reflect over <math>AC \longrightarrow B'</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{BB'}=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>
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~lptoggled  image
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edited by [https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
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== Solution 4 ==
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Using the Pythagorean theorem, we can see that the length of the hypotenuse is <math>2</math>. There are 2024 equally-spaced points on <math>AC</math>, so there are 2025 line segments along that hypotenuse. <math>\frac{2}{2025}</math> is the length of each line segment. We get <math>\frac{2}{2025}+\frac{4}{2025}+...+\frac{4048}{2025} = \frac{2}{2025} \times \frac{2024*2025}{2}=\fbox{(D) 2024}</math>
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Someone please clean this up lol
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~helpmebro
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==Solution 5 (Physics-Inspired)==
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Let <math>B</math> be the origin, and set the <math>x</math> and <math>y</math> axes so that the <math>x</math> axis bisects <math>\angle ABC</math>, and the <math>y</math> axis is parallel to <math>\overline{AC}.</math> Notice that the endpoints of each vector all lie on <math>i=1</math>, so each vector is of the form <math>1i + xj</math>. Furthermore, observe that for each <math>v_k=1i + xj</math>, we have <math>v_{2024-k} = 1i - xj</math>, by properties of reflections about the <math>x</math>-axis: therefore <math>v_k + v_{2024-k} = 2i.</math> Since there are <math>1012</math> pairs, the resultant vector is <math>1012\cdot 2i</math>, the magnitude of which is <math>\boxed{\textbf{(D)\ 2024}}.</math>
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--Benedict T (countmath1)
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== Solution 6 (Complex Number) ==
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[[Image:2024_amc12A_p7_cn.PNG|thumb|center|600px|]]
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Let B be the origin, place C at <math>C= 1+i</math>
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<math>\overrightarrow{CP_{1}} = re^{i\theta}</math>
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<math>\overrightarrow{CP_{n}} = nre^{i\theta}</math>
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Now we'll find <math>re^{i\theta}</math>
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<math>\frac{|\overrightarrow{AC}|}{Number\;of\;Equal\;Segments}</math>
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= <math>\frac{2}{2025} e^{i\pi}</math>
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= <math> - \frac{2}{2025}</math>
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<math>P_{1}</math> to <math>P_{2024}</math> can be written as such:
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<math>P_{1} = C + \overrightarrow{CP_{1}}</math>
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<math>P_{2} = C + \overrightarrow{CP_{2}}</math>
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...
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<math>P_{2024} = C + \overrightarrow{CP_{2024}}</math>
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We want to find the sum of the complex numbers: 
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<math>P_{1}  + P_{2}  + ... + P_{2024}</math>
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<math>= 2024c + re^{i\theta}(1+2+...+2024)</math>
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<math>= 2024c + \frac{2024\cdot2025}{2} \cdot re^{i\theta}</math>
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Now we can plug in our value for <math>C</math> and <math>re^{i\theta}</math>
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<math>2024c + \frac{2024\cdot2025}{2} \cdot re^{i\theta}</math>
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<math>= 2024 (1+i) - 2024</math>
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<math>= 2024i</math>
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So the length is <math>\fbox{(D) 2024}</math>
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~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso]
  
 
==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}}

Latest revision as of 14:06, 17 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 (Pair Sum)

2024 amc12A p7.png

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)

2024 amc12A p7 cn.PNG

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

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. AMC logo.png

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