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

(→‎Solution 3)
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Someone please clean this up lol
 
Someone please clean this up lol
 
~helpmebro
 
~helpmebro
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 +
== Solution 5 (Similar to Solution 4) ==
 +
 +
\text{Let } \mathbf{A} = (0, \sqrt{2}), \mathbf{B} = (0, 0), \mathbf{C} = (\sqrt{2}, 0). \\
 +
\text{We know that } \mathbf{A} \mathbf{P}_1 = \mathbf{P}_1 \mathbf{P}_2 = \dots = \mathbf{P}_{2024} \mathbf{C}, \\
 +
\text{and that the length of each segment is } \frac{1}{2025}. \\
 +
\text{The point } \mathbf{P}_1 \text{ is } \frac{2024}{2025} \text{ of the way from } \mathbf{C} \text{ to } \mathbf{A}, \text{ so we can write:} \\
 +
\mathbf{P}_1 = \frac{2024}{2025} \mathbf{A} + \frac{1}{2025} \mathbf{C}. \\
 +
\text{Similarly, for } \mathbf{P}_2, \text{ we write: } \mathbf{P}_2 = \frac{2023}{2025} \mathbf{A} + \frac{2}{2025} \mathbf{C}, \\
 +
\text{and so on.} \\
 +
\text{The sum of the vectors } \mathbf{B} \mathbf{P}_1 + \mathbf{B} \mathbf{P}_2 + \dots + \mathbf{B} \mathbf{P}_{2024} \text{ is:} \\
 +
\sum_{i=1}^{2024} \mathbf{B} \mathbf{P}_i = \left( \sum_{i=1}^{2024} \frac{2025-i}{2025} \right) \mathbf{A} + \left( \sum_{i=1}^{2024} \frac{i}{2025} \right) \mathbf{C}. \\
 +
\text{We compute the sums:} \\
 +
\sum_{i=1}^{2024} (2025 - i) = \frac{2024 \times 2025}{2}, \quad \sum_{i=1}^{2024} i = \frac{2024 \times 2025}{2}. \\
 +
\text{Thus, the sum is:} \\
 +
\sum_{i=1}^{2024} \mathbf{B} \mathbf{P}_i = \frac{2024 \times 2025}{2025 \times 2} \mathbf{A} + \frac{2024 \times 2025}{2025 \times 2} \mathbf{C} = 1012 (\mathbf{A} + \mathbf{C}). \\
 +
\mathbf{A} + \mathbf{C} = (\sqrt{2}, \sqrt{2}), \\
 +
\text{and the magnitude of } \mathbf{A} + \mathbf{C} \text{ is } \sqrt{2^2 + 2^2} = \sqrt{4} = 2. \\
 +
\text{Thus, the magnitude of the sum is:} \\
 +
\| 1012 (\mathbf{A} + \mathbf{C}) \| = 1012 \times 2 = 2024. \\
 +
\boxed{2024}.
  
 
==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 17:26, 9 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 and edited by ~luckuso

Solution 4

Using Pythagoream theorom, we see the length of the hypotenuse is $2$. There are 2024 points on $AC$, that are equal spaced, 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 (Similar to Solution 4)

\text{Let } \mathbf{A} = (0, \sqrt{2}), \mathbf{B} = (0, 0), \mathbf{C} = (\sqrt{2}, 0). \\ \text{We know that } \mathbf{A} \mathbf{P}_1 = \mathbf{P}_1 \mathbf{P}_2 = \dots = \mathbf{P}_{2024} \mathbf{C}, \\ \text{and that the length of each segment is } \frac{1}{2025}. \\ \text{The point } \mathbf{P}_1 \text{ is } \frac{2024}{2025} \text{ of the way from } \mathbf{C} \text{ to } \mathbf{A}, \text{ so we can write:} \\ \mathbf{P}_1 = \frac{2024}{2025} \mathbf{A} + \frac{1}{2025} \mathbf{C}. \\ \text{Similarly, for } \mathbf{P}_2, \text{ we write: } \mathbf{P}_2 = \frac{2023}{2025} \mathbf{A} + \frac{2}{2025} \mathbf{C}, \\ \text{and so on.} \\ \text{The sum of the vectors } \mathbf{B} \mathbf{P}_1 + \mathbf{B} \mathbf{P}_2 + \dots + \mathbf{B} \mathbf{P}_{2024} \text{ is:} \\ \sum_{i=1}^{2024} \mathbf{B} \mathbf{P}_i = \left( \sum_{i=1}^{2024} \frac{2025-i}{2025} \right) \mathbf{A} + \left( \sum_{i=1}^{2024} \frac{i}{2025} \right) \mathbf{C}. \\ \text{We compute the sums:} \\ \sum_{i=1}^{2024} (2025 - i) = \frac{2024 \times 2025}{2}, \quad \sum_{i=1}^{2024} i = \frac{2024 \times 2025}{2}. \\ \text{Thus, the sum is:} \\ \sum_{i=1}^{2024} \mathbf{B} \mathbf{P}_i = \frac{2024 \times 2025}{2025 \times 2} \mathbf{A} + \frac{2024 \times 2025}{2025 \times 2} \mathbf{C} = 1012 (\mathbf{A} + \mathbf{C}). \\ \mathbf{A} + \mathbf{C} = (\sqrt{2}, \sqrt{2}), \\ \text{and the magnitude of } \mathbf{A} + \mathbf{C} \text{ is } \sqrt{2^2 + 2^2} = \sqrt{4} = 2. \\ \text{Thus, the magnitude of the sum is:} \\ \| 1012 (\mathbf{A} + \mathbf{C}) \| = 1012 \times 2 = 2024. \\ \boxed{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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