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| | + | {{duplicate|[[2024 AMC 12A Problems/Problem 7|2024 AMC 12A #7]] and [[2024 AMC 10A Problems/Problem 9|2024 AMC 10A #9]]}} |
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| | ==Problem== | | ==Problem== |
| − | In <math>\Delta ABC</math>, <math>\angle ABC = 90^\circ</math> and <math>BA = BC = \sqrt{2}</math>. Points <math>P_1, P_2, \dots, P_{2024}</math> lie on hypotenuse <math>\overline{AC}</math> so that <math>AP_1= P_1P_2 = P_2P_3 = \dots = P_{2023}P_{2024} = P_{2024}C</math>. What is the length of the vector sum
| + | Let <math>N</math> be the least positive integer that is divisible by at least <math>3</math> odd primes and at least <math>4</math> perfect squares. What is the sum of the squares of the digits of <math>N</math>? |
| − | <cmath> \overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}? </cmath>
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| − | <math>
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| − | \textbf{(A) }1011 \qquad
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| − | \textbf{(B) }1012 \qquad
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| − | \textbf{(C) }2023 \qquad
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| − | \textbf{(D) }2024 \qquad
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| − | \textbf{(E) }2025 \qquad
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| − | </math>
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| − | ==Solution 1 (technical vector bash)==
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| − | Let us find an expression for the <math>x</math>- and <math>y</math>-components of <math>\overrightarrow{BP_i}</math>. Note that <math>AP_1+P_1P_2+\dots+P_{2023}P_{2024}+P_{2024}C=AC=2</math>, so <math>AP_1=P_1P_2=\dots=P_{2023}P_{2024}=P_{2024}C=\dfrac2{2025}</math>. All of the vectors <math>\overrightarrow{AP_1},\overrightarrow{P_1P_2},</math> and so on up to <math>\overrightarrow{P_{2024}C}</math> are equal; moreover, they equal <math>\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</math>. | |
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| − | We now note that <math>\overrightarrow{AP_i}=i\textbf v=\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle</math> (<math>i</math> copies of <math>\textbf v</math> added together). Furthermore, note that <math>\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.</math>
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| − | We want <math>\sum_{i=1}^{2024}\overrightarrow{BP_i}</math>'s length, which can be determined from the <math>x</math>- and <math>y</math>-components. Note that the two values should actually be the same - in this problem, everything is symmetric with respect to the line <math>x=y</math>, so the magnitudes of the <math>x</math>- and <math>y</math>-components should be identical. The <math>x</math>-component is easier to calculate.
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| − | <cmath>\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.</cmath>
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| − | One can similarly evaulate the <math>y</math>-component and obtain an identical answer; thus, our desired length is <math>\sqrt{\left(1012\sqrt2\right)^2+\left(1012\sqrt2\right)^2}=\sqrt{4\cdot1012^2}=\boxed{\textbf{(D) }2024}</math>.
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| − | ~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]
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| − | == Solution 7 (Extreme Case) ==
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| − | 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>\fbox{(D) 2024}</math>
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| − | ~Emordnilap
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| − | == Video Solution 1 (⚡️3 min solve⚡️) ==
| + | <math>\textbf{(A)}~ 41 \qquad \textbf{(B)}~ 65 \qquad \textbf{(C)}~ 80 \qquad \textbf{(D)}~ 89 \qquad \textbf{(E)}~ 100</math> |
| − | https://youtu.be/abD8gdEI48A
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| − | <i>~Education, the Study of Everything | + | ==Solution== |
| | + | Consider the prime factorization of <math>N</math>. Intuitively, the odd primes that divide <math>N</math> should be as small as possible so we let <math>N</math> be divisible by the three least odd primes: <math>3</math>, <math>5</math>, and <math>7</math>. In order for <math>N</math> to be divisible by at least four perfect squares, it must either be divisible by the sixth power of a prime, or the squares of at least two different primes. These can be obtained by multiplying <math>3 \cdot 5 \cdot 7</math> by either <math>3^{5} = 243</math>, <math>2^{6} = 64</math>, <math>3\cdot 5 = 15</math>, or <math>2^{2}\cdot 3 = 12</math>; any other combination of primes will clearly make <math>N</math> larger. The least of these values is <math>12</math>, giving <math>N = 2^{2} \cdot 3^{2} \cdot 5 \cdot 7 = 1260</math>. The sum of the squares of the digits of <math>N</math> is <math>1^{2}+2^{2}+6^{2}+0^{2} = \boxed{\textbf{(A)}~41}</math>. |
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| | ==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}} |
| | + | {{AMC10 box|year=2024|ab=A|num-b=8|num-a=10}} |
| | {{MAA Notice}} | | {{MAA Notice}} |
be the least positive integer that is divisible by at least 
odd primes and at least 
perfect squares. What is the sum of the squares of the digits of 
, and 
. In order for 
by either 
, 
, 
, or 
; any other combination of primes will clearly make 
, giving 
. The sum of the squares of the digits of 
.
