Difference between revisions of "2024 AMC 12A Problems/Problem 7"
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As a result, <cmath>\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}</cmath> | As a result, <cmath>\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}</cmath> | ||
~lptoggled image | ~lptoggled image | ||
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edited by [https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | edited by [https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
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[[Image:2024_amc12A_p7_cn.PNG|thumb|center|600px|]] | [[Image:2024_amc12A_p7_cn.PNG|thumb|center|600px|]] | ||
| − | + | Let B be the origin, place C at <math>C= 1+i</math> | |
<math>\overrightarrow{CP_{1}} = re^{i\theta}</math> | <math>\overrightarrow{CP_{1}} = re^{i\theta}</math> | ||
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<math>\overrightarrow{CP_{n}} = nre^{i\theta}</math> | <math>\overrightarrow{CP_{n}} = nre^{i\theta}</math> | ||
| − | |||
| − | <math>P_{1}</math> | + | Now we'll find <math>re^{i\theta}</math> |
| + | |||
| + | <math>\frac{|\overrightarrow{AC}|}{Number\;of\;Equal\;Segments}</math> | ||
| + | |||
| + | = <math>\frac{2}{2025} e^{i\pi}</math> | ||
| + | |||
| + | = <math> - \frac{2}{2025}</math> | ||
| + | |||
| + | |||
| + | <math>P_{1}</math> to <math>P_{2024}</math> can be written as such: | ||
| + | |||
| + | <math>P_{1} = C + \overrightarrow{CP_{1}}</math> | ||
| − | <math>P_{2} | + | <math>P_{2} = C + \overrightarrow{CP_{2}}</math> |
... | ... | ||
| − | <math>P_{2024}</math> = | + | <math>P_{2024} = C + \overrightarrow{CP_{2024}}</math> |
| + | |||
| + | |||
| + | We want to find the sum of the complex numbers: | ||
| + | |||
| + | <math>P_{1} + P_{2} + ... + P_{2024}</math> | ||
| + | |||
| + | <math>= 2024c + re^{i\theta}(1+2+...+2024)</math> | ||
| + | |||
| + | <math>= 2024c + \frac{2024\cdot2025}{2} \cdot re^{i\theta}</math> | ||
| + | |||
| + | |||
| + | Now we can plug in our value for <math>C</math> and <math>re^{i\theta}</math> | ||
| + | |||
| − | + | <math>2024c + \frac{2024\cdot2025}{2} \cdot re^{i\theta}</math> | |
| − | + | <math>= 2024 (1+i) - 2024</math> | |
| − | + | <math>= 2024i</math> | |
| − | + | So the length is <math>\fbox{(D) 2024}</math> | |
Latest revision as of 14:06, 17 November 2024
Contents
Problem
In 
, 
and 
. Points 
lie on hypotenuse 
so that 
. What is the length of the vector sum
![\[\overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}?\]](http://latex.artofproblemsolving.com/d/a/a/daafe683a70517576b26bb2a80fdb220002fc49c.png)
Solution 1 (technical vector bash)
Let us find an expression for the 
- and 
-components of 
. Note that 
, so 
. All of the vectors 
and so on up to 
are equal; moreover, they equal 
.
We now note that 
(
copies of 
added together). Furthermore, note that 
We want 
's length, which can be determined from the - and -components. Note that the two values should actually be the same - in this problem, everything is symmetric with respect to the line 
, so the magnitudes of the - and -components should be identical. The -component is easier to calculate.
![\[\sum_{i=1}^{2024}\left(\overrightarrow{BP_i}\right)_x=\sum_{i=1}^{2024}\dfrac{i\sqrt2}{2025}=\dfrac{\sqrt2}{2025}\sum_{i=1}^{2024}i=\dfrac{\sqrt2}{2025}\cdot\dfrac{2024\cdot2025}2=1012\sqrt2.\]](http://latex.artofproblemsolving.com/7/b/7/7b7a0cb72970f4bf95c85216bd0c2340b37b4f8d.png)
One can similarly evaulate the -component and obtain an identical answer; thus, our desired length is 
.
~Technodoggo
Solution 2
Notice that the average vector sum is 1. Multiplying the 2024 by 1, our answer is 
~MC
Solution 3 (Pair Sum)
Let point 
reflect over 
We can see that for all 
,
![\[\overrightarrow{BP_n}+\overrightarrow{BP_{2025-n}}=\overrightarrow{BB'}=2\]](http://latex.artofproblemsolving.com/5/e/7/5e76c2e45b178be64bb0a7b5416b0cb65e212ee6.png)
As a result, ![\[\overrightarrow{BP_1}+\overrightarrow{BP_2 }+ ...+\overrightarrow{BP_{2024}}=2 \cdot 1012=\fbox{(D) 2024}\]](http://latex.artofproblemsolving.com/4/d/7/4d70905305441d6ffba950849169eeaa9a46cf45.png)
~lptoggled image
edited by luckuso
Solution 4
Using the Pythagorean theorem, we can see that the length of the hypotenuse is 
. There are 2024 equally-spaced points on 
, so there are 2025 line segments along that hypotenuse. 
is the length of each line segment. We get 
Someone please clean this up lol
~helpmebro
Solution 5 (Physics-Inspired)
Let be the origin, and set the and axes so that the axis bisects 
, and the axis is parallel to 
Notice that the endpoints of each vector all lie on 
, so each vector is of the form 
. Furthermore, observe that for each 
, we have 
, by properties of reflections about the -axis: therefore 
Since there are 
pairs, the resultant vector is 
, the magnitude of which is 
--Benedict T (countmath1)
Solution 6 (Complex Number)
Let B be the origin, place C at 
Now we'll find 
= 
= 
to 
can be written as such:
...
We want to find the sum of the complex numbers:
Now we can plug in our value for 
and
So the length is 
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. 
