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 | ~lptoggled | ||
| + | |||
| + | == Solution 4 == | ||
| + | Using Pythagoream theorom, we see the length of the hypotenuse is <math>2</math>. There are 2024 points on <math>AC</math>, that are equal spaced, 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> | ||
| + | Someone please clean this up lol | ||
| + | ~helpmebro | ||
==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 01:04, 9 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
Let point 
reflect over 
We can see that for all 
,
![\[\overrightarrow{BP_n}+\overrightarrow{BP_{2025-n}}=\overrightarrow{AA'}=2\]](http://latex.artofproblemsolving.com/1/3/4/134e154effe80acf6ef383428d15bd7d94aed955.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
Solution 4
Using Pythagoream theorom, we see the length of the hypotenuse is 
. There are 2024 points on 
, that are equal spaced, 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
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. 
