Difference between revisions of "2024 AMC 12A Problems/Problem 7"
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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 | 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 | ||
<cmath> \overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}? </cmath> | <cmath> \overrightarrow{BP_1} + \overrightarrow{BP_2} + \overrightarrow{BP_3} + \dots + \overrightarrow{BP_{2024}}? </cmath> | ||
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~Technodoggo | ~Technodoggo | ||
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| + | ==See also== | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:46, 8 November 2024
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
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. 
