2023 AMC 10A Problems/Problem 13
Contents
- 1 Problem
- 2 Solution 1 (Using Trigonometry)
- 3 Solution 2 (Inscribed Angles)
- 4 Solution 3 (Guessing)
- 5 Solution 4
- 6 Solution 5
- 7 Solution 6 (Logic)
- 8 Solution 7
- 9 Video Solution by Little Fermat
- 10 Video Solution by Math-X
- 11 Video Solution 🚀 Under 2 min 🚀
- 12 Video Solution by Power Solve
- 13 Video Solution by SpreadTheMathLove
- 14 Video Solution 1 by OmegaLearn
- 15 Video Solution by CosineMethod
- 16 Video Solution
- 17 Video Solution by MegaMath
- 18 See Also
Problem
Abdul and Chiang are standing feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures . What is the square of the distance (in feet) between Abdul and Bharat?
Solution 1 (Using Trigonometry)
Let and .
By the Law of Sines, we know that . Rearranging, we get that where is a function of . We want to maximize .
We know that the maximum value of , so this yields
A quick check verifies that indeed works.
~Technodoggo ~(minor grammar edits by vadava_lx)
Solution 2 (Inscribed Angles)
We can draw a circle such that the chord AC inscribes an arc of 120 degrees. This way, any point B on the circle not in the inscribed arc will form an angle of 60 degrees with . To maximize the distance between A and B, they must be opposite each other. So, the problem is now finding the length of the diameter of the circle. We know AOC is 120 degrees, so dropping a perpendicular form O to AC gives us the radius as . So, the diameter is which gives us the answer
~AwesomeParrot
Solution 3 (Guessing)
Guess that the optimal configuration is a 30-60-90 triangle, as an equilateral triangle gives an answer of , which is not on the answer choices. Its ratio is , so .
Its square is then
Note: The distance between Abdul and Chiang is constant, so let that be represented as . If we were dealing with an equilateral triangle, the height would be , and if we were dealing with a 30-60-90 triangle, the height would be , which is greater than .
~not_slay
~wangzrpi
Solution 4
We use , , to refer to Abdul, Bharat and Chiang, respectively. We draw a circle that passes through and and has the central angle . Thus, is on this circle. Thus, the longest distance between and is the diameter of this circle. Following from the law of sines, the square of this diameter is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5
We can represent Abdul, Bharat and Chiang as , , and , respectively. Since we have and (from other solutions) , this is a triangle. By the side ratios of a triangle, we can infer that . Squaring AB we get .
~ESAOPS
Solution 6 (Logic)
As in the previous solution, refer to Abdul, Bharat and Chiang as , , and , respectively- we also have . Note that we actually can't change the lengths, and thus the positions, of and , because that would change the value of (if we extended either of these lengths, then we could simply draw such that is perpendicular to , so is unchanged). We can change the position of to alter the values of and , but throughout all of these changes, remains unvaried. Therefore, we can let .
(What is the justification for all of these assumptions??)
It follows that is --, and . is then and the square of is .
-Benedict T (countmath1)
Solution 7
(why?)
We look at the answers and decide: the square of is .
-vvsss
Video Solution by Little Fermat
https://youtu.be/h2Pf2hvF1wE?si=ISeW3ruGd-iLhQZi&t=2819 ~little-fermat
Video Solution by Math-X
https://youtu.be/GP-DYudh5qU?si=unB-KAz2AXgMuLSS&t=3337
~Math-X
Video Solution 🚀 Under 2 min 🚀
~Education, the Study of Everything
Video Solution by Power Solve
https://www.youtube.com/watch?v=jkfsBYzBJbQ
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=nmVZxartc-o
Video Solution 1 by OmegaLearn
Video Solution by CosineMethod
https://www.youtube.com/watch?v=BJKHsHQyoTg
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by MegaMath
https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.