Difference between revisions of "2023 AMC 10A Problems/Problem 13"
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~(minor grammar edits by vadava_lx) | ~(minor grammar edits by vadava_lx) | ||
− | ==Solution 2 ( | + | ==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 <math>\angle{ABC}</math>. 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 <math>16\sqrt{3}</math>. So, the diameter is <math>32\sqrt{3}</math> which gives us the answer <math>\boxed{\textbf{(C) }3072}</math> | |
− | + | ~AwesomeParrot | |
− | + | ==Solution 3 (Guessing)== | |
+ | |||
+ | Guess that the optimal configuration is a 30-60-90 triangle, as an equilateral triangle gives an answer of <math>48^2=2304</math>, which is not on the answer choices. Its ratio is <math>\frac{48}{\sqrt{3}}</math>, so <math>\overline{AB}=\frac{96}{\sqrt{3}}</math>. | ||
Its square is then <math>\frac{96^2}{3}=\boxed{\textbf{(C) }3072}</math> | Its square is then <math>\frac{96^2}{3}=\boxed{\textbf{(C) }3072}</math> | ||
+ | |||
+ | Note: The distance between Abdul and Chiang is constant, so let that be represented as <math>{x}</math>. If we were dealing with an equilateral triangle, the height would be <math>{{x\sqrt3}/2}</math>, and if we were dealing with a 30-60-90 triangle, the height would be <math>{x\sqrt3}</math>, which is greater than <math>{{x\sqrt3}/2}</math>. | ||
~not_slay | ~not_slay | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 5 | + | ==Solution 5 == |
We can represent Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively. | We can represent Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively. | ||
− | Since we have <math>\angle ABC=60^\circ</math> and <math>\angle BCA=90^\circ</math>, this is | + | Since we have <math>\angle ABC=60^\circ</math> and (from other solutions) <math>\angle BCA=90^\circ</math>, this is a <math>30-60-90</math> triangle. |
By the side ratios of a <math>30-60-90</math> triangle, we can infer that <math>AB=\frac{48\times 2}{\sqrt{3}}</math>. | By the side ratios of a <math>30-60-90</math> triangle, we can infer that <math>AB=\frac{48\times 2}{\sqrt{3}}</math>. | ||
Squaring AB we get <math>\boxed{\textbf{(C) 3072}}</math>. | Squaring AB we get <math>\boxed{\textbf{(C) 3072}}</math>. | ||
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==Solution 6 (Logic)== | ==Solution 6 (Logic)== | ||
− | As in the previous solution, refer to Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively- we also have <math>\angle ABC=60^\circ</math>. Note that we actually can't change the lengths, and thus the positions, of <math>AB</math> and <math>BC</math>, because that would change the value of <math>\angle ABC</math> (if we extended either of these lengths, then we could simply draw <math>AC'</math> such that <math>BC'</math> is perpendicular to <math>AC'</math>, so <math>AB</math> is unchanged). We can change the position of <math>AC</math> to alter the values of <math>AC</math> and <math>BC</math>, but throughout all of these changes, <math>AB</math> remains unvaried. Therefore, we can let <math>\angle ACB = 90^\circ</math>. | + | As in the previous solution, refer to Abdul, Bharat and Chiang as <math>A</math>, <math>B</math>, and <math>C</math>, respectively- we also have <math>\angle ABC=60^\circ</math>. Note that we actually can't change the lengths, and thus the positions, of <math>AB</math> and <math>BC</math>, because that would change the value of <math>\angle ABC</math> (if we extended either of these lengths, then we could simply draw <math>AC'</math> such that <math>BC'</math> is perpendicular to <math>AC'</math>, so <math>AB</math> is unchanged). We can change the position of <math>AC</math> to alter the values of <math>AC</math> and <math>BC</math>, but throughout all of these changes, <math>AB</math> remains unvaried. |
+ | Therefore, we can let <math>\angle ACB = 90^\circ</math>. | ||
+ | |||
+ | (What is the justification for all of these assumptions??) | ||
It follows that <math>\triangle ABC</math> is <math>30</math>-<math>60</math>-<math>90</math>, and <math>BC = \frac{48}{\sqrt{3}}</math>. <math>AB</math> is then <math>\frac{96}{\sqrt{3}},</math> and the square of <math>AB</math> is <math>\boxed{\textbf{(C) 3072}}</math>. | It follows that <math>\triangle ABC</math> is <math>30</math>-<math>60</math>-<math>90</math>, and <math>BC = \frac{48}{\sqrt{3}}</math>. <math>AB</math> is then <math>\frac{96}{\sqrt{3}},</math> and the square of <math>AB</math> is <math>\boxed{\textbf{(C) 3072}}</math>. | ||
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-Benedict T (countmath1) | -Benedict T (countmath1) | ||
− | ==Video Solution by | + | ==Solution 7 == |
+ | <math>\angle BAC = 90^\circ - 60^\circ = 30^\circ</math> (why?) | ||
+ | |||
+ | <cmath>\implies BC = \frac {AB}{2} \implies AB^2 = BC^2+ AC^2 \implies</cmath> | ||
+ | |||
+ | <cmath>AB^2 = \frac {4}{3} \cdot 48^2 = 4 \cdot 48 \cdot 16 \approx 200 \cdot 16 = 3200.</cmath> | ||
+ | We look at the answers and decide: the square of <math>AB</math> is <math>\boxed{\textbf{(C) 3072}}</math>. | ||
+ | |||
+ | -vvsss | ||
+ | |||
+ | ==Video Solution by Math-X == | ||
+ | https://youtu.be/GP-DYudh5qU?si=unB-KAz2AXgMuLSS&t=3337 | ||
+ | |||
+ | ~Math-X | ||
+ | ==Video Solution 🚀 Under 2 min 🚀== | ||
+ | |||
+ | https://youtu.be/d5XeBKZvTGQ | ||
+ | |||
+ | <i>~Education, the Study of Everything </i> | ||
− | https://www.youtube.com/watch?v= | + | ==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 1 by OmegaLearn == | ||
https://youtu.be/mx2iDUeftJM | https://youtu.be/mx2iDUeftJM | ||
+ | |||
+ | == Video Solution by CosineMethod== | ||
+ | |||
+ | https://www.youtube.com/watch?v=BJKHsHQyoTg | ||
==Video Solution== | ==Video Solution== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ==Video Solution by MegaMath== | ||
− | + | https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s | |
− | https:// | ||
− | |||
− | |||
− | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2023|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2023|ab=A|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:15, 28 October 2024
Contents
- 1 Problem
- 2 Solution 1
- 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 Math-X
- 10 Video Solution 🚀 Under 2 min 🚀
- 11 Video Solution by Power Solve
- 12 Video Solution by SpreadTheMathLove
- 13 Video Solution 1 by OmegaLearn
- 14 Video Solution by CosineMethod
- 15 Video Solution
- 16 Video Solution by MegaMath
- 17 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
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 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.