Difference between revisions of "2023 AMC 10A Problems/Problem 13"
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<math>\textbf{(A) } 1728 \qquad \textbf{(B) } 2601 \qquad \textbf{(C) } 3072 \qquad \textbf{(D) } 4608 \qquad \textbf{(E) } 6912</math> | <math>\textbf{(A) } 1728 \qquad \textbf{(B) } 2601 \qquad \textbf{(C) } 3072 \qquad \textbf{(D) } 4608 \qquad \textbf{(E) } 6912</math> | ||
+ | ==Video Solution by MegaMath== | ||
+ | |||
+ | https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s | ||
==Solution 1== | ==Solution 1== | ||
Line 20: | Line 23: | ||
==Solution 2 (no law of sines)== | ==Solution 2 (no law of sines)== | ||
− | Let us begin by circumscribing the two points A and C so that the arc it determines has measure <math>120</math>. Then the point B will lie on the circle, which we can quickly find the radius of by using the 30-60-90 triangle formed by the radius and the midpoint of segment <math>\overline{AC}</math>. We will find that <math>r=16 | + | Let us begin by circumscribing the two points A and C so that the arc it determines has measure <math>120</math>. Then the point B will lie on the circle, which we can quickly find the radius of by using the 30-60-90 triangle formed by the radius and the midpoint of segment <math>\overline{AC}</math>. We will find that <math>r=16\times\sqrt3</math>. Due to the triangle inequality, <math>\overline{AB}</math> is maximized when B is on the diameter passing through A, giving a length of <math>32\times\sqrt3</math> and when squared gives <math>\boxed{\textbf{(C) }3072}</math>. |
− | ==Solution 3== | + | ==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> | ||
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~wangzrpi | ~wangzrpi | ||
− | ==Solution== | + | ==Solution 4== |
We use <math>A</math>, <math>B</math>, <math>C</math> to refer to Abdul, Bharat and Chiang, respectively. | We use <math>A</math>, <math>B</math>, <math>C</math> to refer to Abdul, Bharat and Chiang, respectively. | ||
Line 47: | Line 50: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 5 == | ||
+ | |||
+ | 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 (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>. | ||
+ | Squaring AB we get <math>\boxed{\textbf{(C) 3072}}</math>. | ||
+ | |||
+ | ~ESAOPS | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | (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>. | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==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 🚀 Under 2 min 🚀== | ||
+ | |||
+ | https://youtu.be/d5XeBKZvTGQ | ||
+ | |||
+ | <i>~Education, the Study of Everything </i> | ||
+ | |||
+ | ==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== | ||
Line 57: | Line 107: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution by Math-X == | ||
+ | https://youtu.be/N2lyYRMuZuk?si=_Y5mdCFhG-XD7SaG&t=631 | ||
+ | |||
+ | ~Math-X | ||
==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 14:21, 14 August 2024
Contents
- 1 Problem
- 2 Video Solution by MegaMath
- 3 Solution 1
- 4 Solution 2 (no law of sines)
- 5 Solution 3 (Guessing)
- 6 Solution 4
- 7 Solution 5
- 8 Solution 6 (Logic)
- 9 Solution 7
- 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 Math-X
- 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?
Video Solution by MegaMath
https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s
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 (no law of sines)
Let us begin by circumscribing the two points A and C so that the arc it determines has measure . Then the point B will lie on the circle, which we can quickly find the radius of by using the 30-60-90 triangle formed by the radius and the midpoint of segment
. We will find that
. Due to the triangle inequality,
is maximized when B is on the diameter passing through A, giving a length of
and when squared gives
.
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
~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 🚀 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 Math-X
https://youtu.be/N2lyYRMuZuk?si=_Y5mdCFhG-XD7SaG&t=631
~Math-X
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.