Difference between revisions of "2023 AMC 10A Problems/Problem 13"
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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 | + | 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>. |
− | It follows | + | 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) | -Benedict T (countmath1) |
Revision as of 10:07, 27 November 2023
Contents
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 (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
It is quite clear that this is just 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 (Straightforward)
We can represent Abdul, Bharat and Chiang as ,
, and
, respectively.
Since we have
and
, this is obviously a
triangle, and it would not matter where
is.
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
.
It follows that is
-
-
, and
.
is then
and the square of
is
.
-Benedict T (countmath1)
Video Solution by MegaMath
https://www.youtube.com/watch?v=ZsiqPRWCEkQ&t=3s
~megahertz13
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Math-X (First understand the problem!!!)
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.