2023 AMC 10A Problems/Problem 13
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
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
Video Solution 1 by OmegaLearn
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
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