Difference between revisions of "2023 AMC 10A Problems/Problem 13"
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==Problem== | ==Problem== | ||
− | Abdul and Chiang are standing <math>48</math> 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 | + | Abdul and Chiang are standing <math>48</math> 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 <math>60^\circ</math>. What is the square of the distance (in feet) between Abdul and Bharat? |
<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> |
Revision as of 00:05, 10 November 2023
Contents
[hide]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 checks verifies that indeed works.
~Technodoggo
Solution 2 (no law of sines)
Help with the diagram please?
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. Its ratio is , so .
Its square is then
~not_slay
Video Solution 1 by OmegaLearn
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.