2021 AMC 12B Fall Problem 10

Attempt Problem 10 of the 2021 AMC 12B Fall below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Fall solutions, or check the answer key.

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10.

What is the sum of all possible values of tt between 00 and 360360 such that the triangle in the coordinate plane whose vertices are (cos40,sin40), (\cos 40^\circ, \sin 40^\circ),\ (cos60,sin60),(\cos 60^\circ, \sin 60^\circ), and   (cost,sint)\ \ (\cos t^\circ, \sin t^\circ) is isosceles?

100100

150150

330330

360360

380380

Answer: E
Concepts:circlechordisosceles triangle
Difficulty rating: 1820
Solution:

The three points lie on the unit circle at angles 40,40^\circ, 60,60^\circ, and t.t^\circ. A chord's length depends only on the angular separation of its endpoints.

If the third point is equidistant from the other two, it lies on the perpendicular bisector: t=50t = 50 or t=230.t = 230.

If its distance to 4040^\circ equals the fixed chord (separation 2020^\circ), then t=20t = 20 (since t=60t = 60 is degenerate). If its distance to 6060^\circ matches, then t=80t = 80 (since t=40t = 40 is degenerate).

The valid values are 50,230,20,80,50, 230, 20, 80, summing to 380.380.

Thus, the correct answer is E.

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