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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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 (cos⁡40∘,sin⁡40∘), (\cos 40^\circ, \sin 40^\circ),\ (cos⁡60∘,sin⁡60∘),(\cos 60^\circ, \sin 60^\circ), and   (cos⁡t∘,sin⁡t∘)\ \ (\cos t^\circ, \sin t^\circ) is isosceles?
100100
150150
330330
360360
380380
Answer: E
Concepts:circlechordisosceles triangle
Small hint:
All three points lie on the unit circle, so equal sides mean equal chords, i.e. equal arc separations
Big hint:
Consider the three cases where the third point matches distance to A,A, to B,B, or lies on their perpendicular bisector
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 40∘40^\circ equals the fixed chord (separation 20∘20^\circ), then t=20t = 20 (since t=60t = 60 is degenerate). If its distance to 60∘60^\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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