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AMC 12 B

Geometry Difficulty 3.5 AMC 10/12 Find the answer United States

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?

Pick one

Solution

Let A=(cos⁡40∘,sin⁡40∘)A = (\cos 40^\circ, \sin 40^\circ), B=(cos⁡60∘,sin⁡60∘)B = (\cos 60^\circ, \sin 60^\circ), C=(cos⁡t∘,sin⁡t∘)C = (\cos t^\circ, \sin t^\circ), and O=(0,0)O = (0, 0). Then acute ∠AOB=20∘\angle AOB = 20^\circ, so △ABC\triangle ABC will be isosceles with vertex at AA or BB if ∠AOC=20∘\angle AOC = 20^\circ or ∠BOC=20∘\angle BOC = 20^\circ, respectively, which occurs when t=20t = 20 or t=80t = 80.

The vertex will be CC if OC→\overrightarrow{OC} bisects either the acute or the reflex angle AOBAOB, which is 340∘340^\circ. This will occur when t=50t = 50 or t=230t = 230, respectively.

The requested sum is 20+80+50+230=38020 + 80 + 50 + 230 = 380.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.