∠ A B C , and I draw out an angle bisector of ∠ A B C , such that ∠ A B A 1 = ∠ C B A 1 as described above.
Suppose I haveNext, I make an angle bisector of ∠ C B A 1 , such that ∠ A 1 B A 2 = ∠ A 2 B C
Next, I make an angle bisector of ∠ A 1 B A 2 , such that ∠ A 1 B A 3 = ∠ A 3 B A 2
Next, I make an angle bisector of ∠ A 3 B A 2 , such that ∠ A 3 B A 4 = ∠ A 4 B A 2
If I am to repeat this algorithm forever, find n → ∞ lim ∠ A B C ∠ A n B C
Give your answer to 3 decimal places.
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2 a n + 2 = a n + 1 + a n for all natural numbers n, where a 1 = 2 1 and a 2 = 4 1 . If we write these equations for n = 1, 2, 3, . . . . . to infinity and add them all L.H.S. and R.H.S. terms cancel out giving 3a_\inf = 2a_2 + a_1 . Hence the answer 0.3333 . . .
Why can't we just set 1 - (1 - 1/2 + 1/4 - 1/8 + 1/16 - ...) = 1/3?
Noticed we just trisected an angle with ruler and compass?
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If you mean that after an infinite number of steps, then yes you trisected an angle. But it's actually impossible to trisect arbitrary angles .
AnBc=ABC-ABAn-2 So lim->infinity AnBc/ABC =limn->infinityABC/ABC -ABNn-2/Abc =1-0 =1
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Why is ∠ A B C ∠ A B A n − 2 zero?
Why is ∠ A n B C = ∠ A B C − ∠ A B A n − 2 ? Shouldn't it be ∠ A n B C = ∠ A B C − ∠ A B A n ?
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This is a geometric series. In terms of the original angle, the series is
2 1 − 4 1 + 8 1 − . . . = 1 − ( − 2 1 ) 2 1 = 3 1