No, its impossible

Geometry Level 4

Suppose I have A B C \angle ABC , and I draw out an angle bisector of A B C \angle ABC , such that A B A 1 = C B A 1 \angle ABA_{1} = \angle CBA_{1} as described above.

Next, I make an angle bisector of C B A 1 \angle CBA_{1} , such that A 1 B A 2 = A 2 B C \angle A_{1}BA_{2} = \angle A_{2} BC

Next, I make an angle bisector of A 1 B A 2 \angle A_{1}BA_{2} , such that A 1 B A 3 = A 3 B A 2 \angle A_{1}BA_{3} = \angle A_{3}BA_{2}

Next, I make an angle bisector of A 3 B A 2 \angle A_{3}BA_{2} , such that A 3 B A 4 = A 4 B A 2 \angle A_{3}BA_{4} = \angle A_{4}BA_{2}

If I am to repeat this algorithm forever, find lim n A n B C A B C \lim _{ n\rightarrow \infty }{ \frac { \angle { A }_{ n }BC }{ \angle { A }BC } }

Give your answer to 3 decimal places.


The answer is 0.33333333.

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2 solutions

Marta Reece
Jul 23, 2017

This is a geometric series. In terms of the original angle, the series is

1 2 1 4 + 1 8 . . . = 1 2 1 ( 1 2 ) = 1 3 \frac12-\frac14+\frac18-...=\dfrac{\frac12}{1-(-\frac12)}=\boxed{\dfrac13}

Rajen Kapur
Jul 16, 2015

2 a n + 2 = a n + 1 + a n 2a_{n+2} = a_{n+1} + a_n for all natural numbers n, where a 1 = 1 2 a_1 = \frac{1}{2} and a 2 = 1 4 a_2 = \frac{1}{4} . 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?

Pi Han Goh - 5 years, 11 months ago

Noticed we just trisected an angle with ruler and compass?

Julian Poon - 5 years, 11 months ago

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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 .

Pi Han Goh - 5 years, 11 months ago

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Which explains the title of this question :D

Julian Poon - 5 years, 11 months ago

AnBc=ABC-ABAn-2 So lim->infinity AnBc/ABC =limn->infinityABC/ABC -ABNn-2/Abc =1-0 =1

Avi c - 5 years, 11 months ago

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Why is A B A n 2 A B C \displaystyle\frac{\angle ABA_{n-2}}{\angle ABC} zero?

Kenny Lau - 5 years, 11 months ago

Why is A n B C = A B C A B A n 2 \angle A_{n}BC = \angle ABC - \angle ABA_{n-2} ? Shouldn't it be A n B C = A B C A B A n \angle A_{n}BC = \angle ABC - \angle ABA_{n} ?

Julian Poon - 5 years, 11 months ago

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