Hello! I have a doubt which came in my mind while solving a geometry question
Suppose we have 2 rays and with same end point . Angle is any acute angle with value .
another ray is drawn such that it divides acute Angle into two parts, not necessarily equal.
Suppose there is a point on the ray .
My question is that, will a point always exist on ray such that bisect ?
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
There will exist such a point E only if ∠BOA+∠BOC is obtuse.
Log in to reply
Can you give proof of your statement please
Log in to reply
My proof is via complex numbers.
Without loss of generality assume that OB represents the real axis.
Let D=λeix;λ∈R+.
Let D′(=λ′;λ′∈R+) be any point on OB.
Now, D′=E if Arg(2D+D′)=y where y=∠BOC
Simplifying the above condition, we get λλ′=sinx(tany−cotx)
As the LHS is positive, so is the RHS.
Therefore,
tany>cotx
Now, as 0<y<x<90∘, we get (by using the cosine - sum and difference formulas)
0>cos(x+y)
Due to the constraints on the angles x and y, this again simplifies to 180∘>x+y>90∘⟺180∘>∠BOA+∠BOC>90∘
Log in to reply
@Sarthak Singla
Was this helpful?Log in to reply
I think E would also exist if angleBOA+angleBOC=90
Log in to reply
It would be degenerate as E=O in that case.