A geometry problem- I feel it is one

Hello! I have a doubt which came in my mind while solving a geometry question

Suppose we have 2 rays OAOA and OBOB with same end point OO. Angle AOBAOB is any acute angle with value xx.

another ray OCOC is drawn such that it divides acute Angle AOB AOB into two parts, not necessarily equal.

Suppose there is a point DD on the ray OAOA.

My question is that, will a point EE always exist on ray OBOB such that OCOC bisect DEDE?

Note by Sarthak Singla
3 years, 3 months ago

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1 vote

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There will exist such a point EE only if BOA+BOC\angle BOA + \angle BOC is obtuse.

Deeparaj Bhat - 3 years, 3 months ago

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Can you give proof of your statement please

Sarthak Singla - 3 years, 3 months ago

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My proof is via complex numbers.

Without loss of generality assume that OBOB represents the real axis.

Let D=λeix;  λR+D=\lambda e^{ix} ; \; \lambda \in \mathbb{R}^+ .

Let D(=λ;  λR+)D' (=\lambda' ; \; \lambda' \in \mathbb{R}^+) be any point on OBOB.

Now, D=ED'=E if Arg(D+D2)=y \text{Arg} \left( \frac{D+D'}{2} \right) = y where y=BOCy = \angle BOC

Simplifying the above condition, we get λλ=sinx(tanycotx) \frac{\lambda'}{\lambda} = \sin x ( \tan y - \cot x)

As the LHS is positive, so is the RHS.

Therefore,

tany>cotx \tan y > \cot x

Now, as 0<y<x<900<y<x<90^\circ, we get (by using the cosine - sum and difference formulas)

0>cos(x+y) 0>\cos (x+y)

Due to the constraints on the angles xx and yy, this again simplifies to 180>x+y>90    180>BOA+BOC>90180^\circ > x+y > 90^\circ \iff 180^\circ > \angle BOA + \angle BOC >90^\circ

Deeparaj Bhat - 3 years, 3 months ago

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@Deeparaj Bhat Was this helpful? @Sarthak Singla

Deeparaj Bhat - 3 years, 3 months ago

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@Deeparaj Bhat Thanks for the solution. Can it not be proved by some simpler method? I feel it would take me quite a few years to decipher this solution as I am unaware of complex numbers.

Sarthak Singla - 3 years, 3 months ago

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I think E would also exist if angleBOA+angleBOC=90

Sarthak Singla - 3 years, 3 months ago

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It would be degenerate as E=OE=O in that case.

Deeparaj Bhat - 3 years, 3 months ago

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