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 $OA$ and $OB$ with same end point $O$. Angle $AOB$ is any acute angle with value $x$.

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

Suppose there is a point $D$ on the ray $OA$.

My question is that, will a point $E$ always exist on ray $OB$ such that $OC$ bisect $DE$?

Note by Sarthak Singla
4 years, 11 months ago

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

- 4 years, 11 months ago

- 4 years, 11 months ago

My proof is via complex numbers.

Without loss of generality assume that $OB$ represents the real axis.

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

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

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

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

As the LHS is positive, so is the RHS.

Therefore,

$\tan y > \cot x$

Now, as $0, 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 $180^\circ > x+y > 90^\circ \iff 180^\circ > \angle BOA + \angle BOC >90^\circ$

- 4 years, 11 months ago

- 4 years, 11 months ago

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.

- 4 years, 11 months ago

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

- 4 years, 11 months ago

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

- 4 years, 11 months ago