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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
3 months ago

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There will exist such a point \(E\) only if \(\angle BOA + \angle BOC\) is obtuse. Deeparaj Bhat · 3 months ago

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@Deeparaj Bhat Can you give proof of your statement please Sarthak Singla · 3 months ago

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@Sarthak Singla 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 \(0<y<x<90^\circ\), 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 \[180^\circ > x+y > 90^\circ \iff 180^\circ > \angle BOA + \angle BOC >90^\circ\] Deeparaj Bhat · 3 months ago

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@Deeparaj Bhat Was this helpful? @Sarthak Singla Deeparaj Bhat · 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 months ago

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I think E would also exist if angleBOA+angleBOC=90 Sarthak Singla · 3 months ago

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@Sarthak Singla It would be degenerate as \(E=O\) in that case. Deeparaj Bhat · 3 months ago

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