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Question Number 191305 by Mingma last updated on 22/Apr/23

Commented by mr W last updated on 22/Apr/23

$$thereisnouniquesolution.butthere$$$$isa{r}_{min}.$$

Answered by mr W last updated on 23/Apr/23

Commented by mr W last updated on 23/Apr/23

$$OA=a=12$$$$OB=b=5$$$$sayP(p,r),Q(r,q)$$$$p=r+2r\cos \theta$$$$q=r+2r\sin \theta$$$$M=midpointofPQ$$$$M(\frac{p+r}{2},\frac{q+r}{2})=[(1+\cos \theta )r,(1+\sin \theta )r]$$$$QM=\sqrt{3}r$$$$x_R=(1+\cos \theta +\sqrt{3}\sin \theta )r$$$$y_R=(1+\sin \theta +\sqrt{3}\cos \theta )r$$$$eqn.ofAB:$$$$\frac{x}{a}+\frac{y}{b}=1\Rightarrow bx+ay-ab=0$$$$distancefromRtoABisr:$$$$\frac{\mid b(1+\cos \theta +\sqrt{3}\sin \theta )r+a(1+\sin \theta +\sqrt{3}\cos \theta )r-ab\mid }{\sqrt{a^2+b^2}}=r$$$$r=\frac{ab}{a+b+\sqrt{a^2+b^2}+(a+\sqrt{3}b)\sin \theta +(\sqrt{3}a+b)\cos \theta }$$$$r=\frac{ab}{a+b+\sqrt{a^2+b^2}+2\sqrt{a^2+b^2+\sqrt{3}ab}\sin (\theta +{\tan }^{-1}\frac{\sqrt{3}a+b}{a+\sqrt{3}b})}$$$$at\theta +{\tan }^{-1}\frac{\sqrt{3}a+b}{a+\sqrt{3}b}=\frac{\pi }{2},i.e.\theta =\frac{\pi }{2}-{\tan }^{-1}\frac{\sqrt{3}a+b}{a+\sqrt{3}b}:$$$$r_{min}=\frac{ab}{a+b+\sqrt{a^2+b^2}+2\sqrt{a^2+b^2+\sqrt{3}ab}}$$$$witha=12,b=5:$$$$r_{min}=\frac{30}{15+\sqrt{169+60\sqrt{3}}}\approx 0.952$$

Commented by Mingma last updated on 23/Apr/23

Great Work, sir!

Commented by mr W last updated on 23/Apr/23

$$with{r}_{min}:$$

Commented by mr W last updated on 23/Apr/23

Commented by mr W last updated on 23/Apr/23

$$examplesforotherpossibilities:$$

Commented by mr W last updated on 23/Apr/23

Commented by mr W last updated on 23/Apr/23

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