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`DeltaABC` is equilateral triangle of side a. P lies on AB such that A is midpoint of PB. If `r_(1)` is inradius of PAC and `r_(2)` is ex radius of PBC opposite to P, then `r_(1) + r_(2) =`

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The correct Answer is:
`00000.50`
`/_T_(1)O_(1)R=60^(@)` and `/_AO_(1)R=0^(@)`
similarly `/_BO_(2)S=30^(@)`
`T_(1)T_(2)=T_(1)A+AB+BT_(1)=RA+RB+SB`
`=r_(1) tan 30^(@)+a+r_(2)tan30^(@)=(r_(1)+r_(2))/(sqrt(3))+a`
and `T_(1)^(')T_(2)^(')=T_(1)^(')C+T_(2)^(')C=a-RA+a-BS=2a-(r_(1)+r_(2))/(sqrt(3))`
Since common external tangents are equal `T_(1)T_(2)=T_(1)^(')T_(2)^(')`
`implies r_(1)+r_(2)=(sqrt(3)a)/2`
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