If the incircle of the triangle ABC(AB≠BC≠CA), passes through it's circumcentre, then the (cosA+cosB+cosC)2 is
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Solution
From the given information it is evident that distance between I (incentre) and O (circumcentre) should be equal to inradius of triangle.
If ′r′ and ′R′ be the inradius and circumradius respectivley. then AI=r cosec A2=4RsinB2sinC2
and AO=R,∠OAI=A2−(π2−C)=C−B2
Now, IO2=OA2+AI2−2(OA)(AI)cosC−B2 ⇒r2=R2+16R2sin2B2sin2C2−8R2sinB2sinC2cos(C−B2) ⇒r2=R2(1+8sinB2sinC2(2sinB2−cos(C−B2))) =R2(1+8sinB2sinC2(sinB2sinC2−cosB2cosC2)) =R2(1−8sinB2sinC2cos(B+C2)) =R2(1−8sinA2sinB2sinC2) =R2(1−8r4R)=R2−2rR ⇒(rr)2+2(rR)−1=0 ⇒rR=−2±√4+42=√2−1, as rR>0 ⇒1+rR=√2⇒cosA+cosB+cosC=√2