If f, g, and h are the lengths of the perpendiculars from the circumcentre on the sides a,b, and c of a triangle ABC respectively then #a/f + b/g + c/h = K((abc)/(fgh))#. Then the value of K is?

Question

2833 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-13T10:10:56

enter image source here

Let O be the circumcentre of the triangle #ABC# in which #BC=a,CA=b and AB=c#.

The lengths of perpendiculars drawn from #O# on three sides are #OM_a=f,OM_b=g and OM_c=h#

Let

#/_OBC=OCB=alpha#

#/_OCA=OAC=beta#

#/_OBA=OAB=gamma#

Now

#cotalpha=(a/2)/f=a/(2f)#

Similarly

#cotbeta=(b/2)/g=a/(2g)#

and

#cotgamma=(c/2)/h=a/(2h)#

Now #A+B+C=pi#

#=>(beta+gamma)+(gamma+alpha)+(alpha+beta)=pi#

#=>2alpha+2beta+2gamma=pi#

#=>alpha+beta+gamma=pi/2#

#=>alpha+beta=pi/2-gamma#

#=>cot(alpha+beta)=cot(pi/2-gamma)#

#=>(cotalphacotbeta-1)/(cosbeta+cotalpha)=tangamma=1/cotgamma#

#=>cotalpha+cotbeta+cotgamma=cotalphacotbetacotgamma#

Inserting the established values we get

#=>a/(2f)+b/(2g)+c/(2h)=a/(2f)xxb/(2g)xxc/(2h)#

#=>a/f+b/g+c/h=1/4((abc)/(fgh))#

Comparing this relation with the given one we get

#K=1/4#