# Question regarding circumcentre?

• Jul 29th 2008, 06:53 AM
fardeen_gen
Question regarding circumcentre?
Let f, g, h be the lengths of the perpendiculars from the circumcentre of the triangle ABC on the sides a,b and c respectively. If (a/f) + (b/g) + (c/h) = k.(abc)/(fgh), then the value of k is:

A)1/4
B)1/2
C)1
D)2
• Jul 29th 2008, 03:58 PM
red_dog
Let O be the circumcenter of the triangle ABC, $OM\perp BC, \ ON\perp AC, \ OP\perp AB$
We have $BC=a, AC=b, AB=c, OM=f, ON=g, OP=h$
Let $\widehat{BOM}=\alpha, \ \widehat{NOC}=\beta, \ \widehat{POA}=\gamma$. Then $\alpha+\beta+\gamma=\pi$

$\frac{a}{f}=2\tan\alpha, \ \frac{b}{g}=2\tan\beta, \ \frac{c}{g}=\tan\gamma$

Then the equality becomes $2\tan\alpha+2\tan\beta+2\tan\gamma=8k\tan\alpha\ta n\beta\tan\gamma$

But is known that if $\alpha+\beta+\gamma=\pi$ then $\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\bet a\tan\gamma$

So, $k=\frac{1}{4}$