Question

If x, y, z are the distances of the vertices of triangle ABC from its orthocenter, then x+y+z = .

• A. r+R
• B. 2(r+R)
• C. 2r+R
• D. r+2R

Answer 1

The correct option is B 2(r+R)

We have,
x+y+z = 2R cos A +2R cos B +2R cos C = 2R(cos A + cos B +cos C)
$=2R\{2\cos \frac{A+B}{2}\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}\}=2R\{2\sin \frac{C}{2}\cos \frac{A-B}{2}-2{\sin }^2\frac{C}{2}\}+2R=4R\sin \frac{C}{2}\{\cos \frac{A-B}{2}-\cos \frac{A+B}{2}\}+2R=8R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}+2R=2r+2R=2(r+R)$