Romanian Mathematical Olympiad

Question

A straight line passing through the incenter

I

of a triangle

A B C

meets the sides

A B

and

A C

at points

P

and

Q

respectively. Let

B C = a

,

A C = b

,

A B = c

and

\frac{P B}{P A} = p

,

\frac{QC}{Q A} = q

.

a) Prove that

a (1 + p) I P = (a - p b) I B - c p I C

.

b) Prove that

a = b p + c q

.

c) Prove that if

a^{2} = 4 b c pq

, then the straight lines

A I

,

B Q

and

C P

are concurrent.

Accepted Answer

a) The hypothesis yields (p+1)IP = IB + pIA. The conclusion follows now from aIA + bIB + cIC = 0. b) Analogously, a(1+q)IQ = (a-cq)IC - bqIB. The points P, I, Q are collinear, hence (a - pb)(a - cq) = bcpq, whence the conclusion. c) Squaring the relation from b), a^2 4bcpq, with equality if and only if bp = cq. In this case, if D = AI BC, then PBPA QAQC DCDB = p 1q bc = 1 and the converse of Ceva's Theorem guarantees the conclusion.