Problem 5
An equilateral pentagon AMNPQ is inscribed in triangle ABC such that M∈AB,Q∈AC, and N,P∈BC. Let S be the intersection of MN and PQ Denote by ℓ the angle bisector of ∠MSQ.
Prove that OI is parallel to ℓ, where O is the circumcenter of triangle ABC, and I is the incenter of triangle ABC.
Solution
First solution (complex) In fact, we only need AM=AQ=NP and MN=QP.
We use complex numbers with ABC the unit circle, assuming WLOG that A, B, C are labeled counterclockwise. Let x, y, z be the complex numbers corresponding to the arc midpoints of BC, CA, AB, respectively; thus x+y+z is the incenter of △ABC.
Finally, let s>0 be the side length of AM=AQ=NP.
Then, since MA=s and MA⊥OZ, it follows that
m−a=i⋅sz.
Similarly, p−n=i⋅sy and a−q=i⋅sx, so summing these up gives
i⋅s(x+y+z)=(p−q)+(m−n)=(m−n)−(q−p).
Since MN=PQ, the argument of (m−n)−(q−p) is along the external angle bisector of the angle formed, which is perpendicular to ℓ. On the other hand, x+y+z is oriented in the same direction as OI, as desired.
Second solution (trig, Danielle Wang) .Let δ and ϵ denote ∠MNB and ∠CPQ.
Also, assume AMNPQ has side length 1.
In what follows, assume AB<AC. First, we note that
BN=(c−1)cosB+cosδ,
CP=(b−1)cosC+cosϵ, and
a=1+BN+CP
from which it follows that
cosδ+cosϵ=cosB+cosC−1
Also, by the Law of Sines, we have sinδc−1=sinB1 and similarly on triangle CPQ, and from this we deduce