59th Ukrainian National Mathematical Olympiad

Question

The inscribed circle of an acute triangle ABC touches sides AB and BC in points

C_{1}

and

A_{1}

respectively. Let M be the midpoint of the side AC, and N be the midpoint of an arc ABC of the circumcircle of ABC. Also, let P be the projection of a point M to the segment

A_{1} C_{1}

. Prove the points I, P, and N lie on the same line. (Anton Trygub)

Fig. 35

Accepted Answer

Let's denote the second intersection point of the line BI with the circumcircle of

△ A B C

as W (fig. 35). Clearly, W is a midpoint of the smaller arc AC, and points M, W, and N lie on the same line.

Hence, it is enough to prove that triangles

P M N

and

I W N

are similar (which will imply that

∠ P N M = ∠ I N W

and the desired collinearity). In order to prove the similarity, we are going to show that

\frac{N M}{N W} = \frac{P M}{I W}

. Let

∠ B = 2 β

, then

∠ M N A = β

and

\frac{N M}{N W} = \frac{P M}{I W} \cdot \frac{N A}{N W} = cos^{2} β .

Considering a segment PM yields that PM is a midline of a trapezoid AXYC, where X, Y are the projections of A and C onto the line

A_{1} C_{1}

respectively. Going further, we get

2 PM = CY + XA = CA_{1} sin YA_{1} C + C_{1} sin XC_{1} A = (CA_{1} + AC_{1}) cos β = AC cos β,

because

∠ YA_{1} C = ∠ XC_{1} A = 9 0^{\circ} - β

, and

CA_{1} + AC_{1} = AC

, because

A_{1}

and

C_{1}

are the points in which inscribed circle touches the sides of a triangle

△ A B C

. Finally, from the trillium theorem,

I W = C W

, meaning that

\frac{P M}{I W} = \frac{A C}{2 C W} cos β = \frac{C M}{C W} cos β = cos^{2} β,

finishing the proof.