Track / Stage 6 / 253 of 400 #1253 of 2000

Problem 1253

National olympiad, first round

Geometry Difficulty 6.4 Prove it

Zaslavsky A.A.

Given an acute-angled triangle $ABC$ and a point $P$ that does not coincide with the orthocenter of the triangle. Prove that the circles passing through the midpoints of the sides of triangles $PAB$ , $PAC$ , $PBC$ , and $ABC$ , as well as the circle passing through the projections of point $P$ onto the sides of triangle $ABC$ , intersect at one point.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Let the midpoint of $AP$ be denoted by $A_{1}$ , the midpoint of $BC$ by $A_{2}$ , and the projection of point $P$ onto $BC$ by $A_{3}$ . Points $B_{1}, B_{2}, B_{3}$ and $C_{1}, C_{2}, C_{3}$ are defined similarly. Let the intersection of the circumcircles of triangles $B_{1} C_{2} A_{1}$ and $C_{1} B_{2} A_{1}$ be denoted by $Q$ . We need to prove that the circumcircle of triangle $C_{1} B_{1} A_{2}$ also contains point $Q$ . Note that $\angle A_{1} \mathrm{QC} C_{1}=180^{\circ}-\angle A_{1} B_{2} C_{1}=180^{\circ}-\angle A_{1} P C_{1}$ . Similarly, $\angle A_{1} \mathrm{QB}_{1}=180^{\circ}-\angle A_{1} \mathrm{~PB}_{1}$ . Therefore, $\angle B_{1} \mathrm{QC}_{1}=\angle A_{1} \mathrm{QB}_{1}+\angle A_{1} \mathrm{QC}_{1}=360^{\circ}-\angle A_{1} \mathrm{P} C_{1}-\angle A_{1} \mathrm{~PB}_{1}=\angle B_{1} \mathrm{PC}_{1}=B_{1} A_{2} C_{1}$ .

Thus, the circumcircle of triangle $C_{1} B_{1} A_{2}$ also contains point $Q$ .

Similarly, it can be shown that the circumcircle of triangle $A_{2} B_{2} C_{2}$ also contains point $Q$ .

It remains to prove that the circumcircle of triangle $A_{3} B_{3} C_{3}$ also contains point $Q$ . For this, it is sufficient to show that $\angle A_{3} C_{3} B_{3}=\angle A_{3} Q B_{3}$ . Point $C_{3}$ is symmetric to point $P$ with respect to $A_{1} B_{1}$ . Therefore, $\angle A_{1} C_{3} B_{1}=\angle A_{1} \mathrm{~PB}_{1}=\angle A_{1} C_{2} B_{1}$ . Hence, point $C_{3}$ lies on the circumcircle of triangle $A_{1} C_{2} B_{1}$ .

Fig. 12

Fig. 13

Since points $A_{1}, B_{3}, B_{2}, C_{1}, Q$ lie on the same circle and quadrilateral $A B_{3} \mathrm{PC}_{3}$ is cyclic (since angles $B_{3}$ and $C_{3}$ in it are right angles), we have $\angle B_{3} Q_{1}=\angle B_{3} A_{1} C_{1}=\angle C_{1} A_{1} P=\angle \mathrm{PAB} 3=\angle B_{3} C_{3} P$ .

Similarly, $\angle A_{3} C_{3} P=\angle A_{3} Q C_{1}$ . Therefore,

$\angle A_{3} C_{3} B_{3}=\angle A_{3} C_{3} P+\angle B_{3} C_{3} P=\angle B_{3} Q C_{1}+\angle A_{3} Q C_{1}=\angle A_{3} Q B_{3},$

which is what we needed to prove.

Other cases of point placement are considered similarly.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.