Eötvös 1923/1 solution

27th Eötvös 1923

Problem 1

The circles OAB, OBC, OCA have equal radius r. Show that the circle ABC also has radius r.

Solution

Let P, Q, R be the centers of the circles OBC, OCA, OAB. Then OQAR, ORBP, OPCQ are rhombi side r. Hence AR and CP are equal and parallel to OQ and hence to each other. So ARPC is a parallelogram. Hence AC = PR. Similarly, AB = PQ and BC = QR, so the triangles ABC and PQR are congruent. Hence the circumradius of ABC is the same as the circuradius of PQR. But O lies on a circle center P radius r, so OP = r. Similarly OQ = OR = r, so O is the circumcenter of PQR, and r is the circumradius of PQR.

27th Eötvös 1923