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Geometry Difficulty 5.2 AIME, harder Prove it Estonia

Inside a circle cc with the center OO there are two circles c1c_1 and c2c_2 which go through OO and are tangent to the circle cc at points AA and BB respectively. Prove that the circles c1c_1 and c2c_2 have a common point which lies in the segment ABAB.

Solution

The radius AOAO of the circle cc is perpendicular to the common tangent to circles cc and c1c_1 at the point AA, hence AOAO is a diameter of the circle c1c_1. Similarly BOBO is a diameter of the circle c2c_2.

If the circles c1c_1 and c2c_2 are tangent at the point OO (Fig. 1), then the diameters AOAO and BOBO are both perpendicular to the common tangent to c1c_1 and c2c_2 at the point OO, whence the lines AOAO and BOBO coincide, i.e. OO lies in the segment ABAB.

If the circles c1c_1 and c2c_2 intersect at OO (Fig. 2), then let MM be the other intersection point of the circles. Since AMO=90\angle AMO = 90^\circ and BMO=90\angle BMO = 90^\circ (angles at the circumference supported by a diameter), the lines AMAM and BMBM coincide and MM lies in the segment ABAB.

Figure 1
Figure 1

Figure 2
Figure 2

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