Problem 6 · 2009 Math Kangaroo
Medium
Geometry & Measurement
pythagorean-triplecasework
The circles \(k_1\) (with centre \(M_1\) and radius 13) and \(k_2\) (with centre \(M_2\) and radius 15) intersect each other in the points P and Q. The length of \(PQ\) is 24. What possible value could the distance \(M_1M_2\) be?
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Answer: D — 14
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Hint 1 of 2
Drop a perpendicular from each centre to the common chord PQ; it bisects PQ.
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Hint 2 of 2
Each centre’s distance to the chord is a leg of a right triangle with the radius.
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Approach: right triangles from each centre to the chord
- Half the chord is 12, so the distance from M₁ to PQ is √(13² − 12²) = 5, and from M₂ it is √(15² − 12²) = 9.
- If the centres lie on opposite sides of PQ, M₁M₂ = 5 + 9 = 14.
- That value, 14, is among the options.
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