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Circle (Posted on 2005-07-28) Difficulty: 5 of 5
C is a circle with center O. AB is a chord not passing through O. M is the midpoint of AB. C' is the circle with diameter OM. T is a point on C'. The tangent to C' at T meets C at P. Show that PA² + PB² = 4 PT².

No Solution Yet Submitted by nilshady    
Rating: 3.6667 (3 votes)

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Hints/Tips Thoughts | Comment 1 of 5

First of all it doesn't matter which of the two intersection points of the circle C with the tangent to C' is chosen as being point P. Choose either one of them.

The stated equality that's to be shown is equivalent to showing that if PA is one leg of a right triangle and PB is another, then the hypotenuse is 2*PT.

If a circle is constructed with center at T, passing through P, its diameter will be 2*PT.  On this constructed circle, construct chords from point P equal in length to PA and PB.  Call the other endpoints A' and B' respectively. What needs to be shown is that the chord formed by these endpoints is in fact a diameter of the circle centered at T.  It may help to consider the fact that isosceles triangles are formed: BPB' and APA'.  Once this is shown to be a diameter, angle B'PA' is shown to be a right angle and the triangle B'PA' is a right triangle with sides of the length that shows what the puzzle asks.

These relationships all seem to hold when constructing this with the Geometer's Sketchpad, but I don't know a proof.

  Posted by Charlie on 2005-07-28 19:41:13
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