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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 PA + PB = 4 PT.

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

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Solution Ugly solution | Comment 3 of 5 |
Set up a Cartesian coordinate system with the origin at O. We choose our axes such that the x-axis is parallel to the chord AB, and our units such that circle C has radius 1.

Our coordinates are then:
O - (0,0)
A - (-cos x, sin x)
B - (cos x, sin x)
M - (0, sin x)
P - (cos y, sin y)
X - (0, sin x)
where X is the centre of the smaller circle

Then
PA = (cos y + cos x) + (sin y - sin x)
PB = (cos y - cos x) + (sin y - sin x)
PT = PX - XT = (cos y) + (sin y - sin x) - OX (since XT and OX are the radii of the smaller circle)
= (cos y) + (sin y - sin x) - ( sin x)

Just expand and it's easy to see how the equality holds...



  Posted by Tan Kiat Chuan on 2005-07-31 18:20:31
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