Here's a little mathematics puzzle.
Suppose points A, B, C are all on a circle
and we have a point P outside the circle such that
the line that passes through A and P is a tangent to the
circle ( at A )
the line that passes through B and P is a tangent to the
circle ( at B)
also the line segments PA and BC are parallel.
Suppose |AP| = 3
and |BC| = 2
What is |AC| ?
Background level of mathematics required: high-school
Difficulty level: by high-school standards it is a bit tricky.
In a while I'll post the answer ...
The answer is sqrt(6):
ReplyDelete________________________
Proof 1:
Use cartesian co-ordinates:
let:
P = (0,0)
B= ( 3, 0) , PAB is isosceles
A = ( 3 cos x, 3 sin x ) , where x is the angle at APB
C = ( 3 + 2 cos x, 2 sin x )
Q = ( 3 , r ) , where Q is the center of the circle and r is the circle radius,
r ^ 2 = | QC | ^ 2
= 4 + r^2 - 4 r sin x
so r = 1/ sin x (i)
and looking at BPQ we see that:
tan( x / 2 ) = r / 3 (ii)
so, combining (i) and (ii) :
sin x / ( 1 + cos x ) = 1 / ( 3 * sin x )
thus cos x = 2 /3
and so | AC | = sqrt (6)
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Proof 2:
Outline:
By looking at the angles round the centre of the circle we can show that the triangles PAB and ABC are similar isoscelese triangles.
( that takes a bit of work).
Once we know that we look at the ratio of the lengths, we now know that:
3/L = L/2
where L = |AC| = |AB|
hence |AC| = sqrt(6)