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 ...

## 1 comment:

The answer is sqrt(6):

________________________

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)

________________________

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)

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