Approach: Based on the first and second pieces of information, we can form this relationship easily:

(

) =π

a r10 − 1

(10 ) 2

r −1 ⇒ a r10 − 1 = 100π ( r − 1) --- (1).

(

)

The third piece of information gives us this equation: a + ar 2 + ar 4 + ar 6 + ar 8 = 10π + ar + ar 3 + ar 5 + ar 7 + ar 9 . Using the GP sum formula, we may condense it to: 5 5 a r 2 − 1 ar r 2 − 1 = 10π + 2 2 r −1 r −1 10 2 ⇒ a r − 1 = 10π r − 1 + ar r10 − 1 --- (2).

( ) (

( )

)

(

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)

(

)

To obtain the desired equation, we may consider two ways:

(

)

•

replace the terms a r10 − 1 in (2), or

•

replace the term 10π r 2 − 1 = 10π ( r − 1)( r + 1) in (2).

(

)

If we choose the first way, we obtain:

(

)

100π ( r − 1) = 10π r 2 − 1 + r 100π ( r − 1) . Upon simplification (you are encouraged to do it!), we arrive at a nice quadratic equation: 11r 2 − 20r + 9 = 0 . Factorisation leads to: giving r =

( r − 1)(11r − 9 ) = 0 ,

9 , since the sectors must become smaller progressively. 11

With the second way we get:

(

)

a r −1 = 10

(

) ( r + 1) + ar

a r10 − 1 10

(r

10

)

−1 .

Simplifying (again, you are encouraged to get your hands dirty!), we arrive at: 11r11 − 9r10 − 11r + 9 = 0 , which looks more intimidating than the equation of the first way but still remains easy for us to factorise:

(r

With r =

10

)

− 1 (11r − 9 ) = 0 .

9 , a can be found from (1) and the value of ar 9 can then be established. 11

From this example, we see the need to reduce our problem-solving effort to arrive at a solution as efficiently as possible. It is essential that we make mathematics work favourably for us, rather than have it stump us unwittingly.

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