# Netball!!

phgao
Posers and Puzzles 04 May '05 06:57
1. 04 May '05 06:57
Alice's netball squad warms up by spacing themselves equally around a circle, facing inwards, and doing the following exercises.

In the first exercise, each girl passes the ball to the first player on her left, starting and ending with Alice. In the second exercise, they pass to the second girl to their left; in the third exercise to the third girl to their left , and so on.

Each exercise starts and ends with Alice. The angle of an exercise is the between the lines along which a player receives and passes on the ball.

Eg. With 8 players, throwing to the person on the left, the angle is 135 degrees.

With 8 players, throwing to the 2nd person on the left the angle is 90 degrees.

With 8 players, throwing to the 3rd person on the left the angle is 45 degrees.

With 8 players, throwing to the 4th person on the left the angle is 0 degrees. (The throw is just between two people.)

So, the path which the ball follows for exercise 5 is the same as the path for exercise 3, but with the direction of the ball movement reversed (first-left then right). The same applies to the paths for exercises 2 and 6 and for exercises 1 and 7.

Q: Find, with proof, the angle of the fourth exercise (throwing the ball to the fourth person on the left) with 9 players, note: with 8 players angle=0 degrees.

Q: At one training session the coach suggests an exercise with an angle of 17.5 degrees. Alice protests that too many players would be needed. What is the smallest number of players required, and which exercise uses an angle of 17.5 degrees?
2. 04 May '05 22:19
Originally posted by phgao
Alice's netball squad warms up by spacing themselves equally around a circle, facing inwards, and doing the following exercises.

In the first exercise, each girl passes the ball to the first player on her left, starting and ending with Alice. In the second exercise, they pass to the second girl to their left; in the third exercise to the third girl to thei ...[text shortened]... at is the smallest number of players required, and which exercise uses an angle of 17.5 degrees?
(n-2)180 degrees in the sum of the interior angles each angle is

(n-2)180/n This is the angle for throwing to the first person left

Now consider the angles formed by drawing all diagonals of the polygon to
one specific vertex (Alice). These angles are all equal because they are
all inscribed angles cutting equal arcs. n-2 such angles at each vertex.
The sum of these is the interior angle SO

(n-2)A=(n-2)180/n OR
A=180/n

Each time you skip a player you subtract 180/n from this initial angle.

3. 05 May '05 09:40
Cmon guys, give it a go!
4. 14 May '05 22:19
wazzup phago..

nice 'intermediate problem'

its kinda hard...
try providing a diagram ðŸ˜‰
5. 15 May '05 00:11
really quick got 60 for #1
6. AThousandYoung
All My Soldiers...
19 May '05 09:101 edit
Originally posted by phgao
Alice's netball squad warms up by spacing themselves equally around a circle, facing inwards, and doing the following exercises.

In the first exercise, each girl passes the ball to the first player on her left, starting and ending with Al ...[text shortened]... layers required, and which exercise uses an angle of 17.5 degrees?
OK, got the proof.

Draw the circle and the 9 girls (g1, g2, etc) equally spaced around it. Draw in the center point C.

If the ball is passed 4 to the left twice, it will end up at the girl 1 right (g9). The angle from the center to g1 and g9 is 360 degrees divided by 9, or 40 degrees.

Now draw a diameter from the second girl (the vertex of the unknown angle, or g5) to the point midway between g1 and g9 through the center (m). Now we can describe a triangle Cmg1. Angle g1Cm is half of 40 degrees, or 20 degrees.

Angle g1Cg5 is 180 degrees minus angle g1Cm, or 160 degrees.

Triangle g1Cg5 is an isoscoles triangle (is that the right word? It has two sides the same and one different). The "different" angle is g1Cg5 = 160 degrees. The other two must be identical, and must sum up with 160 degrees to make 180 degrees; so each is ten degrees.

So, angle g1g5m is ten degrees. But m is midway around the circle from g1 to g9; clearly angle mg5g9 is also ten degrees. Sum these two and you get angle g1g5g9, which is twenty degrees.

Did I make that clear and rigorous enough?
7. 19 May '05 21:49
Originally posted by AThousandYoung
OK, got the proof.

Draw the circle and the 9 girls (g1, g2, etc) equally spaced around it. Draw in the center point C.

If the ball is passed 4 to the left twice, it will end up at the girl 1 right (g9). The angle from the center to g1 and g9 is 360 degrees divided by 9, or 40 degrees.

Now draw a diameter from the second girl (the vertex of ...[text shortened]... and you get angle g1g5g9, which is twenty degrees.

Did I make that clear and rigorous enough?
the answer is correct but u can get it though the theorem of triangles, ie that the 2 isoceles triangles share the same chord base. Therefore, they are congrunt, and the top angles are the same. Then u can get 20 degrees.