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Math Help - Star Polygons

  1. #1
    Junior Member phgao's Avatar
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    Star Polygons

    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.

    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?

    I see this Q, as taking into account the formula (n-2)180 = sum of angles inside a regular polyhedra.

    (n-2)180 degrees in the sum of the interior angles each angle is (n-2)180/n...

    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
    A=180/n

    But my way could be completly wrong, any thoughts would be appreciated.
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  2. #2
    MHF Contributor
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    If there are are 8 players, the circumference of the circle is divided into 8 equal arcs. Each one measures 360/8 = 45 degrees.

    If there are are 9 players, the circumference of the circle is divided into 9 equal arcs. Each one is 360/9 = 40 degrees.

    If there are are n players, the circumference of the circle is divided into n equal arcs. Each arc is 360/n degrees.

    An angle is inscribed in a circle if the vertex lies on the circumference of the circle.
    The measure of this inscribed angle is half the measure of the arc subtended by the angle.

    --------
    Q: Find, with proof, the angle of the fourth exercise (throwing the ball to the fourth person on the left) with 9 players.)

    9 players. 9 equal arcs of 40 degrees each.
    The ball is thrown to the 4th player, then to the next 4th. That means the last player to catch the ball is the 8th player from Alice. That means there is still one more 40-degree arc before Alice, or, between the 8th player and Alice, there is one 40-degree arc.
    This last 40-degree arc is the arc subtended by the angle in question.
    Therefore, the angle of the fourth exercise is 40/2 = 20 degrees. ----answer.

    ------
    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?

    17.5 degrees
    So, 17.5 *2 = 35 degrees ---the measure of the arc subtended by the angle of the exercise.

    Then,
    360/n = 35
    360 = 35*n
    n = 360/35 = 10.286 ---not an integer.
    That means, the arc subtended by the angle in question is not one equal arc.
    The arc subtended by the 17.5 degrees, which is 35 degrees, is a combination of more than one equal spaces between players.

    How many equal spaces can there be in a 35-degree arc?
    >>>Seven of 5-degree spaces.
    >>>Five of 7-degree spaces.

    360/7 = 51.428 players ---cannot be.

    360/5 = 72 players ---could be.

    Hence, there are 72 players,and each is separated by a 5-degree arc.

    The desired angle, 17.5 degrees, needs 35 degrees, and that is the total arc for 7 of 5-degree spaces.

    Umm, there is no way to throw the ball to two players and end up with 7 equal spaces left.
    72 - 7 = 65
    65/2 = 32.5 ---there is no player between the 32nd and 33rd players.

    If we multiply that 32.5 by 2, we get 65.

    ....that is it!

    Then we divide the 5-degree space by 2. We get 2.5 degrees per space.
    Then, 360/2.5 = 144 equal spaces = 144 players.

    Alice throws the ball to the 65th, then the 65th throws the ball to the, (65*2), 130th player.
    144 -130 = 14 spaces left.

    14*(2.5 degrees) = 35 degrees
    The inscribed angle subtended by this 35-degree arc is
    35/2 = 17.5 degrees ---the desired angle for the exercise.

    Therefore, for this exercise,
    >>>144 is the smallest number of players.
    >>>throwing to the 65th player to the left is the pattern.
    Last edited by ticbol; May 21st 2005 at 10:29 AM.
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  3. #3
    Junior Member phgao's Avatar
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    Thumbs up

    Thanks for your detailed solution!
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