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Magoosh Brain Twister: Sidesplitting Fun


This is a really tricky one, and it’s Numeric Entry. I’m wondering, can at least three people answer this correctly, making sure to provide a quick explanation. You’ll have till Thursday.

Polygon X has r sides, and each vertex has an angle measure of s, an integer. If Polygon Q has r/4 sides, what is the greatest possible of t, the angle measure of each vertex of Polygon Q?

Numeric Entry: _______________________


Don’t forget to check back on Thursday for the answer and explanation! 🙂

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5 Responses to Magoosh Brain Twister: Sidesplitting Fun

  1. Steve January 27, 2015 at 4:08 pm #

    Since the angle in the interior polygon cannot exceed 180 degrees (would mean it’s a straight line)

    (r-2)*180/r = s(179,178,177,176,175…) – since s needs to be an integer

    Trying out 179r = 180r-360
    r= 360 sides

    (90 -2)*180/90 = 176 degrees (t)

  2. Nitish January 27, 2015 at 10:00 am #

    The answer is 160 degrees.

    The numbers of polygon for which the ‘r/4’ values will be valid are 3, 5, 6, 9, 10, 12, 15, and 18. Any polygon above this value will have a value of ‘s’ which is not an integer.

    Thus the greatest value of t is the value of the angle of the polygon with 18 equal sides, which is 160 degrees

  3. Nitish January 27, 2015 at 9:46 am #

    I guess the answer is 2.

    Unfortunately this is a guess, I couldn’t generalize the solution, but did found that the only possible values for r/4 would be 3,5,6,9,10,15..

    Somehow, this didn’t gave a common solution, I then did it with trial and error, the answer lied between 2 and 2.5 and 2 seemed to be a better choice of the two.

  4. Amit January 27, 2015 at 1:19 am #

    I think the answer should be 176.

    Concept being used here is that more the no. of sides..bigger the angle measure of each vertex will be. (can be derived from the total angle formula below)

    Now the biggest angle can be 179 (largest integer) for the polygon X (It cant be 180 or more). for this angle, no. of sides will be 360…Consequently no .of sides for the polygon Q will be 90..following which angle of each vertex will be 176. using the logic below:-

    Total angle = No. of sides * vertex angle = (No. of sides-2) * 180

    To check for the error, we can try making angle 177 in that case no. of sides will be 120 for polygon Q and 480 for polygon X. for which the corresponding angle is 179.5–> not an integer.

  5. Jenil January 26, 2015 at 6:23 pm #

    #) Doubt – Is this with respect to convex or concave polygon ?

    For convex polygon: following is my approach –

    Polygon X => r sides ; with each angle measure “s” – an integer
    n (no of sides) = 360 – no of sides should be multiple of 4
    Sum of interior angles = 180 x (n-2) = 180 x (360-2) = 180 x (358) = 64,440
    Individual angle = 64,440/360 = 179. (The max integer value for each angle could be 179)

    Polygon Q => r/4 sides ; need to find the greatest angle measure of each vertex
    Polygon Q = 90 sides
    Sum of interior angles = 180 x (90-2) = 180 x (88) = 15,840

    Greatest possible angle measure for each vertex in polygon Q will be 176. (answer)
    Greatest possible angle measure for one individual vertex will be 179.9

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