Geometric Puzzle
1) To solve this problem, I used a technique that mathematicians love: I turned the problem into a simpler problem.
I started with circles split into 2s,3s,4s,5s, 6s, 8s, and 10ths and noticed that the circles that were split into an odd number of parts did not have diametrically opposite points but the even ones always did. My understanding of this is that the even ones can always be split into halves at any pair of antipodal points (the diametrically opposite points) and then further split until they reach the appropriate fraction. Since 30 is even, I noted that this was not a trick question and did in fact have a solution!
From this understanding, I also noticed that the antipodal points were always half a cycle away. By that, for a circle split into 2n parts, then the cycle would be 2n and antipodal points would always be n away. So, for 30, the antipodal point of 7 must be 15 away which is 22!
To confirm my suspicions, I also attempted this problem algebraically. Each of the points was mapped onto the unit circle by the formula
x = cos( a*2*pi/n ); y = sin( a*2*pi/n ), where n = 30 and a = 0, 1, 2, .... , 29, 30
Then, the coordinates for the 7th point is at
x = cos( 7*pi/15 ); y = sin( 7*pi/15 )
and it's diametrically opposite point must be
x = cos( 7*pi/15 + pi ); y = sin( 7*pi/15 + pi).
But, 7*pi/15 + pi = 7pi/15 + 15pi/15 = 22 pi/15.
So a = 22 for the diametrically opposite point to 7.
2) From my understanding of the problem, I would also present it with different and more familiar numbers for circles like 12, 60, and 360 and then present an odd example so that it won't work. The impossible example is intended to scaffold students to discover this property of circles and fractions.
3) I'm not sure this problem was truly geometric actually. Although I approached this example geometrically at first, I generalized the problem logically through my understanding of odd and even numbers as well as how fractions worked.
I started with circles split into 2s,3s,4s,5s, 6s, 8s, and 10ths and noticed that the circles that were split into an odd number of parts did not have diametrically opposite points but the even ones always did. My understanding of this is that the even ones can always be split into halves at any pair of antipodal points (the diametrically opposite points) and then further split until they reach the appropriate fraction. Since 30 is even, I noted that this was not a trick question and did in fact have a solution!
From this understanding, I also noticed that the antipodal points were always half a cycle away. By that, for a circle split into 2n parts, then the cycle would be 2n and antipodal points would always be n away. So, for 30, the antipodal point of 7 must be 15 away which is 22!
To confirm my suspicions, I also attempted this problem algebraically. Each of the points was mapped onto the unit circle by the formula
x = cos( a*2*pi/n ); y = sin( a*2*pi/n ), where n = 30 and a = 0, 1, 2, .... , 29, 30
Then, the coordinates for the 7th point is at
x = cos( 7*pi/15 ); y = sin( 7*pi/15 )
and it's diametrically opposite point must be
x = cos( 7*pi/15 + pi ); y = sin( 7*pi/15 + pi).
But, 7*pi/15 + pi = 7pi/15 + 15pi/15 = 22 pi/15.
So a = 22 for the diametrically opposite point to 7.
2) From my understanding of the problem, I would also present it with different and more familiar numbers for circles like 12, 60, and 360 and then present an odd example so that it won't work. The impossible example is intended to scaffold students to discover this property of circles and fractions.
3) I'm not sure this problem was truly geometric actually. Although I approached this example geometrically at first, I generalized the problem logically through my understanding of odd and even numbers as well as how fractions worked.
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