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Masterclass Mentorship Team C

Masterclass Mentorship Team C. 1 dimension and 0 dimensions: A research on the relationship between the number of points and number of lines. A research and presentation by Cai Yi Zhan, Tan Wei Chuan, Darryll Chong, Lim Jan Jay and Ryan Wee. An Introduction. Points and lines: Simple

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Masterclass Mentorship Team C

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  1. Masterclass Mentorship Team C 1 dimension and 0 dimensions: A research on the relationship between the number of points and number of lines A research and presentation by Cai Yi Zhan, Tan Wei Chuan, Darryll Chong, Lim Jan Jay and Ryan Wee

  2. An Introduction • Points and lines: • Simple • Fundamental • Elementary • Profound • Our research – relationship between number of points and number of lines

  3. Our Research Question What is the minimum number of points needed to form n straight lines? Conditions: • Line passes through exactly 2 points • n≥1

  4. Illustrations A minimum of 3 points are needed to form 3 straight lines A minimum of 4 points are needed to form 6 straight lines A minimum of 4 points are also needed to form 5 straight lines

  5. Table of data collection • Let us start by showing the results if n= 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10.

  6. Our findings • To find the minimum number of points needed to form nstraight lines: • First obtain the greatest possible value of k such that: • n-(1+2+3… …+(k-1))>0 • Then the minimum number of points is: • k+1 • Where • n: no. of lines • K: any number

  7. Proof • Conjecture: • To draw 1+1+2+3+…+(k-1), 1+1+2+3…+(k-1)+1, 1+1+2+3+…+(k-1)+2 to 1+1+2+3…+(k-1)+(k-1) lines, the minimum no. of points is k+1

  8. Proof • Look at table: true if k is 4. • E.g.: k = 4. k-1= 3. 1+1+2+3 = 7. 1+1+2+3+1 = 8. 1+1+2+3+3 = 9. 1+1+2+3+3 = 10. • All (7, 8, 9, 10) made by 5 points. 5 is 4+1. • Shown to be true by table. • Mathematical Induction: • Know 1 example of k true. That is, to draw 1+1+2+3+...+(k-1) to 1+1+2+3+…+(k-1)+(k-1) lines uses k+1 points is true for 1 value of k.

  9. Proof • Prove the conjecture true for k+1: • As we know 1 example of k is true, that k numbers are all made by k+1 points, imagine k+1 points on a plane. • By adding 1 point, points can join each of k+1 existing points to form a line. • Thus, k+2 points can produce: • 1+1+2+3+…+(k-1)+(k-1)+1, 1+1+2+3+…+(k-1)+(k-1)+2, 1+1+2+3+…+(k-1)+(k-1)+3 to 1+1+2+3+…+(k-1)+(k-1)+k+1 lines

  10. Proof • As seen in previous slide, 1,2 …., k, k+1 is a total of k+1 numbers. • So, with k+2 points, k+1 different number of lines can be drawn. • Hence shown that conjecture true, as: • We have a true example for k. • Then, as proved k+1 true, means next value true. • But as k is any number, means that new value can be k. Next value true as k+1 proven. • So on and so forth.

  11. Proof - How we found the formula • To get a formula: • Utilise fact: proven that over a period of 1+1+2+3+…+(k-1) lines to 1+1+2+3+…+(k-1)+(k-1) lines, or k values of n, when k≥1, minimum number of points is k+1. • So when k was 1, only 1 value of a number of lines can be drawn, (1) with 2 lines. When k is 2, 2 different values of a number of lines can be drawn (2, 3) with 3 lines. When k was 3, 3 values of a number of lines could be drawn (4, 5, 6) with 4 lines. So on and so forth.

  12. Proof - How we found the formula • Considering that: • start by subtracting 1, then 2, so on, until result is the lowest integer. In that instance, assume that min. number of points for n will be x. • Then, as proven, k is the number of different values of n that can be made with x points. • Have proven, that for 1+1+2+3+…+(k-1) lines to 1+1+2+3+…+(k-1)+(k-1) lines, or for k diff. values of n, min. no. of points: k+1. • So, minimum number of points: k+1 • That was how we got our formula.

  13. Proof – finding the formula • Why is it the minimum? • Maximum number of lines to be drawn with given set of n points: • Must use all points • (n-1)+(n-2)+… … +2+1 • = (n-1)n/2 • As 2nd line connects to 1st point (1 line), 3rd line connects to 2nd and 1st points (2 lines), 4th line connects to 3rd, 2nd and 1st points (3 lines), so on. nth point connects to (n-1) points before it. • E.g. 4 points. 1+2+3=6. • Similarly, (4-1)4/2= 2(4-1) =3x2=6

  14. Proof – finding the formula • Our formula also obtains maximum, as: • In formula, n=k+1, so max. lines would be k(k+1)/2 when given k+1 points. • 1 value of n must equal k(k+1)/2 = k(k-1)/2 +k • In formula: • Already said n-(1+2+… … +(k-1)) >0 = n- k(k-1)/2 >0.

  15. Proof – finding the formula • k values of n allow this statement to be true with same value of k. • Biggest value of n - result in total being k. • Simplify to n- k(k-1)/2 -k =0. • Simplify even more to: n = k(k-1)/2 +k • Hence obtained maximum • As we obtained maximum, that means that for each value for a number of points, we will always end up using all the points.

  16. Proof – finding the formula • We will always obtain the maximum value of n for any number of points. • Therefore, have minimum, as: • Assume x1 is any value of the minimum number of points needed . • Assume n1 is a value of n, which is the maximum number of lines that can be drawn with x1 points, which our formula will obtain. • E.g. if x1 is 3, then n1 would be 3.

  17. Proof – finding the formula • Assume n2 is the next value of n, after n1, which has to be the minimum number of lines to be drawn for x2, the next value of the minimum no. of points after x1. • n3 is next consecutive value of n after n2, and serves the same relationship to x2that n1 serves to x1. • n4, next consecutive value of n after n3,serves same relationship to x3 , next consecutive value of x after x2,that n2 serves x2. • So on and so forth. • As we have proven that we will have the maximum, this process will repeat forever.

  18. Proof – finding the formula • For any value of x, a given set of the minimum number of points, we will always start with the minimum possible value of n, the minimum number of lines that can be drawn with it, and end with the maximum possible value of n, the maximum number of points that can be drawn with it. • That means that we are always utilizing all the points. • Therefore, with our formula, we have the minimum.

  19. Further Research • How else to add value to what we have done? • Changing the conditions • Line passes through 3 points • Intersection point counted as a point • All provide: • Different experience • Different perspective

  20. Further Research • Changing the question • Minimum number of lines to connect n points • However: • Experience something simple first

  21. Acknowledgements • Dr. Soon, the expert mentor • Mr. Goh for helping us • Families of members who encouraged them • Our Math Teachers who guided us • Our fellow friends in the Mentorship programme

  22. Thank You for your kind attention! Any questions? A research and presentation by Cai Yi Zhan, Tan Wei Chuan, Darryll Chong, Lim Jan Jay and Ryan Wee

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