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TRIANGLES, PROBABILITY, AND AMAZEMENT A CONNECTED EXPERIENCE FOR THE CLASSROOM JIM RAHN WWW.JAMESRAHN.COM [email protected] Question: If we place six(6) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? Smith, Richard J., “Equal Arcs, Triangles, and Probability, Mathematics Teacher, Vol. 96, No. 9, December 2003, pp. 618-621. Make a List of ALL Possible Triangles that can be formed using three of these points. MAKE a List! Triangle Vertices Triangle Vertices ABC DEF ABD DEA ABE DEB ABF DEC BCD EFA BCE EFB BCF EFC BCA EFD CDE FAB CDF FAC CDA CDB Remove all duplicates FAD FAE TOTAL: 20 Triangles. How many are Right Triangles? Triangle Vertices ABC What is necessary to be guaranteed a right triangle? Triangle Vertices BCD ABD BCE ABE BCF ABF BDE ACD BDF ACE CDE ACF CDF ADE CEF ADF DEF AEF EFB There are three diameters Line Segments AD, BE and CF TOTAL: 20 Triangles. How many are Right Triangles? Triangle Vertices Triangle Vertices ABC BCD ABD BCE ABE BCF ABF BDE ACD BDF ACE CDE ACF CDF ADE CEF ADF DEF AEF EFB How many right triangles can be formed with diameter AD? TOTAL: 20 Triangles. How many are Right Triangles? Triangle Vertices ABC ABD ABE ABF ACD ACE ACF ADE ADF AEF Triangle Vertices BCD BCE BCF BDE BDF CDE CDF CEF DEF EFB How many right triangles can be formed with diameter BE? TOTAL: 20 Triangles. How many are Right Triangles? Triangle Vertices Triangle Vertices ABC BCD ABD BCE ABE BCF ABF BDE ACD BDF ACE CDE ACF CDF ADE CEF ADF DEF AEF EFB How many right triangles can be formed with diameter CF? TOTAL: 20 Triangles. There are 12 right triangles Triangle Diameter Vertices End Points? Triangle Diameter Vertices End Points? ABC BCD ABD AD BCE BE ABE BE BCF CF BDE BE ABF ACD AD BDF ACE CDE ACF CF ADE AD ADF AD AEF Which triangles use diameters AD, BE, or CF? CDF CF CEF CF DEF EFB BE TOTAL: 20 Triangles. There are twelve right triangles Triangle Diameter Vertices End Points? Triangle Diameter Vertices End Points? ABC Obtuse BCD Obtuse ABD AD BCE BE ABE BE BCF CF ABF Obtuse BDE BE ACD AD BDF Equilateral ACE Equilateral CDE Obtuse ACF CF CDF CF ADE AD CEF CF ADF AD DEF Obtuse AEF Obtuse EFB BE Which triangles are equilateral triangles? Which triangles are obtuse? If we place six(6) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? 12 3 20 5 Smith, Richard J., “Equal Arcs, Triangles, and Probability, Mathematics Teacher, Vol. 96, No. 9, December 2003, pp. 618-621. Using simulation to determine the probability that the vertices of a right triangle is form by randomly selecting three points from six(6) evenly spaced points around the circumference of a circle. Place SIX Cubes (two of three different colors) into a bag. Draw out three cubes. If two cubes are of the same color, the triangle is a right triangle! (Repeat 100 times) Experimental Results Right Triangle Non Right Triangle Total Using simulation to determine the probability that the vertices of a right triangle is form by randomly selecting three points from six(6) evenly spaced points around the circumference of a circle. Compare your results. Gather the results from the class. What does it show? Experimental Results Right Triangle 58 Non Right Triangle 42 Total 100 Using simulation to determine the probability that the vertices of a right triangle is form by randomly generating three numbers from three numbers. Opposite vertices will have the same numbers. =1 =3 =2 =2 =3 =1 Using your graphing calculator: Type randint(1, 3, 3). This means you will be selecting three numbers from 1,2, and 3. If two digits are the same number, the triangle is a right triangle! (Repeat 100 times) Experimental Results Right Triangle Non Right Triangle Total Using simulation to determine the probability that the vertices of a right triangle is form by randomly generating three numbers from three numbers. Opposite vertices will have the same numbers. =1 =3 =2 =2 Compare your results. Gather the results from the class. What does it show? =3 =1 Experimental Results Right Triangle 58 Non Right Triangle 42 Total 100 If we place three(3) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? There is only 1 possible triangle and NO Diameters, A Probability three points form a right triangle is 0 B C If we place four(4) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? A There are 4 possible triangles BUT There are TWO Diameters, thus 4 Right Triangles D B C Probability three points form a right triangle is 4/4 =1 If we place five(5) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? There are 10 possible triangles BUT There are NO Diameters, thus NO Right Triangles Probability three points form a right triangle is 0 Number of Total Number of Number of equally spaced Triangles Right Triangles points (T) (R) (N) Probability a Triangle is Right (R/T) 3 1 0 0/1 = 0 4 4 4 4/4 = 1 5 10 0 0/10 = 0 6 20 12 12/20 = 3/5 Odd > 3 0 8 10 12 : What patterns do you see in the Total Number of Triangles (T)? If we place eight(8) evenly spaced points around the circumference of a circle and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? We will need to determine the total number of triangles that can be formed by using three points. Number of Total Triangles How can we determine the total number of triangles? Method 1 for finding the total number of triangles: Number of equally spaced points (N) Total Number of Triangles (T) 3 1 1 4 4 1+3 5 10 1+3+6 6 20 1+3+6+10 8 1+3+6+10+? The total number of triangles is the sum of the first n-2 triangular numbers. The nth triangular number is n 2 n 1 2 Method 2: Use combinations because we are choosing 3 points at random from 8 points. As long as we have the same three points selected there is only one triangle that can be formed. 8 C3 8! 8 7 6 5! 8 7 6 56 3!5! 3 2 1 5! 3 2 How many right triangle can be found? What is necessary for the triangle to be a right triangle? One side of the triangle must be a diameter. How many diameters can be drawn? 4 Diameters How many right triangles can be formed with each diameter? A Each diagonal forms 6 right triangle with the remaining vertices H B How many right triangles can be formed? What is the probability of forming a right triangle when three points are selected at random from 8 points equally spaced around a circle? C G Probability three points form a right triangle is 24/56 = 3/7 F D Number (N) of equally spaced points Total Number (T) of Triangles Number (R) of Right Triangles Probability a Triangle is Right (R/T) 3 1 0 0/1 = 0 4 4 4 4/4 = 1 5 10 0 0/10 = 0 6 20 12 12/30 = 3/5 Odd > 3 8 0 56 24 24/56 = 3/7 10 12 : How can we generalize how many right triangles will be formed? What must be true about the number of points equally spaced around the circle? If n= number of points is an even number, can we determine the number of right triangles formed? n(n 2) n ( n 2) 2 2 Number (N) of equally spaced points Total Number (T) of Triangles Number (R) of Right Triangles Probability a Triangle is Right (R/T) 3 1 0 0/1 = 0 4 4 4 4/4 = 1 5 10 0 0/10 = 0 6 20 12 12/30 = 3/5 Odd > 3 8 0 56 24 24/56 = 3/7 10 12 : Complete the chart for 10 and 12 points equally spaced around the circle. 10 points equally spaced around a circle Number of right triangles Number of total triangles Probability of forming a right triangle 12 points equally spaced around a circle Number of right triangles Number of total triangles Probability of forming a right triangle Number (N) of equally spaced points Total Number (T) of Triangles Number (R) of Right Triangles Probability a Triangle is Right (R/T) 3 1 0 0/1 = 0 4 4 4 4/4 = 1 5 10 0 0/10 = 0 6 20 12 12/30 = 3/5 Odd > 3 0 8 56 24 24/56 = 3/7 10 40 120 40/120=3/9 12 80 220 60/220=3/11 : What patterns do you observe? What conjecture would you like to make? If n points are equally spaced on the circumference of a circle and if three points are chosen at random, the probability that the three points will form a right triangle is 3 if n is even n-1 0 if n is odd If we place n evenly spaced points around the circumference of a circle, where n is an even number greater than 3, and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? What do you notice about the number of possible triangles? Number Number Using Triangular Using of of Numbers Combinations points Triangles around the n! n (n 1) (n 2) (n 3)! n (n 1) (n 2) circle n C3 3!(n 3 2 11+3 (n 3)! 64C3=4 4 4 3)! 6 20 1+3+6+10 6C3=20 8 56 1+3+6+10+15+21 8C3=56 10 120 1+3+6+10+15+21+28+36 10C3=120 12 220 1+3+6+10+15+21+28+36+ 45+55 12C3=220 Number of points around the circle Number of Triangles Using Triangular Numbers Using Combinations 4 4 1+3 4C3=4 6 20 1+3+6+10 6C3=20 8 56 1+3+6+10+15+21 8C3=56 10 120 1+3+6+10+15+21+28+36 10C3=120 12 220 1+3+6+10+15+21+28+36+ 45+55 (n 2)(n 1) 1 3 6 2 n-2 terms 12C3=220 n n C3 n C3 n! 3!(n 3)! n! n (n 1)(n 2)(n 3)! n (n 1)(n 2) 3!(n 3)! 3!(n 3)! 6 If we place n evenly spaced points around the circumference of a circle, where n is an even number greater than 3, and then randomly select three points to form the vertices of a triangle, what is the probability that the triangle formed is a RIGHT TRIANGLE? How many right triangles will there be? There are n/2 diameters. Each forms with (n - 2) right triangles with the remaining vertices. Thus the number of right triangles is: n (n 2) 2 n(n-2) # of rightΔ 3 2 P(right triangle being formed)= = = n(n-1)(n-2) # of totalΔ n-1 6 Number (N) of equally spaced points Total Number (T) of Triangles Number (R) of Right Triangles Probability a Triangle is Right (R/T) 3 1 0 0/1 = 0 4 4 4 4/4 = 1 5 10 0 0/10 = 0 6 20 12 12/30 = 3/5 Odd > 3 0 8 56 24 24/56 = 3/7 10 120 40 40/120 = 3/9 12 220 60 60/220 = 3/11 : Even (E) number>3 3/(E – 1)