BEYOND PYTHAGORAS By: Megan Garibian 0A What this coursework has asked me to do is to investigate and find a generalisation, for a family of Pythagorean triples. This will include odd numbers and even numbers. I am going to investigate a family of right-angled triangles for which all the lengths are positive integers and the shortest is an odd number. I am going to check that the Pythagorean triples (5,12,13) and (7,24,25) cases work; and then spot a connection between the middle and longest sides. The first case of a Pythagorean triple I will look at is: The numbers 5, 12 and 13 satisfy the connection. 5² + 12² = 13² 25 + 144 = 169 69 = 13 The second case of a Pythagorean triple I will look at is: The numbers 7, 24 and 25 satisfy the connection. 7² + 24² = 25² 49 + 576 = 625 625 = 25 There is a connection between the middle and longest side. This is that there is a one number difference. So if M= middle and L= longest L = M + 1 I am going to use the triples, (3,4,5), (5,12,13) and (7,24,25) to find other triples. Then I will put my results in a table and look for a pattern that will occur. I will then try and predict the next results in the table and prove it. n Smallest ( Middle ( Longest ( 3 4 5 2 5 2 3 3 7 24 25 There is a clear pattern between the middle and longest side. There is also a sequence forming. n = 1 S = 3 M
Beyond Pythagoras The aim of this investigation is to investigate Pythagoras theorem and to find a formula for the shortest side, middle length, hypotenuse, area and perimeter. Because I have typed this up on a computer... ^ is squared * is times / is divided Pythagoras is A2 + B2 = C2 I am going to prove this theory be finding out if the following numbers adhere to the rule. Triangle 1 5 12 13 5^ + 12^ = 13^ 5^ = 5*5 = 25 2^ = 12*12 = 144 3^ = 13*13 = 169 So 5^ + 12^ = 25 + 144 = 169 = 25^ The perimeter of the triangle is All the lengths of the side added up 5 + 24 + 25 = 30 The area of the triangle is /2 base * height /2 * 12 * 5 = 30 Triangle 2 7 24 25 7^ + 24^ =25^ 7^ = 7*7 = 49 24^ = 24*24 = 576 25^ = 25*25 = 625 So 7^ + 24^ = 49 + 576 = 625 = 25^ The perimeter of the triangle is All the lengths of the side added up 7 + 24 + 25 = 56 The area of the triangle is /2 base * height /2 * 24 * 7 = 84 Length of shortest side Length of middle side Length of longest side Perimeter Area 3 4 5 2 6 5 2 3 30 30 7 24 25 56 84 9 40 41 90 80 1 60 61 32 330 3 84 85 82 546 5 12 12 239 840 7 44 44 305 224 Please find enclosed "sheet 1" To create this I used excel to find the Pythagorean triangles Basically I created one horizontal line of numbers going up one at a time and another vertical line the same.
Pythagoras was a famous Greek mathematician. He produced a formula (a² + b² = c²) called the 'Pythagoras Theorem.' This allows you to correctly calculate the length of a right-angled triangles' hypotenuse, when you know the length of the other two sides. A Pythagorean triple is when a right-angled triangles three lengths (a = shortest, b = medium, c = hypotenuse) are all whole numbers- integers. At the start of this course, I was given a set of 3 Pythagorean triples. These were: A B C 3 4 5 5 2 3 7 24 25 I was asked to look for patterns between the numbers in each Pythagorean triple. This is what I found: * The difference between b and c is one * a was odd * b was even and a multiple of four * c was odd * c = b + 1 * (a² + 1)/2 = c * (a² - 1)/2 = b The two formulae that I am going to focus on, and use to discover more Pythagorean Triples are: * (a² + 1)/2 = c * (a² - 1)/2 = b Using these two rules I can use some more odd numbers to expand my table to: A B C 9 40 41 1 60 61 3 84 85 5 12 13 To test that these were possible Pythagorean triples, I put them through the original Pythagoras theory: * 9² + 40² = 1,681 V1,681 = 41 * 11² + 60² = 3,721 V3,721 = 61 * 13² + 84² = 7,225 V7,225 = 85 * 15² + 112² = 12,769
Investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers.
Beyond Pythagoras The aim of this piece of my coursework is to investigate the area of triangle studies including the Pythagorean Theorem and in particular Pythagorean Triples, sets of numbers where the shortest side is an odd value and all three are positive whole integers. I will collate the relevant data and formulae for the nth terms by using a grid and from this data I should be able to make predictions on the nth terms of Pythagorean Triples. I will keep a narrative of what I am doing and discovering, referring within it to each process of my investigation. The numbers 3, 4 and 5 satisfy the condition 32+42=52, Because 32= 3x3 =9 42= 4x4 =16 52= 5x5 =25 And so... 32+42=9+16=25=52 I now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) 2+ (middle number) 2= (largest number) 2. * 5, 12, 13 52+122 = 25+144 = 169 = 132. * 7, 24, 25 72+242 = 49+576 = 625 +252 Here is a table containing the results: a b c area perimeter 3 5 7 9 1 4 2 24 40 60 5 3 25 41 61 6 3 84 80 330 2 30 56 90 32 2n +1 2n² + 2n 2n³ + 2n + n² +n 2n³ + 2n + n² + n (2n + 1) + (2n (n+1)) + (2n² + (2n + 1) I looked at the table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side. I already know that the (smallest number)² + (middle
Investigating families of Pythagorean triples.
I was set the task of investigating families of Pythagorean triples. I started off by investigating families of Pythagorean triples where the shortest side is an odd number and c is always equal to b+1. I then extended my research into other families of triples. Finally, I came up with a formula that can calculate all Pythagorean triples. Pythagoras's formula is as follows: a2 + b2 = c2 This is referring to such a triangle: The triangle always has one right angle and c is always the longest side, also known as the hypotenuse. An example of this is with the triple 3,4 and 5: Here, a is 3, b is 4 and c is 5. 52 = 32 + 42 25 = 9 + 16 I will now investigate the Pythagorean triples where c is always equal to b+1. The following table shows the first ten cases, along with the areas and perimeters: N a b c P A a2 b2 c2 3 4 5 2 6 9 6 25 2 5 2 3 30 30 25 44 69 3 7 24 25 56 84 49 576 625 4 9 40 41 90 80 81 600 681 5 1 60 61 32 330 21 3600 3721 6 3 84 85 82 546 69 7056 7225 7 5 12 13 240 840 225 2544 2769 8 7 44 45 306 224 289 20736 21025 9 9 80 81 380 710 361 32400 32761 0 21 220 221 462 2310 441 48400 48841 By looking at the numbers for a, I noticed that they are all odd numbers. The basic formula for odd numbers is: 2n + 1 When I applied this formula to the numbers in the
Investigating Towers of Hanoi Mathematics coursework
Investigating Towers of Hanoi Mathematics coursework Surname: Thomas First name: Jawhari Candidate no: 7160 Centre no: 14368 The legend In an ancient city in a temple monks were given the task of moving a pile of 64 sacred disks from one location to another. The disks are fragile; only one can be carried at a time. A disk may not be placed on top of a smaller, less valuable disk. And, there is only one other location in the temple, (besides the original and destination locations) sacred enough that a pile of disks can be placed there. The monks start moving disks back and forth, between the original pile, the pile at the new location, and the intermediate location, always keeping the piles in order (largest on the bottom, smallest on the top). The legend is that, before the monks make the final move to complete the new pile in the new location, the temple will turn to rubble and the world will end. The Game. There's a game based on this legend. You have collection of disks and three piles into which you can place them. The left most pile is the starting pile you need to move the discs to the right most pile, never putting a disk on top of one, which is smaller. The middle pile is there for intermediate storage. The piles work like a triangle and a disc can be moved from one pile to any other pile. Looking for the Pattern. To figure out how many turns it
For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician.
BEYOND PYTHAGORAS Introduction For this piece of work I am investigating Pythagoras. Pythagoras was a Greek mathematician. Pythagoras lived on the island of Samos and was born around 569BC. He did not write anything but he is regarded as one of the world's most important characters in maths. His most famous theorem is named after him and is called the Pythagoras Theorem. It is basically a²+b²=c². This is what the coursework is based on. I am going to look at the patterns, which surround this theorem and look at the different sequences that can be formed. The coursework The numbers 3, 4 and 5 satisfy the condition 3²+4²=5² because 3²=3x3=9 4²=4x4=16 5²=5x5=25 And so 3²+4²=9+16=25=5² I will now check that the following sets of numbers also satisfy the similar condition of (smallest number) ²+(middle number) ²=(largest number) ² a) 5, 12, 13 5²=5x5=25 12²=12x12=144 25+144=169 V169 = 13 This satisfies the condition as 5²+12²=25+144=169=13² b) 7, 24, 25 7²=7x7=49 24²=24x24=576 49+576=625 V625=25 This satisfies the condition as 7²+24²=49+576=625=25² The numbers 3,4 and 5 can be the lengths - in appropriate units - of the side of a right-angled triangle. 5 3 The perimeter and area of this triangle are: Perimeter = 3+4+5=12 units Area = 1/2x3x4=6 units ² The numbers 5,12,13 can
Dice Game Three players A, B and C play a game with a single dice. The rules of the game are: Player A ALWAYS goes first. A rolls the dice. If the dice lands showing a 1 then A wins the game. If A does not role a 1 then B takes a turn. B rolls the dice. If the dice lands showing a 2 or 3 then B wins the game. If B does not roll a 2 or 3 then C takes a turn. C rolls the dice. If the dice lands showing a 4, 5 or a 6 then C wins the game. If C does not roll a 4,5 or a 6 then A starts again. This procedure continues until there is a winner. Investigate any or all of: The probabilities of each of A, B or C winning the game. Who will be the most likely winner? 3. The most likely length of the game in terms of the number of rolls of the dice to produce a winner. Playing the game. Before we started to work out all the probabilities for the questions we played the game to give us an idea of what th outcome should roughly be. Here are the results : ROLLABCROLLABC1WIN21WIN2WIN22WIN3WIN23WIN4WIN24WI N5WIN25WIN6WIN26WIN7WIN27WIN8WIN28WIN9WIN29WIN10WIN 30WIN11WIN31WIN12WIN32WIN13WIN33WIN14WIN34WIN15WIN35 WIN16WIN36WIN17WIN37WIN18WIN38WIN19WIN39WIN20WIN40WI N4115398 ABC72013 As you can see from this pie chart, in practice it suggests that B is the most likely player yo win the game. From looking at this I would predict that when I have worked
Maths Number Patterns Investigation
The numbers 3, 4 and 5 satisfy the condition 32+42=52, Because 32= 3x3 =9 42= 4x4 =16 52= 5x5 =25 And so... 32+42=9+16=25=52 I now have to find out if the following sets of numbers satisfy a similar condition of (smallest number) 2+ (middle number) 2= (largest number) 2. a) 5, 12, 13 52+122 = 25+144 = 169 = 132. b) 7, 24, 25 72+242 = 49+576 = 625 +252 Here is a table containing the results: I looked at the table and noticed that there was only a difference of 1 between the length of the middle side and the length of the longest side. I already know that the (smallest number) 2+ (middle number) 2= (largest number) 2. So I know that there will be a connection between the numbers written above. The problem is that it is obviously not: (Middle number) 2+ (largest number) 2= (smallest number) 2 Because, 122 + 132 = 144+169 = 313 52 = 25 The difference between 25 and 313 is 288 which is far to big, so this means that the equation I want has nothing to do with 3 sides squared. I will now try 2 sides squared. (Middle)2 + Largest number = (smallest number)2 = 122 + 13 = 52 = 144 + 13 = 25 = 157 = 25 This does not work and neither will 132, because it is larger than 122. There is also no point in squaring the largest and the smallest or the middle number and the largest number. I will now try 1 side squared. 22 + 13 = 5 This couldn´t work because 122 is already
Beyond Pythagoras .
Beyond Pythagoras Pythagoras of Samos is often described as the first pure mathematician. He is an extremely important figure in the development of mathematics yet we know relatively little about his mathematical achievements. Unlike many later Greek mathematicians, where at least we have some of the books, which they wrote, we have nothing of Pythagoras's writings. The society which he led, half religious and half scientific, followed a code of secrecy which certainly means that today Pythagoras is a mysterious figure. As a child Pythagoras spent his early years in Samos but travelled widely with his father although little is known of his childhood. All accounts of his physical appearance are likely to be fictitious except the description of a striking birthmark that Pythagoras had on his thigh. It is probable that he had two brothers although some sources say that he had three. He was well educated, learning to play the lyre, learning poetry and to recite Homer, a poet who wrote the Iliad and Odyssey among his work. There were, among his teachers, three philosophers who were to influence Pythagoras while he was a young man. One of the most important was Pherekydes who many describe as the teacher of Pythagoras. Another teacher of his was Thales who was said to have first introduced him to mathematical ideas. Although he created a strong impression on Pythagoras, he