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I found out that the first point connects to all of the points, then the rest of the points went down 1 until there is none to connect.
On a 6 point mystic rose, you would need to add 5, 4, 3, 2 and 1 to make how many lines there are.
6+5+4+3+2+1 for a 7 point one, and so on. All the answers are triangular numbers.
I discovered that the first set of lines that you do will have 1 less than the amount of points in the mystic rose and then you just keep on taking away one.
I worked out that every number of lines that come up at the same time decreases by one each time. e.g 7+6+5+4+3+2+1
Both Alison and Charlie are right with their calculations.
Alison would work out other mystic roses by starting with (n-1) lines and then decrease by 1 each time and adding them up. Charlie would use the formula. They should always get the same result.
The advantage for Charlie is the formula is a lot quicker to do but Alison's method is more obvious.
James from Wilson's School agreed that both Alison and Charlie were right:
Alison's theory relates to the diagram because each point has to join every other point once, so if we consider a 10 pointed mystic rose, the first point connects to all the other 9 points, the second joins up with 8 points (because it's all ready joined with the first), etc, etc.
Charlie's method relates to the diagram in the way that there are 10 points around the circle, and each one must be connected with a line to the other 9. But he must not count every line twice, so after multiplying 10 by 9, Charlie divides by 2.
Alison would work them out by taking the number of points on the circle, then adding all the consecutive numbers before that number (not including the number of points on the circle). Charlie would work them out by, every time, multipling the number of points on a circle with one less than its number, then dividing the outcome by 2. They will always get the same result.
A student from Mearns Castle explained this well:
Allison and Charlie are both correct (both equal 45).
It seems that Alison saw that n-1 lines were drawn from point 1, n-2 lines were drawn from point 2 and so on, which in this case is 9+8+7+6+5+4+3+2+1.
It seems Charlie looked at the completed mystic rose, and saw that there were n-1 (9) lines drawn from each of the n (10) points, so multiplying them together would give the right answer. However, this includes each line twice, thus you must divide by 2.
1+2+3+4+5...+(n-1) = n(n-1)/2 so they will both be right for any case.
Jonathan from Wilson's School agreed:
Alison and Charlie are both right, and their methods relate to the diagram:
Alison is counting the lines from 9 lines at the start to 8 then 7 and so on, whereas Charlie has devised a method which counts the points and the amount of "first lines", times them and then divides by two.
For a mystic rose of 25 points Alison would go 24+23+22+21+20+19+18...until she got to 1. Whereas Charlie would do 25x24 divided by 2, so they will always get the same result, but Charlie's method is quicker.
For a 100 pointed mystic rose you need 4950 lines as 4950 = 100x99 divided by 2.
The advantage of alternative methods is that you can compare them to see if you always get the same result and if that answer is right.
Craig wrote:
Alison and Charlie have two working methods which will correctly show the number of lines for any completed mystic rose.
Charlie's has an actual mathematic formula which allows faster calculation of the answer, and is what is used by mathematicians. As the number of points increases, Alison would take longer to work out the number of lines, as the number of calculations needed will increase. Charlie would simply be working with two larger numbers.
Janahan, also from Wilson's School, connected the number of lines to triangle numbers:
The way in which you know how many lines you need for every mystic rose depends on how many points it has.
If it has 5 points then you find the 4th triangular number.
If the mystic rose has 12 points then you find the 11th triangular number.
Example: 5 pointed mystic rose: 4+3+2+1=10 and 10 is the 4th triangular number.
Similar solutions with similar good explanations were sent by Matthew and Amar from Wilson's School and Adam from Bradon Forest School. Well done to you all.
Jack from Athol Road Primary School in Springvale South, Victoria, Australia sent us the solution to the Final Challenge about a Chess tournament. Here is his solution:
Mystic Rose
Age 14 to 16
Challenge Level 
Several students from Orchard Junior School noticed how the mystic rose was being constructed:
I found out that the first point connects to all of the points, then the rest of the points went down 1 until there is none to connect.
On a 6 point mystic rose, you would need to add 5, 4, 3, 2 and 1 to make how many lines there are.
6+5+4+3+2+1 for a 7 point one, and so on. All the answers are triangular numbers.
I discovered that the first set of lines that you do will have 1 less than the amount of points in the mystic rose and then you just keep on taking away one.
I worked out that every number of lines that come up at the same time decreases by one each time. e.g 7+6+5+4+3+2+1
Thomas from Wilson's School wrote:
Both Alison and Charlie are right with their calculations.
Alison would work out other mystic roses by starting with (n-1) lines and then decrease by 1 each time and adding them up. Charlie would use the formula. They should always get the same result.
The advantage for Charlie is the formula is a lot quicker to do but Alison's method is more obvious.
Lucie from Munich came to the same conclusion.
James from Wilson's School agreed that both Alison and Charlie were right:
Alison's theory relates to the diagram because each point has to join every other point once, so if we consider a 10 pointed mystic rose, the first point connects to all the other 9 points, the second joins up with 8 points (because it's all ready joined with the first), etc, etc.
Charlie's method relates to the diagram in the way that there are 10 points around the circle, and each one must be connected with a line to the other 9. But he must not count every line twice, so after multiplying 10 by 9, Charlie divides by 2.
Alison would work them out by taking the number of points on the circle, then adding all the consecutive numbers before that number (not including the number of points on the circle). Charlie would work them out by, every time, multipling the number of points on a circle with one less than its number, then dividing the outcome by 2. They will always get the same result.
A student from Mearns Castle explained this well:
Allison and Charlie are both correct (both equal 45).
It seems that Alison saw that n-1 lines were drawn from point 1, n-2 lines were drawn from point 2 and so on, which in this case is 9+8+7+6+5+4+3+2+1.
It seems Charlie looked at the completed mystic rose, and saw that there were n-1 (9) lines drawn from each of the n (10) points, so multiplying them together would give the right answer. However, this includes each line twice, thus you must divide by 2.
1+2+3+4+5...+(n-1) = n(n-1)/2 so they will both be right for any case.
Jonathan from Wilson's School agreed:
Alison and Charlie are both right, and their methods relate to the diagram:
Alison is counting the lines from 9 lines at the start to 8 then 7 and so on, whereas Charlie has devised a method which counts the points and the amount of "first lines", times them and then divides by two.
For a mystic rose of 25 points Alison would go 24+23+22+21+20+19+18...until she got to 1. Whereas Charlie would do 25x24 divided by 2, so they will always get the same result, but Charlie's method is quicker.
For a 100 pointed mystic rose you need 4950 lines as 4950 = 100x99 divided by 2.
The advantage of alternative methods is that you can compare them to see if you always get the same result and if that answer is right.
Craig wrote:
Alison and Charlie have two working methods which will correctly show the number of lines for any completed mystic rose.
Charlie's has an actual mathematic formula which allows faster calculation of the answer, and is what is used by mathematicians. As the number of points increases, Alison would take longer to work out the number of lines, as the number of calculations needed will increase. Charlie would simply be working with two larger numbers.
Janahan, also from Wilson's School, connected the number of lines to triangle numbers:
The way in which you know how many lines you need for every mystic rose depends on how many points it has.
If it has 5 points then you find the 4th triangular number.
If the mystic rose has 12 points then you find the 11th triangular number.
Example: 5 pointed mystic rose: 4+3+2+1=10 and 10 is the 4th triangular number.
Esther from St Bede's interchurch school in Cambridge used these results for the final part of the problem:
No. of points x (No. of points - 1)/2 = No. of lines
So a 100 point rose would have:
(100 x 99) /2 = 9900/2 = 4950 lines.
To work out if the later numbers could be the lines of a mystic rose, you need to use the method backwards.
4851 x 2 = 9,702
98 x 99 = 9,702
So this number is the number of lines for a 99 point mystic rose.
6214 x 2 = 12,428
This however, does not have two consecutive numbers that multiply to make it, so it cannot make a rose.
3655 x 2 = 7310
85 x 86 = 7310
So this number is the number of lines for a 86 point mystic rose.
7626 x 2 = 15252
123 x 124 = 15252
So this number is the number of lines for a 124 point mystic rose.
8656 x 2 = 17312
This however, does not have two consecutive numbers that multiply make it, so it cannot make a rose.
Mrs Dillon's class from Ashville College in Harrogate agreed:
We think 4851 comes from a mystic rose with 99 points (98 x 99/2)
6214 isn't one - the nearest is 6216 from a 112 pointed rose (111 x 112/2)
3655 is 86 points (85 x 86/2)
7626 is 124 points (123 x 124/2)
8656 isn't one - the nearest is 8646 from a 132 pointed rose (131 x 132/2)
At first we tried multiplying random numbers and dividing by 2 to see if we could get the answer. Then to save time, we doubled the answer and tried to find the consecutive numbers that gave this. We then used our square root button to help us. We square rooted double the answer and then tried the 2 whole numbers either side. This method worked and was quick to do.
Similar solutions with similar good explanations were sent by Matthew and Amar from Wilson's School and Adam from Bradon Forest School. Well done to you all.
Jack from Athol Road Primary School in Springvale South, Victoria, Australia sent us the solution to the Final Challenge about a Chess tournament. Here is his solution:
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