6703
/www/nrich/html/content/id/6703/
1
2011-02-01T00:00:01
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<mdoxml version="1.0"><p>The circle below has seven points spread equally around its circumference. Press start to watch the construction of a seven pointed mystic rose. You can construct different sized roses by using the slider.<br></br>
<br></br>
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<param name="allowFullScreen" value="true"></param>
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<br></br>
Watch the animation for some different sized mystic roses.<br></br>
What did you see? Describe how to construct a mystic rose.<br></br>
<br></br>
Now describe what a completed mystic rose looks like.<br></br>
<br></br>
<strong>Alison and Charlie have been wondering how many lines are needed to draw a 10 pointed mystic rose.</strong><br></br>
<br></br>
Alison wrote down the calculation $9+8+7+6+5+4+3+2+1$.<br></br>
<br></br>
Charlie wrote down the calculation $\frac{10 \times 9}{2}$<br></br>
<br></br>
Who is right? Can you explain how the calculations relate to the diagram?<br></br>
<br></br>
Investigate the number of lines needed in mystic roses of different sizes.<br></br>
How would Alison work them out? How would Charlie do it?<br></br>
Will they always get the same result?<br></br>
<br></br>
<strong>What are the advantages of the alternative methods?</strong><br></br>
<br></br>
How many lines are needed for a 100 pointed mystic rose?<br></br>
<br></br>
Which of the numbers below could be the number of lines needed to draw a very large mystic rose? How many points would each mystic rose have around its circumference?</p>
<ul>
<li>4851</li>
</ul>
<ul>
<li>6214</li>
</ul>
<ul>
<li>3655</li>
</ul>
<ul>
<li>7626</li>
</ul>
<ul>
<li>8656</li>
</ul>
<br></br>
<br></br>
<p><span style="font-style: italic;">You may wish to try the problems</span> <a href="http://nrich.maths.org/public/viewer.php?obj_id=2274&part=" style="font-style: italic;">Picturing Triangle Numbers</a> <span style="font-style: italic;">and</span> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6708&part=" style="font-style: italic;">Handshakes</a><span style="font-style: italic;">. Can you see why we chose to publish these three problems together?</span><br></br>
<br></br>
<span style="font-style: italic;">You may also be interested in reading the article</span> <a href="http://nrich.maths.org/public/viewer.php?obj_id=2478&part=" style="font-style: italic;">Clever Carl</a><span style="font-style: italic;">, the story of a young mathematician who came up with an efficient method for adding lots of consecutive numbers.</span><br></br>
</p>
<p><a href="http://nrich.maths.org/8013">Click here for a poster of this problem</a>.</p></mdoxml>
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<p><span class="editorial">Several students from Orchard Junior
School noticed how the mystic rose was being
constructed:</span></p>
<br></br>
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.<br></br>
<br></br>
On a 6 point mystic rose, you would need to add 5, 4, 3, 2 and 1 to
make how many lines there are.<br></br>
6+5+4+3+2+1 for a 7 point one, and so on. All the answers are
triangular numbers.<br></br>
<br></br>
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.<br></br>
<br></br>
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<br></br>
<br></br>
<p><span class="editorial">Thomas from Wilson's School
wrote:</span></p>
<br></br>
Both Alison and Charlie are right with their calculations.<br></br>
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.<br></br>
The advantage for Charlie is the formula is a lot quicker to do but
Alison's method is more obvious.<br></br>
<br></br>
<p><span class="editorial">Lucie from Munich came to</span> <a href="/content/id/6703/Mystic%20Roses.rtf" class="editorial">the
same conclusion</a><span class="editorial">.</span></p>
<br></br>
<span class="editorial">James from Wilson's School agreed that both
Alison and Charlie were right:</span><br></br>
<br></br>
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.<br></br>
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.<br></br>
<br></br>
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.<br></br>
<br></br>
<span class="editorial">A student from Mearns Castle explained this
well:</span><br></br>
<br></br>
Allison and Charlie are both correct (both equal 45).<br></br>
<br></br>
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.<br></br>
<br></br>
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.<br></br>
<br></br>
1+2+3+4+5...+(n-1) = n(n-1)/2 so they will both be right for any
case.<br></br>
<br></br>
<br></br>
<span class="editorial">Jonathan from Wilson's School
agreed:</span><br></br>
<br></br>
Alison and Charlie are both right, and their methods relate to the
diagram:<br></br>
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.<br></br>
<br></br>
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.<br></br>
<br></br>
For a 100 pointed mystic rose you need 4950 lines as 4950 = 100x99
divided by 2.<br></br>
<br></br>
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.<br></br>
<br></br>
<br></br>
<span class="editorial">Craig wrote:</span> <br></br>
<br></br>
Alison and Charlie have two working methods which will correctly
show the number of lines for any completed mystic rose. <br></br>
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.
<br></br>
<br></br>
<br></br>
<span class="editorial">Janahan, also from Wilson's School,
connected the number of lines to triangle numbers:</span><br></br>
<br></br>
The way in which you know how many lines you need for every mystic
rose depends on how many points it has.<br></br>
If it has 5 points then you find the 4th triangular number.<br></br>
If the mystic rose has 12 points then you find the 11th triangular
number.<br></br>
<br></br>
Example: 5 pointed mystic rose: 4+3+2+1=10 and 10 is the 4th
triangular number.<br></br>
<p><span class="editorial">Esther from St Bede's interchurch school
in Cambridge used these results for the final part of the
problem:</span></p>
<div>No. of points x (No. of points - 1)/2 = No. of lines</div>
<br></br>
<div>So a 100 point rose would have:</div>
<div>(100 x 99) /2 = 9900/2 = 4950 lines.</div>
<br></br>
<div>To work out if the later numbers could be the lines of a
mystic rose, you need to use the method backwards.</div>
<br></br>
<div>4851 x 2 = 9,702</div>
<div>98 x 99 = 9,702</div>
<div>So this number is the number of lines for a 99 point mystic
rose.</div>
<br></br>
<div>6214 x 2 = 12,428</div>
<div>This however, does not have two consecutive numbers that
multiply to make it, so it cannot make a rose.</div>
<br></br>
<div>3655 x 2 = 7310</div>
<div>85 x 86 = 7310</div>
<div>So this number is the number of lines for a 86 point mystic
rose.</div>
<br></br>
<div>7626 x 2 = 15252</div>
<div>123 x 124 = 15252</div>
<div>So this number is the number of lines for a 124 point mystic
rose.</div>
<br></br>
<div>8656 x 2 = 17312</div>
<div>This however, does not have two consecutive numbers that
multiply make it, so it cannot make a rose.</div>
<br></br>
<p><span class="editorial">Mrs Dillon's class from Ashville College
in Harrogate agreed:</span></p>
<div>We think 4851 comes from a mystic rose with 99 points (98 x
99/2)</div>
<div>6214 isn't one - the nearest is 6216 from a 112 pointed rose
(111 x 112/2)</div>
<div>3655 is 86 points (85 x 86/2)</div>
<div>7626 is 124 points (123 x 124/2)</div>
<div>8656 isn't one - the nearest is 8646 from a 132 pointed rose
(131 x 132/2)</div>
<br></br>
<div>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.</div>
<br></br>
<br></br>
<span class="editorial">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.</span><br></br></mdoxml>
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<h3>Why do this problem?</h3>
<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6703&part=">
This problem</a> offers students an opportunity to relate numerical
ideas to spatial representation, and vice versa.</div>
<div>Thinking about different ways of counting the number of lines
in a mystic rose can lead to a better understanding of the general
formula for triangle numbers.</div>
<h3>Possible approach</h3>
<div>This problem works very well in conjunction with <a href="http://nrich.maths.org/public/viewer.php?obj_id=2274&part=">
Picturing Triangle Numbers</a> and <a href="http://nrich.maths.org/public/viewer.php?obj_id=6708&part=">
Handshakes</a>. The whole class could work on all three problems
together, or small groups could be allocated one of the three
problems to work on, and then report back to the rest of the
class.</div>
<br></br>
<div>Start by showing the animation of the seven point mystic rose.
Then reset it and ask the students to describe to their partners
what they saw. Choose a different mystic rose and show the
animation, pausing it as it plays. Ask the class to predict what
will happen at each stage. Can they predict how many lines will be
drawn in total?</div>
<br></br>
<div>Set the class the challenge of working out how many lines are
needed to draw 8, 9, and 10 point mystic roses. Allow
them some time to work on this. Bring the class together to discuss
their answers and methods, and more importantly, how their methods
relate to the construction of the mystic rose.</div>
<br></br>
<div>Set the class another challenge, this time to work out how
many lines are needed to draw a much larger mystic rose
(e.g. a 161 point mystic rose). Allow them
some time to work on this. When they report back, discuss the need
for efficient ways of working this out. Draw attention to
"Charlie's method" in the problem, if no-one has suggested it, and
encourage students to think about how this method relates to the
image of the completed mystic rose.</div>
<br></br>
<div>For a class that has been introduced to algebra, students
could express "Alison's method" and "Charlie's method"
algebraically. </div>
<br></br>
<div>Finally, ask them to work out which of the following numbers
of lines could be used to draw mystic roses:</div>
<ul>
<li>4851</li>
<li>6214</li>
<li>3655</li>
<li>7626</li>
<li>8656 </li>
</ul>
<h3></h3>
<h3>Key questions</h3>
<div>What is special about the numbers of lines needed for
different sizes of mystic rose?<br></br>
How do the different ways of working out the number of lines relate
to the construction and final image of the mystic rose?<br></br>
</div>
<br></br>
<h3>Possible extension</h3>
<div>Can you draw a mystic rose using 9, 19, 29, 39, ... lines? Are
these impossible? How do you know?</div>
<br></br>
<div>Will there ever be a mystic rose constructed from a multiple
of 1000 lines?</div>
<h3>Possible support</h3>
<div>Students could construct their own mystic roses using
different colours for the lines from each point, to build up an
understanding of their structure. Circle templates with dots evenly
spaced on the circumference can be found <a href="http://nrich.maths.org/public/viewer.php?obj_id=6676&part=#CircleTemplates">
here</a>.</div>
<br></br></mdoxml>
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<mdoxml version="1.0">How many lines are drawn at each stage of construction of the
mystic rose?<br></br>
Now look at the completed mystic rose - can you describe how many
lines are drawn from each point? <br></br>
How do these relate to "Alison's method" and "Charlie's method"?
<br></br>
<br></br></mdoxml>
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Mystic Rose
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
Numbers and the Number System
Triangle numbers
Using, Applying and Reasoning about Mathematics
Visualising
Using, Applying and Reasoning about Mathematics
Generalising
Information and Communications Technology
Animations
Secondary Mapping Document
Patterns and sequences US
Secondary Mapping Document
DisplayCabinet