2670
/www/nrich/html/content/id/2670/
1
2011-02-01T00:00:01
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<mdoxml version="1.0"><br></br>
An arithmagon is a polygon with numbers at its vertices which determine the numbers written on its edges.<br></br>
<br></br>
Check you understand how the values at the vertices determine the values on the edges in the arithmagons generated by the interactivity: <br></br>
<div style="position:relative;"><mdo:image id="arithmagon" src="arith.jpg"></mdo:image>
<div id="dcircle1"> </div>
<div id="dcircle2"> </div>
<div id="dcircle3"> </div>
<div id="drect1"> </div>
<div id="drect2"> </div>
<div id="drect3"> </div>
<input id="reveal" onclick="reveal()" type="button" value="Reveal"></input> <input id="dnewquestion" onclick="dgenerate()" type="button" value="New Numbers"></input></div>
<br></br>
<br></br>
<div>If you are given the values on the edges, can you find a way to work out what values belong at the vertices?</div>
<div>Use the interactivity below to test out your strategies.</div>
<div>There are four different challenge levels to try.</div>
<div> </div>
<div style="position:relative;"><mdo:image id="arithmagon" src="arith.jpg"></mdo:image> <select id="gamechoice">
<option value="level1">Level 1</option>
<option value="level2">Level 2</option>
<option value="level3">Level 3</option>
<option value="level4">Level 4</option>
</select> <input id="circle1" type="text"></input> <input id="circle2" type="text"></input> <input id="circle3" type="text"></input>
<div id="rect1"> </div>
<div id="rect2"> </div>
<div id="rect3"> </div>
<input id="check" onclick="check()" type="button" value="Check"></input>
<div id="feedback"> </div>
<input id="newquestion" onclick="generate()" type="button" value="New Numbers"></input></div>
<script src="2670-script.js" type="text/javascript">
</script>
<div> </div>
<div> </div>
<div> </div>
<div style="font-weight: bold;">Once you are confident that you can work out the values at the vertices efficiently, here are some questions you might like to consider:</div>
<div style="font-weight: bold;"> </div>
<ul>
<li style="font-weight: bold;"><span style="font-weight: 400;">Can you describe a strategy to work out the values at the vertices irrespective of the values given for the edges?</span><br></br>
</li>
<li style="font-weight: bold;"><span style="font-weight: 400;">By looking at the numbers on the edges, how can you predict whether the numbers at the vertices will all be positive?</span><br></br>
</li>
<li style="font-weight: bold;"><span style="font-weight: 400;">By looking at the numbers on the edges, how can you predict whether the numbers at the vertices will all be whole numbers?</span><br></br>
</li>
<li style="font-weight: bold;"><span style="font-weight: 400;">What happens to the numbers at the vertices if you increase one or more of the numbers on the edges by 1? By 2? ... By n?</span><br></br>
</li>
<li style="font-weight: bold;"><span style="font-weight: 400;">What relationships can you find in square arithmagons? Or pentagonal arithmagons?</span></li>
</ul>
<br></br>
<div>A poster of this resource is available <a href="http://nrich.maths.org/public/viewer.php?obj_id=2803&part=">here</a>.</div>
<br></br></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><span class="editorial"></span><br></br>
<span class="editorial">Esther did some wonderful "detective work"
and made a number of very useful observations:</span><br></br>
<br></br>
1) The totals in the squares are double the totals in the circles.
I think this is because each circle number is used twice to solve
the puzzle.<br></br>
<br></br>
2) If the numbers in the squares are all even, the numbers in the
circles are either all odd or all even.<br></br>
<br></br>
If the numbers in the squares are all odd, I cannot solve the
puzzle with whole numbers. This is because an even number minus an
odd number equals an odd number.<br></br>
<br></br>
If you have two odd and one even in the squares then the numbers
in the circles are either two odd and one even or two even and one
odd.<br></br>
<br></br>
However, if you have two even and one odd number in the squares you
cannot solve the puzzle using whole numbers.<br></br>
<br></br>
3) If you draw the lines of symmetry through a circle and the
opposite square, the difference in the numbers in the squares on
either side equals the difference in the numbers in the circles on
either side.<br></br>
<br></br>
Also the sum of the numbers on each line of symmetry is the same.
This also works for decimals and negative numbers.<br></br>
<br></br>
4) I found a rule for working out the numbers in the circles. You
add the numbers in the squares next to it, subtract the number in
the opposite square, and divide the answer by two. This is why you
sometimes get decimals and negative numbers. I have only ever found
one solution to each puzzle using my rule.<br></br>
<br></br>
I think you can always get answers as long as you can use decimals
and negative numbers using my rule as zero can always be made of +2
and -2 for example.<br></br>
<br></br>
<span class="editorial">Harriet and Laura from The Mount School
offered the following explanation:</span><br></br>
<br></br>
We found that when you added together the numbers in the squares,
(the ones already given) it totalled twice the accumulated amount
of the numbers in the circles. The reason for this can be shown
using algebra and is illustrated in the diagram below. <br></br>
<br></br>
<mdo:image width="250" height="224" src="triarith1.gif" alt="arithmagon"></mdo:image><br></br>
<br></br>
You can find the numbers being represented by $a$, $b$ or $c$ if
you have numbers in the squares.<br></br>
We will show this by finding b using an example found on the
website:<br></br>
<br></br>
Using the equation $(b + a) + (b + c) = 2b + a + c $<br></br>
you can take away $a$ and $c$ to find $2b$:<br></br>
$(2b + a + c) - (a + c) = 2b $<br></br>
So $(b + a) + (b + c) - (a + c) = 2b $<br></br>
<br></br>
<mdo:image width="250" height="224" alt="arithmagon" src="triarith2.gif"></mdo:image><br></br>
<br></br>
$2b = 12 + 4 - 10 = 6 $<br></br>
$b = 3 $<br></br>
<br></br>
This also works with negative numbers.<br></br>
<br></br>
<span class="editorial">Robert from Leventhorpe School used a
similar analysis:</span><br></br>
<br></br>
To work out the formula for working out these arithmagons, we must
first substitute the numbers as such: <br></br>
<br></br>
<mdo:image width="250" height="224" alt="arithmagon" src="triarith3.gif"></mdo:image><br></br>
<br></br>
If we know $b, c$ and $e$, then we can say that<br></br>
$a + d = b$,<br></br>
$d + f = e$ and<br></br>
$a + f = c$<br></br>
<br></br>
In the first one, $b = 12, c =10$ and $e = 4$,<br></br>
so $a + d = 12$, $d + f = 4$ and $a + f = 10$<br></br>
<br></br>
If we then add all of these together,<br></br>
we get $(a+d) + (d+f) + (a+f) = 12 + 4 + 10 $<br></br>
<br></br>
We do not need the brackets,<br></br>
so we can make it $a + d + d + f + a + f = 26 $<br></br>
<br></br>
This can be cancelled down to $2a + 2d + 2f = 26 $<br></br>
<br></br>
If we then divide everything by $2$ we get $a + d + f = 13$.<br></br>
<br></br>
Because we know that $a + d = 12$, $f$ must be $1$.<br></br>
Since $d + f = 4$, $d$ must be $3$.<br></br>
Finally, we can say that a must be $9$.<br></br>
<br></br>
Double checking this, it works. Take a look:<br></br>
<br></br>
<mdo:image width="250" height="224" src="triarith4.gif" alt="arithmagon"></mdo:image><br></br>
<br></br>
<span class="editorial">Tom from Colyton Grammar School also used
some algebraic thinking to analyse the problem:</span><br></br>
<br></br>
If the circles were $z, y$ and $x$ from the top clockwise and the
squares are $q, r$ and $p$ from the left clockwise then<br></br>
$x + z = q$,<br></br>
$z + y = r$<br></br>
and $x + y = p$.<br></br>
$q - r = x - y$<br></br>
so $q + p - r = 2x$<br></br>
so $x = (q + p - r) / 2$.<br></br>
<br></br>
This means that $y = ( p + r - q ) / 2$<br></br>
and $z = (q + r - p) / 2$, by symmetry.<br></br>
<br></br>
<span class="editorial">Shaun from Nottingham High School came to
the same conclusion:</span><br></br>
<br></br>
It struck me this problem would be most easily investigated using
simultaneous equations as a means of deriving a common rule for all
arithmagons.<br></br>
I will call the three circles of the arithmagon $a, d$ and
$f$.<br></br>
Hence, each squares can be expressed as the sum of two of the
variables.<br></br>
Since we have three variables, and three equations involving two of
them each, we know it can be solved this way.<br></br>
Some preliminary doodlings proved this to be true.<br></br>
Let the squares be defined as follows:<br></br>
(1) $a + d = b$<br></br>
(2) $d + f = e$<br></br>
(3) $f + a = c$<br></br>
<br></br>
<mdo:image width="250" height="224" src="triarith3.gif" alt="arithmagon"></mdo:image><br></br>
<br></br>
Therefore:<br></br>
(4) $b + f = c + d$<br></br>
(5) $b + f = e + a$<br></br>
(6) $e + a = c + d$<br></br>
From (4): $f = d + c - b$<br></br>
Combining this result and (2):<br></br>
$e - d = d + c - b$<br></br>
$2d = e - c + b$<br></br>
$d = (b + e - c)/2$<br></br>
And so, using this formula, the value in a circle, when the three
squares in the arithmagon are known, can be found.<br></br>
Of course, the variable $b$ can be used to denote any of the three
circles, and $x$ and $y$ the adjacent squares (does not matter
which), and $z$ the opposite.<br></br>
Following these rules, the formula can also be written as:<br></br>
$a = (b + c - e) / 2$<br></br>
$c = (c + e - b) / 2$<br></br>
I think the process I have gone through above explains why all
arithmagons can be completed - they all exhibit these attributes,
and so can all be solved in the same way.<br></br>
<br></br>
<br></br>
<span class="editorial">Lindsay's solution is</span> <a class="editorial" href="/content/id/2670/LindsaySolution.doc">here</a><span class="editorial">.</span><br></br>
<br></br>
<span class="editorial">It explains why Tom from Cottenham Village
College found that:</span><br></br>
<br></br>
If you add up all of the numbers in the squares and halve it, then
subtract the number on the opposite side to a circle (the number in
the square) you will get the number that goes in the circle.<br></br>
<br></br>
<span class="editorial">Aurora, from the British School in Manila,
explained her strategy clearly:</span> <br></br>
<br></br>
Firstly, let's call the bottom left vertex z, the top one x, and
the bottom right y.<br></br>
<br></br>
Assume:<br></br>
x+y= 17<br></br>
y+z= 20<br></br>
z+x= 15<br></br>
<br></br>
If x+y is 17, and z+x is 15, the difference between z and y must be
2:<br></br>
z+2=y<br></br>
<br></br>
y+z= 20<br></br>
z+z+2 =20<br></br>
z=9<br></br>
<br></br>
If z=9, y is 11 and x is 6<br></br>
<br></br>
<a class="editorial" href="/content/id/2670/JohnKarlSolution.docx">Here</a><span class="editorial"> is how John and Karl, from King's School
Grantham, explained how they derived a formula for finding the
values of the vertices.</span><br></br>
<br></br>
<a class="editorial" href="/content/id/2670/CharlotteSolution.doc">Here</a> <span class="editorial">is how Charlotte, from Llandovery College,
summarised two of the most popular strategies, and</span> <a class="editorial" href="/content/id/2670/KrystofSolution.pdf">here</a> <span class="editorial">is how Krystof, from Uhelny Trh in Prague,
applied his strategy.</span><br></br>
<br></br>
<br></br>
<span class="editorial">We also received good solutions to this
problem from Leighton, Sia Jia Rui from Raffles Institution in
Singapore, Bhavik, Brenna from Bream Bay College, Heer, Sarah &
Heledd from St Stephen's Carramar in Perth, Alex and Luke from
Llandovery College in Wales, Sam from Shrewsbury House School,
Robert from West Hoathly Primary School, Amy from Hanham High
School, Ryan and Alisha from Lacon Childe School, Lauren from
St.Peters Primary School and Michael, Alexander, Jake, Hussein and
Charlie from Wilson's School.</span><br></br>
<br></br></mdoxml>
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<h3>Why do this problem?</h3>
<div><a href="http://nrich.maths.org/2670&part=">This
problem</a> offers students the opportunity to explore number
patterns with or without the use of symbols, and offers several
routes to generality.</div>
<div> </div>
<div>Relating to <a href="http://nrich.maths.org/7472&part=">this month's
theme</a>, think of the process of putting numbers in the vertices
and then calculating the edge numbers as an action. Is it possible
to undo that action uniquely, that is, to 'solve' the
arithmagon?</div>
<br></br>
<h3>Possible approach</h3>
<div>Suggest that the students watch you in silence and then
describe what you have done.</div>
<div>Draw the skeleton of an arithmagon, fill in numbers at the
vertices and then the numbers on the edges.</div>
<div>Do this again, completing the numbers at the vertices and
waiting to be prompted by the class before completing the numbers
on the edges. You could introduce fractions, decimals and/or
negative numbers.</div>
<br></br>
<div>Once you are convinced that they know what's happening, draw
another skeleton and fill in the numbers on the edges. Can they
find the numbers at the vertices?</div>
<div> </div>
<div>"Do you think you will be able to complete any arithmagon if I
give you the numbers on the edges?"</div>
<div> </div>
<div>"Try it out yourselves - draw a skeleton and choose some
numbers to go on the edges. Can you now work out the numbers that
must go at the vertices?"</div>
<div> </div>
<div><span style="font-style: italic;">If a computer room is
available, students could work on the problem using the
interactivity.</span></div>
<div> </div>
<div>Any arithmagon they can't complete could be displayed for
whole class consideration.</div>
<div> </div>
<div>After some time draw the class together and share findings.
The following questions could be used as a focus for
discussion:</div>
<div> </div>
<ul>
<li>Are there any arithmagons that you haven't been able to
complete?</li>
<li>When do you need negative numbers?</li>
<li>When do you need fractions or decimals?</li>
<li>Does anyone have a strategy for completing the
arithmagons?</li>
<li>Can algebra help us?</li>
</ul>
<div>Further questions worth considering can be found at the end of
the problem.<br></br>
</div>
<br></br>
<div>It might be interesting to ask students to try <a href="http://nrich.maths.org/public/viewer.php?obj_id=5533&part=">
Sums of Pairs</a> a few weeks later and see if they notice that it
is essentially the same problem.<br></br>
</div>
<h3>Key questions</h3>
<div>Is it always possible to find numbers to go at the vertices
given any three numbers on the edges? </div>
<div>What is the relationship between the total of the edge numbers
and the total of the vertex numbers?</div>
<br></br>
<h3 style="font-weight: bold; text-decoration: none;">Possible
extension</h3>
<div>Investigate the properties of the quadrilateral arithmagons
generated by putting two arithmagons side by side overlapping along
one edge.</div>
<div> </div>
<div>Investigate square arithmagons.</div>
<div>Can all square arithmagons be solved? Are the solutions
unique?</div>
<div> </div>
<div>What about pentagonal arithmagons?</div>
<div> </div>
<div>What about <a href="http://nrich.maths.org/7447&part=">Multiplication
Arithmagons</a>? </div>
<br></br>
<div>Students might also like to explore the related concept of
<span style="font-style: italic;">magic graphs</span> .</div>
Here are some NRICH magic graph challenges:<br></br>
<br></br>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=470">Magic
W</a> <a href="http://nrich.maths.org/public/viewer.php?obj_id=476"></a> <a href="http://nrich.maths.org/public/viewer.php?obj_id=1835"></a><br></br>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=456"></a><a href="http://nrich.maths.org/460&part=">Olympic Magic</a><br></br>
<h3>Possible support</h3>
To help students get started, use the interactivity on Level 1 to
generate arithmagons which can be solved using only positive whole
numbers.<br></br>
<br></br>
<br></br></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
How does the sum of the numbers at the edges relate to the sum of
the numbers at the vertices? <br></br>
<br></br>
It might help to label the numbers at the vertices $A$, $B$ and $C$
and then express the edge numbers in terms of $A$, $B$ and
$C$.<br></br>
<br></br></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<span class="editorial">We received a number of very good solutions
to this problem. Esther did some wonderful "detective work" and
made a number of very useful observations:</span><br></br>
<br></br>
These are the patterns I found:<br></br>
<br></br>
1) The totals in the squares are double the totals in the
circles.<br></br>
I think this is because each circle number is used twice to solve
the puzzle.<br></br>
<br></br>
2) If the numbers in the squares are all even, the numbers in the
circles are either all odd or all even.<br></br>
<br></br>
If the numbers in the squares are all odd, I cannot solve the
puzzle with whole numbers. This is because an even number minus an
odd number equals an odd number. You can solve these puzzles using
decimals and negative numbers.<br></br>
<br></br>
If you have two odd and one even in the squares then the numbers in
the circles are either two odd and one even or two even and one
odd.<br></br>
<br></br>
However, if you have two even and one odd number in the squares you
cannot solve the puzzle using whole numbers but only with decimals
and negative numbers.<br></br>
<br></br>
3) If you draw the lines of symmetry through a circle and the
opposite square, the difference in the numbers in the squares on
either side equals the difference in the numbers in the circles on
either side.<br></br>
<br></br>
Also the sum of the numbers on each line of symmetry is the same.
This also works for decimals and negative numbers.<br></br>
<br></br>
4) I found a rule for working out the numbers in the circles. You
add the numbers in the squares next to it, subtract the number in
the opposite square, and divide the answer by two. This is why you
sometimes get decimals and negative numbers. I have only ever found
one solution to each puzzle using my rule.<br></br>
<br></br>
I think you can always get answers as long as you can use decimals
and negative numbers using my rule as zero can always be made of +2
and -2 for example.<br></br>
<p><span class="editorial">Harriet and Laura from The Mount School
offered the following explanation:</span></p>
We found that when you added together the numbers in the squares,
(the ones already given) it totalled twice the accumulated amount
of the numbers in the circles.<br></br>
The reason for this can be shown using algebra and is illustrated
in the diagram below.<br></br>
<br></br>
<mdo:image height="224" width="250" src="triarith1.gif" alt="arithmagon using letters a, b, c in vertices"></mdo:image><br></br>
<br></br>
You can find the numbers being represented by $a$, $b$ or $c$ if
you have numbers in the squares.<br></br>
We will show this by finding b using an example found on the
website:<br></br>
<br></br>
Using the equation $(b + a) + (b + c) = 2b + a + c $<br></br>
you can take away $a$ and $c$ to find $2b$:<br></br>
$(2b + a + c) - (a + c) = 2b $<br></br>
So $(b + a) + (b + c) - (a + c) = 2b $<br></br>
<br></br>
<br></br>
<mdo:image height="224" width="250" alt="vertex numbers a, b, c and side numbers 12, 10, 4" src="triarith2.gif"></mdo:image><br></br>
<br></br>
$2b = 12 + 4 - 10 = 6 $<br></br>
$b = 3 $<br></br>
<br></br>
This also works with negative numbers.<br></br>
<p class="editorial">Robert from Leventhorpe School used a similar
analysis:</p>
To work out the formula for working out these arithmagons, we must
first substitute the numbers as such:<br></br>
<br></br>
<mdo:image height="224" width="250" src="triarith3.gif" alt=" from top vertex in clockwise direction a, c, f, e, d, b"></mdo:image><br></br>
<br></br>
If we know $b, c$ and $e$, then we can say that<br></br>
$a + d = b$,<br></br>
$d + f = e$ and<br></br>
$a + f = c$<br></br>
<br></br>
In the first one, $b = 12, c =10$ and $e = 4$,<br></br>
so $a + d = 12$, $d + f = 4$ and $a + f = 10$<br></br>
<br></br>
If we then add all of these together,<br></br>
we get $(a+d) + (d+f) + (a+f) = 12 + 4 + 10 $<br></br>
<br></br>
We do not need the brackets,<br></br>
so we can make it $a + d + d + f + a + f = 26 $<br></br>
<br></br>
This can be cancelled down to $2a + 2d + 2f = 26 $<br></br>
<br></br>
If we then divide everything by $2$ we get $a + d + f = 13$.<br></br>
<br></br>
Because we know that $a + d = 12$, $f$ must be $1$.<br></br>
Since $d + f = 4$, $d$ must be $3$.<br></br>
Finally, we can say that a must be $9$.<br></br>
<br></br>
Double checking this, it works. Take a look:<br></br>
<br></br>
<mdo:image height="224" width="250" alt="from top vertex in clockwise direction: 9, 10, 1, 4, 3, 12" src="triarith4.gif"></mdo:image><br></br>
<br></br>
<p><span class="editorial">Tom from Colyton Grammar School also
used some algebraic thinking to analyse the problem:</span></p>
<div>If the circles were $z, y$ and $x$ from the top clockwise and
the squares are $q, r$ and $p$ from the left clockwise then</div>
<div>$x + z = q$,</div>
<div>$z + y = r$</div>
<div>and $x + y = p$.</div>
<div>$q - r = x - y$</div>
<div>so $q + p - r = 2x$</div>
<div>so $x = (q + p - r) / 2$.</div>
<div>This means that $y = ( p + r - q ) / 2$</div>
<div>and $z = (q + r - p) / 2$, by symmetry.</div>
<p class="editorial">Shaun from Nottingham High School came to the
same conclusion:</p>
<div>It struck me this problem would be most easily investigated
using simultaneous equations as a means of deriving a common rule
for all arithmagons.</div>
<div>I will call the three circles of the arithmagon $a, d$ and
$f$.</div>
<div>Hence, each squares can be expressed as the sum of two of the
variables.</div>
<div>Since we have three variables, and three equations involving
two of them each, we know it can be solved this way.</div>
<div>Some preliminary doodlings proved this to be true.</div>
<div>Let the squares be defined as follows:</div>
<div>(1) $a + d = b$</div>
<div>(2) $d + f = e$</div>
<div>(3) $f + a = c$</div>
<div><mdo:image height="224" width="250" src="triarith3.gif" alt="standard arithmagon labelled a, b, c, d, e, f"></mdo:image></div>
<div>Therefore:</div>
<div>(4) $b + f = c + d$</div>
<div>(5) $b + f = e + a$</div>
<div>(6) $e + a = c + d$</div>
<div>From (4): $f = d + c - b$</div>
<div>Combining this result and (2):</div>
<div>$e - d = d + c - b$</div>
<div>$2d = e - c + b$</div>
<div>$d = (b + e - c)/2$</div>
<div>And so, using this formula, the value in a circle, when the
three squares in the arithmagon are known, can be found.</div>
<div>Of course, the variable $b$ can be used to denote any of the
three circles, and $x$ and $y$ the adjacent squares (does not
matter which), and $z$ the opposite.</div>
<div>Following these rules, the formula can also be written
as:</div>
<div>$a = (b + c - e) / 2$</div>
<div>$c = (c + e - b) / 2$</div>
<div>I think the process I have gone through above explains why all
arithmagons can be completed - they all exhibit these attributes,
and so can all be solved in the same way.</div>
<p class="editorial">Well done to you all, and to Leighton, Jeremy,
Lindsay and Bhavik who also sent us very good solutions.</p>
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Arithmagons
Can you find the values at the vertices when you know the values on
the edges?
Using, Applying and Reasoning about Mathematics
Making and testing hypotheses
Using, Applying and Reasoning about Mathematics
Generalising
Algebra
Simultaneous equations
Secondary Mapping Document
Equations and formulae US
Secondary Mapping Document
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Secondary processes
Reasoning, Justifying, Convincing, Proof
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