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/www/nrich/html/content/03/10/six4/
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2011-02-01T00:00:01
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When the dots on square dotty paper are joined by straight lines the enclosed figures have dots on their perimeter ($p$) and often internal ($i$) ones as well.
<p>Figures can be described in this way: $(p, i)$.<br></br>
For example, the red square has a $(p,i)$ of $(4,0)$, the grey triangle $(3,1)$, the green triangle $(5,0)$ and the blue shape $(6,4)$:</p>
<p><mdo:flash height="320" id="/content/03/10/six4/picks-theorum.swf" width="320"><param name="allowfullscreen" value="true"></param><param name="movie" value="/content/03/10/six4/picks-theorum.swf"></param>
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<p>Each figure you produce will always enclose an area ($A$) of the square dotty paper.</p>
<p>The examples in the diagram have areas of $1$, $1 {1 \over 2}$, and $6$ sq units.</p>
<p>Do you agree?</p>
<p>Draw more figures; tabulate the information about their perimeter points ($p$), interior points ($i$) and their areas ($A$).</p>
<p>Can you find a relationship between all these three variables ($p$, $i$ and $A$)?</p>
<p><a href="http://nrich.maths.org/public/viewer.php?obj_id=5954&part=">Click here for a poster of this problem</a>.</p>
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<p><span class="editorial">Yun from Garden Internation School thought that</span></p>
<div>The same number of dots on the perimeters and on the inside result in the same areas.<br></br>
Red square: $(4, 0)$ $ A=1$</div>
<div>Orange parellelogram: $(4, 0) $ $A=1$</div>
<p class="editorial"><br></br>
Lily from Garden International School noticed that</p>
as the sum of the perimeter increases the sum of the area increase by ½<br></br>
<br></br>
<a href="/content/03/10/six4/Table%20for%20PT.xls">Here</a> are her results.<br></br>
<p><span class="editorial">Kathryn, also from Garden International School, noticed two relationships:</span></p>
As the number of dots on the shape's perimeter increases by one, the area increases by half.<br></br>
As the number of internal dots increases by one, the area also increases by one.<br></br>
<br></br>
Here are her results.<br></br>
<br></br>
<mdo:image alt="solution" height="592" src="Pick%27s.gif" width="581"></mdo:image><br></br>
<p><span class="editorial">Nadia from Melbourn Village College, Yun from Garden Internation School, Simeran and Aaron from Woodfield Junior School, and Kahlia from Merici College all discovered that:</span></p>
When<br></br>
A = the area of the shape,<br></br>
p = the number of dots on the perimetre and<br></br>
i = the number of dots inside the shape,<br></br>
<br></br>
the area is equal to half of p, added to i, minus 1<br></br>
<br></br>
<div style="text-align: center;">A = (p x ½) + (i - 1)</div>
<br></br>
<p><span class="editorial">This solution of Andrei's from School 205 Bucharest, gives a very useful algebraic view of the problem. He thought these were useful questions to be asking at the start:</span></p>
<p>How many different shapes can I draw of the form (4,0)?<br></br>
Do they all have the same area? Can I see why?<br></br>
<br></br>
Look at other shapes of the form (n,0). What do these shapes have in common?<br></br>
What about shapes of the form (3,n)?</p>
<p>By breaking the problem down in this way patterns emerged that gave hints about what was happening.</p>
<p>I took a look at the figures given in the problem, and I calculated the area (A), the number of lattice points on the perimeter (p) and the number of interior lattice points (i). I found the table below:</p>
<table border="1">
<tbody>
<tr>
<td width="25%">
<h4>Figure</h4>
</td>
<td width="22%">
<h4 align="center">A</h4>
</td>
<td width="22%">
<h4 align="center">p</h4>
</td>
<td width="31%">
<h4 align="center">i</h4>
</td>
</tr>
<tr>
<td>Square</td>
<td>
<div align="center">1</div>
</td>
<td>
<div align="center">4</div>
</td>
<td>
<div align="center">0</div>
</td>
</tr>
<tr>
<td>Triangle 1</td>
<td>
<div align="center">3/2</div>
</td>
<td>
<div align="center">3</div>
</td>
<td>
<div align="center">1</div>
</td>
</tr>
<tr>
<td>Hexagon</td>
<td>
<div align="center">6</div>
</td>
<td>
<div align="center">6</div>
</td>
<td>
<div align="center">4</div>
</td>
</tr>
<tr>
<td>Triangle 2</td>
<td>
<div align="center">3/2</div>
</td>
<td>
<div align="center">5</div>
</td>
<td>
<div align="center">0</div>
</td>
</tr>
<tr>
<td>Parallelogram</td>
<td>
<div align="center">1</div>
</td>
<td>
<div align="center">4</div>
</td>
<td>
<div align="center">0</div>
</td>
</tr>
</tbody>
</table>
<p>I started from the idea that if any relation exists between A, p and i, then there must be a linear relationship. Let x be the coefficient of A, y be the coefficient of p, and z be the coefficient of i; t is the free term. So:</p>
<p>xA + yp + zi + t = 0</p>
<p>Replacing the values corresponding to the first four rows (because the last is identical with the first) I found a system of 4 equations with four unknowns:</p>
<p>x + 4y + t = 0<br></br>
3/2x + 3y +z + t = 0<br></br>
4x + 6y + 4z + t = 0<br></br>
3/2x + 5y + t = 0</p>
<p>Except the (0,0,0,0) solution, I found the relation between x, y, z and t either of the type:</p>
<p>x = -i, y = i/2, z = i, t = -i,<br></br>
or:<br></br>
x = i, y = -i/2, z = -i, t = i.</p>
<p>The last two solutions imply that the relation between A, p and i must be:</p>
<p>-A + p/2 + i -1 = 0</p>
<p>or:</p>
<p>A = p/2 + i -1</p>
<p>This is not a demonstration, from what I have found I can't be sure it works always. But I have looked on the web, and I found some derivations of Pick's theorem, that looks as I found it. One is at <a href="http://mathforum.org/trscavo/geoboards/intro4.html">Geoboards in the classroom</a> and another is at <a href="http://www.cut-the-knot.org/ctk/Pick.shtml">Cut the Knot</a><br></br>
<br></br>
This is a method of calculating the area of any polygon on a geoboard quickly and easily. The theorem has been found by Georg Alexander Pick, born in 1859 in Vienna, and was first published in 1899.</p>
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<h3>Why do this problem?</h3>
At its simplest level this problem allows students to consolidate
their understanding of how to calculate the area of irregular
shapes. The extra mathematical demand comes from requiring students
to identify the relationship between three variables.<br></br>
<br></br>
<h3>Possible approach</h3>
Draw a polygon on a square dotty grid on the board. Clarify that we
shall be interested in three variables: the number of dots on the
perimeter, $(p)$, the number of dots in the inside, $(i)$, and the
area $(A)$. Ask students to work out the $(p)$, $(i)$ and $(A)$ of
the shape that you have drawn.<br></br>
<br></br>
Display the five shapes from the problem. For each, ask students to
work in pairs and agree on the values of $(p)$, $(i)$ and
$(A)$.<br></br>
<br></br>
Draw attention to the two shapes that have an area of 1. What do
they notice about their $(p)$ and $(i)$? Is this true for all
shapes that have an area of 1? Allow the students some time to draw
and share results. Confirm that there are an infinite number of
possibilities of shapes which satisfy these conditions.<br></br>
<br></br>
"The size of the shape will determine the area, so $(p)$ and $(i)$
may well determine the area. Your challenge is to draw some more
shapes and find out if there is a relationship between these three
variables."<br></br>
<br></br>
Identify a central place where students can post their conjectures
or other observations and encourage students to check their
validity.<br></br>
<br></br>
At an appropriate time bring the students back together to discuss
the relationships they have discovered. Use this also as an
opportunity to discuss effective strategies for identifying
relationships, eg keeping one variable fixed.<br></br>
<br></br>
You may wish to use the interactive pinboard, found <a href="http://nrich.maths.org/public/viewer.php?obj_id=2883&part=">
here</a> , to support your presentation/discussion of the
problem.<br></br>
<br></br>
<a href="/content/03/10/six4/Pick%27s%20Theorem%20-%20final.pdf" style="font-style: italic;">Here</a> <span style="font-style: italic;">is an account of one teacher's approach
to using this problem.</span><br></br>
<br></br>
<h3>Key questions</h3>
If $(p)$ is fixed and $(i)$ increases by 1, what is the effect on
the area?<br></br>
If $(i)$ is fixed and $(p)$ increases by 1, what is the effect on
the area?<br></br>
<br></br>
<h3>Possible extension</h3>
Does the same relationship hold when shapes are drawn on isometric
dotty paper?<br></br>
<br></br>
<h3>Possible support</h3>
Suggest that students start with shapes with small areas. How many
different shapes can they draw for an area of 2? What possible
values of $(p)$ and $(i)$ can they find? What about an area of
3..4..5...?<br></br>
<br></br>
<br></br>
<br></br>
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* How many different figures can be described as $(4, 0)$?
<p>* What do you notice about $(4,0)$ figures?</p>
<p>* Choose another particular value for $(p,i)$ and explore
different shapes.</p>
<p>* Have you tried drawing shapes with the same area?</p>
<p>* What do you notice about those figures whose areas are the
same?</p>
<p>* What ways are there of increasing the area by $1$ unit?</p>
<p>* Draw more figures; tabulate the information about their
perimeter points ($p$), interior points ($i$) and their areas
($A$).</p>
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<p>This solution of Andrei's from School 205 Bucharest, gives a
very useful algebraic view of the problem and is worth a read.</p>
<p>I would encourage you all to have a go at this problem by
stepping back and beginning by breaking down the problem as
suggested in the hints. How many different shapes can you draw of
the form (4,0)? Do they all have the same area? Can you see why?
Look at other shapes of the form (n,0). What do these shapes have
in common? What about shapes of the form (3,n)?</p>
<p>By breaking the problem down in this way patterns will emerge
that give hints about what is happening.</p>
<p><br></br>
I started from the figures given in the problem, and I calculated
the area (A), the number of lattice points on the perimeter (p) and
the number of interior lattice points (i). I found the table
below:</p>
<table width="50%" border="1" cellpadding="0" cellspacing="0" bordercolor="#000000">
<tr>
<td width="25%">
<h4>Figure</h4>
</td>
<td width="22%">
<h4 align="center">A</h4>
</td>
<td width="22%">
<h4 align="center">p</h4>
</td>
<td width="31%">
<h4 align="center">i</h4>
</td>
</tr>
<tr>
<td>Square</td>
<td>
<div align="center">1</div>
</td>
<td>
<div align="center">4</div>
</td>
<td>
<div align="center">0</div>
</td>
</tr>
<tr>
<td>Triangle 1</td>
<td>
<div align="center">3/2</div>
</td>
<td>
<div align="center">3</div>
</td>
<td>
<div align="center">1</div>
</td>
</tr>
<tr>
<td>Hexagon</td>
<td>
<div align="center">6</div>
</td>
<td>
<div align="center">6</div>
</td>
<td>
<div align="center">4</div>
</td>
</tr>
<tr>
<td>Triangle 2</td>
<td>
<div align="center">3/2</div>
</td>
<td>
<div align="center">5</div>
</td>
<td>
<div align="center">0</div>
</td>
</tr>
<tr>
<td>Parallelogram</td>
<td>
<div align="center">1</div>
</td>
<td>
<div align="center">4</div>
</td>
<td>
<div align="center">0</div>
</td>
</tr>
</table>
<p>I started from the idea that if any relation exists between A, p
and i, then there must be a linear relationship. Let x be the
coefficient of A, y be the coefficient of p, and z be the
coefficient of k; t is the free term. So:</p>
<p>xA + yp + zi + t = 0</p>
<p>Replacing the values corresponding to the first four rows
(because the last is identical with the first) I found a system of
4 equations with four unknowns:</p>
<p>x + 4y + t = 0<br></br>
3/2x + 3y +z + t = 0<br></br>
4x + 6y + 4z + t = 0<br></br>
3/2x + 5y + t = 0</p>
<p>Except the (0,0,0,0) solution, I found the relation between x,
y, z and t either of the type:</p>
<p>x = -i, y = i/2, z = i, t = -i,<br></br>
or:<br></br>
x = i, y = -i/2, z = -i, t = i.</p>
<p>The last two solutions imply that the relation between A, p and
i must be:</p>
<p>-A + p/2 + i -1 = 0</p>
<p>or:</p>
<p>A = p/2 + i -1</p>
<p>This is not a demonstration, from what I have found I can't be
sure it works always. But I have looked on the web, and I found
some derivations of Pick's theorem, that looks as I found it. One
is at the address:
http://mathforum.org/trscavo/geoboards/intro4.html , another is:
www.cut-the-knot.org/ctk/Pick.shtml and some others could be found
as well.<br></br>
This is a method of calculating the area of any polygon on a
geoboard quickly and easily. The theorem has been found by Georg
Alexander Pick, born in 1859 in Vienna, and was first published in
1899.</p>
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2
3
0
0
1
0
0
Pick's Theorem
Polygons drawn on square dotty paper have dots on their perimeter
(p) and often internal (i) ones as well. Find a relationship
between p, i and the area of the polygons.
Measures and Mensuration
Perimeters
Measures and Mensuration
Area
2D Geometry, Shape and Space
Other polygons
Using, Applying and Reasoning about Mathematics
Generalising
Algebra
Creating expressions/formulae
Mathematics Tools
Pinboard/geoboard
Admin
Stage 3&4 Investigation
Secondary Mapping Document
Straight line graphs
Secondary Mapping Document
Area and volume LS
Secondary Mapping Document
DisplayCabinet
Secondary processes
PM - Posing Questions. Making Conjectures
Secondary processes
PM - Exploring and noticing structure
Secondary processes
PM - Visualising