Some(?) of the Parts

A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle

Baby Circle

A small circle fits between two touching circles so that all three circles touch each other and have a common tangent? What is the exact radius of the smallest circle?

Logosquares

Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares.

Picturing Pythagorean Triples

Stage: 4 and 5

 Somewhat surprisingly every Pythagorean triple $( a, b, c)$, where $a, b$ and $c$ are positive integers and $a^2+b^2 = c^2$, can be illustrated by this diagram, in which the L shaped region has area $b^2$, and the areas of the larger and smaller squares are $c^2$ and $a^2$. With this clue you can find some triples for yourself right away. With an L strip of width 1 unit you get the whole class of Pythagorean triples with $a$ and $c$ as consecutive integers, that is $c = a+1.$
 Diagram 2 This diagram (extended as far as required) illustrates the fact that, for all $n$, the sum of the first $n$ odd numbers gives $n^2$. As we can see from the diagram to the left: \begin{eqnarray}1+3&=&2^2&=&4 \\ 1+3+5 &=& 3^2 &=& 9 \\ 1+3 +5 + 7 &=& 4^2 &=& 16 \\ 1+3+5+7+9 &=& 5^2 &=& 25 \end{eqnarray}

Taking $a= 4$ and $c= 5$, the area of the outer L'' strip is $b^2= 9$ and this gives the first Pythagorean triple $4^2+ 3^2= 5^2$.

Before reading on, can you use more strips of unit width and extend the diagram to find Pythagorean triples with $b^2 = 25, 49, 81, 121$ etc.?

For every odd square number $b^2$, we can pick out a Pythagorean triple.

As the Diagram 2 shows, $c = a + 1$ and hence \begin{eqnarray} b^2 = c^2 - a^2 \\ b^2 = (a+1)^2 - a^2 \\ b^2 = a^2 + 2a +1 -a^2\\ = 2a + 1 \end{eqnarray} The next odd square after $3^2 = 9$ is $5^2 = 25$
Let $$25 = b^2$$
$$25 = 2*12 + 1$$
giving $a$ = 12 and from there the triple $5^2+12^2=13^2$

The next one in this class is $b=7$ giving $a=24 = \frac{(49-1)}{2}$ and $c=25$

Giving the triple $24^2+7^2=25^2$, and so on.

This method gives us infinitely many Pythagorean triples, but what about others where the width of the L strip (the difference between $a$ and $c$) is more than 1?
 To find other triples with L strips of different widths try this for yourself: choose any two positive integers $p$ and $q$ that have no common factors and with $p > q$; \item calculate $2p q$ and $p^2+q^2$; \item draw two squares with sides $a=2p q$ and $c = p^2 + q^2$ as shown in Diagram 3. The width of the L strip is $w = (p-q)^2$ and its area is $b^2 = (p^2 -q^2)^2$, and if you wish you can check this algebra for yourself. For example, taking $p= 5$ and $q= 2$, the inner square has dimensions $20$ by $20$ and the outer square has dimensions $29$ by $29$, the width of the L strip is $9$, and the area of the L strip is $(25-4)^2=21^2$, giving the Pythagorean triple $(20, 21, 29)$.
The process given by these instructions produces what is called a parameterisation and $p$ and $q$ are called parameters. It is easy to show by simple algebra that, by choosing positive integers $p$ and $q$, with $p> q$, and taking
$$b = p^2-q^2$$
$$a = 2p q$$
$$c = p^2 + q^2$$
we get a Pythagorean triple.

This does not give all triples, for example it does not give ($9,12,15)$, but it does give all triples for which $a, b$ and $c$ have no common factor (primitive' triples). Every other triple can then be found by calculating $( k a, k b, k c)$ for some integer $k$.

These results are proved in the article Pythagorean Triples II .

We shall concentrate on finding all the triples where $a, b$ and $c$ have no common factor. If $p$ and $q$ have a common factor, or if $p$ and $q$ are both odd, then $a, b$ and $c$ will have a common factor.

We only need the cases where one of $p$ and $q$ is even and the other is odd. In such cases the width of the L strip $= c-a=(p-q)^2$ is always an odd square.

The table below (extended in the obvious way) shows all the primitive' Pythagorean triples, and some non-primitive triples as well.

 Pythagorean Triples $c-b=1^2$ ($p=q+1$) $c-b=3^2$ ($p=q+3$) $c-b=5^2$ ($p=q+5$) $c-b=7^2$ ($p=q+7$) $c-b=9^2$ ($p=q+9$) $c-a=2 \times 1^2$ $(q=1)$ 3, 4, 5 15, 8, 17 35, 12, 37 63, 16, 65 99, 20, 101 $c-a=2 \times 2^2$ $(q=2)$ 5, 12, 13 21, 20, 29 45, 28, 53 77, 36, 85 117, 44, 125 $c-a=2 \times 3^2$ $(q=3)$ 7, 24, 25 27, 36, 45 55, 48, 73 91, 60, 109 135, 72, 153 $c-a=2 \times 4^2$ $(q=4)$ 9, 40, 41 33, 56, 65 65, 72, 97 105, 88, 137 153, 104, 185 $c-a=2 \times 5^2$ $(q=5)$ 11, 60, 61 39, 80, 89 75, 100, 125 119, 120, 169 171, 140, 221 $c-a=2 \times 6^2$ $(q=6)$ 13, 84, 85 45, 108, 117 85, 132, 157 133, 156, 205 189, 180, 261 $c-a=2 \times 7^2$ $(q=7)$ 15, 112, 113 51, 140, 149 95, 168, 193 147, 196, 245 207, 224, 305 $c-a=2 \times 8^2$ $(q=8)$ 17, 144, 145 57, 176, 185 105, 208, 233 161, 240, 289 225, 272, 353 $c-a=2 \times 9^2$ $(q=9)$ 19, 180, 181 63, 216, 225 115, 252, 277 175, 288, 337 243, 324, 405

There is a dual illustration of each triple where the smaller square has side $p^2 - q^2$ and the width of the L strip is $2q^2$. In the dual case for $p=5$ and $q =2$ the inner square has dimensions $21$ by $21$ and the width of the L strip is $8$ units giving the triple $(20,21,29)$.

Given a Pythagorean triple $(a, b, c)$ you can always `work back' to find $p$ and $q$, and given $p$ and $q$ you can always calculate $a, b$ and $c$. You could compete with a friend, both choose a value for $p$ and $q$ and compute a Pythagorean triple $(a, b, c)$ which you give to the other person. Then without using calculators, see who can be the first to find the values of $p$ and $q$. You may need to keep $p$ and $q$ small at first to make this easy for yourselves but as you get better you can take bigger values of $p$ and $q$.
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