<?xml version="1.0" encoding="ISO-8859-1" ?>
  <resource>
  <id>376</id>
  <path>/www/nrich/html/content/01/09/15plus4/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;p&gt;The square $ABCD$ has sides of length 1 unit and it is split into three triangles by the lines $BP$ and $CP$. If $P$ is the midpoint of $AD$, find the radii of the inscribed circles of these triangles.&lt;/p&gt;
&lt;p&gt;Now suppose the lengths $AP$ and $PD$ are $(1- p)$ and $p$ respectively. Find the radii of the three circles $r_1$, $r_2$and $r_3$ in terms of $p$ and plot, on the same axes, the graphs of $r_1$, $r_2$ and $r_3$ as $p$ varies from 0 to 1. Can the ratio of the radii $r_1 : r_2 : r_3$ ever take the value $1:2:3$?&lt;/p&gt;
&lt;p&gt;&lt;mdo:image alt=&quot;square ABCD with three circles within it.&quot; src=&quot;pic2.gif&quot;&gt;&lt;/mdo:image&gt;&lt;/p&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div class=&quot;framework&quot;&gt;&lt;strong&gt;Notes and Background&lt;/strong&gt;&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Sangaku in Japanese means a mathematics tablet. During the Edo period (1603-1867) when Japan was cut off from the western world, people of all classes produced theorems in Euclidean geometry as beautifully coloured drawings on wooden tablets to be hung in a temple. Proofs were rarely given. The tablets challenged other geometers: &amp;quot;See if you can prove this.&amp;quot;&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
This solution is from Etienne of Parramatta Highschool, NSW
Australia &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
I need to prove the area of a triangle $ A $ is given by $ A = rs $
where $ r $ is the radius of the incircle (inscribed circle) and $
s$ is the semi-perimeter (half the perimeter $ (a+b+c)/2$). Let $ a
$, $ b $, $ c $ be the lengths of the sides of triangle $ ABC $.
Join the incentre I of the triangle to the 3 corners. From I drop 3
perpendiculars to each of the sides, each of these has a length of
$ r $. The areas of $ ABI $, $ BCI $, $ CAI $ are $ cr/2 $, $ ar/2
$ and $ br/2 $ respectively. They sum up to give the area of
triangle $ ABC $ &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Area of $ ABC = cr/2 + ar/2 + br/2 = r(a+b+c)/2 = rs $. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Back to the question! &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
When $ P $ is the midpoint of $ AD, r_1 = r_2 $ because they sit on
congruent triangles. The area of triangle $ ABP = AP \times AP/2 =
1/4. PB = \sqrt(1+1/4) = \sqrt5/2. $ &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
The semi perimeter of $ ABP = (1 + 1/2 + \sqrt5/2)/2 =
[3+\sqrt5]/4. $ Using the formula $ A = rs $ for the area of the
triangle, the area of $ ABP = r_1[3+\sqrt5]/4 = 1/4. $ This gives $
r_1 = 1/[3 + \sqrt5] = [3 - \sqrt5]/4 $ so $ r_1 = r_2 = [3 -
\sqrt5]/4. $ &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
The area of BPC is 1 - 1/4 - 1/4 = 1/2 and $ PB = PC = \sqrt5/2. $
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
The semi-perimeter of $ BPC = [1+ \sqrt5/2 + \sqrt5/2]/2 = [1 +
\sqrt5]/2. $ From the area of triangle $ BPC $ we get $ r_3[1 +
\sqrt5]/2 = 1/2 $ so $ r_3 = 1/[1 + \sqrt5] = [\sqrt5 - 1]/4. $
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Now suppose the lengths AP and PD are $ 1-p $ and $ p $. The area
of $ APB = (1-p)/2. $ The length $ PB = \sqrt(1 + p^2 - 2p + 1) =
\sqrt(p^2 - 2p + 2) $ and the semi perimeter of $ APB = [1 + (1-p)
+ \sqrt(p^2 - 2p + 2)]/2.$ So, using the area formula again, &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$$ r_2 = {1-p\over 2} \times {2\over [2 - p + \sqrt(p^2 - 2p + 2)]}
= {1\over 2}[2 - p - \sqrt(p^2 - 2p + 2)]. $$ &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Similary with $ DPC $. The area of $ DPC = p/2 $ and the length $
PC = \sqrt(p^2 + 1). $ The semi perimeter of $ PC = [1 + p +
\sqrt(p^2 + 1)]/2. $ So&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$$ r_1 = {p\over 2} \times {2\over [1 + p + \sqrt(p^2 + 1)]} =
{1\over 2}[1 + p - \sqrt(p^2 + 1)]. $$ &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Similarly $$ r_3 = {1\over 2} \times {2\over [1 + \sqrt(p^2 + 1) +
\sqrt(p^2 -2p + 2)]} = {1\over 2}[1 + \sqrt(p^2 + 1) - \sqrt(p^2
-2p +2)][\sqrt(p^2 +1)- p]. $$ &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
To finish this question you need to use the formulae found above
which give the radii $ r_1 $, $ r_2$ and $ r_3 $ in terms of $ p $
and then consider how these values change as $ p $ varies from $0$
to $1$. For example $ r_1 $ varies from $0$ to approximately
$0.293$ as $ p $ increases from $0$ to $1$. In order to discover
whether the ratio of the radii $ r_1 : r_2 : r_3 $ can ever take
the value $1 : 2 : 3$ you could plot on the same axes the graphs of
$ r_1 $, $ r_2 $ and $ r_3 $ as $ p $ varies from $0$ to $1$.
&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML/>
  <canonXML/>
  <end_user_role>2</end_user_role>
  <difficulty>3</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>0</keystage3>
  <keystage4>0</keystage4>
  <keystage4plus>1</keystage4plus>
  <title>Sangaku</title>
  <description>The square ABCD is split into three triangles by the lines BP and
CP. Find the radii of the three inscribed circles to these
triangles as P moves on AD.</description>
  <spec_group>Measures and Mensuration
    <specifier>Area</specifier>
  </spec_group>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Inscribed circle</specifier>
  </spec_group>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Radius (radii) &amp; diameters</specifier>
  </spec_group>
  <spec_group>Sequences, Functions and Graphs
    <specifier>Graphs</specifier>
  </spec_group>
</resource>