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{\bf July  2000 }
{\bf Tough Nut Solution  15+ Challenges}

{\bf So Big}

Sue Liu of Madras College, St Andrew's sent a solution to this problem which 
depends on the use of Heron's Formula for the area of a triangle. 
Here is Sue's method.
\vskip 6 true cm

The incircle divides the sides of the triangle into lengths $c' = a + b$, 
$b' = b + x$ and $a' = x + a$ as shown in the diagram. The semi-perimeter 
of the triangle is given by $s = x + a + b$ and from Heron's formula 
the area of the triangle is 
$$\eqalign{
A &= \sqrt{(s(s - a')(s - b')(s - c'))} \cr
  &= \sqrt{((a + b + x)abx)}}$$

Also, the triangle is divided into three smaller triangles and the total area is given by
$$\eqalign{
A &= {1\over 2}(a + b)r + {1\over 2}(b + x)r + {1\over 2}(x + a)r  \cr
  &=(a + b + x)r .}$$
Equating the two answers

$$\eqalign {
\sqrt{abx(a + b + x)} &= (a + b + x)r \cr
abx(a + b + x) &= (a + b + x)^2 r^2  \cr
abx &= (a + b + x)r^2  \cr
x &= {(a + b)r^2\over (ab - r^2).}}$$

Hence
$$\eqalign{
A &= (a + b + x)r  \cr
  &= {r((a + b)(ab - r^2) + (a + b)r^2) \over ab - r^2}    \cr
  &= {abr(a + b)\over ab - r^2.}}$$

An alternative method, not using Heron's formula, is based on finding x in 
terms of $a$ and $b$ using the tangents of the angles at the centre of the 
circle.
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We see that
As an
alternative method you may like to work out the areas of the six right 
angled triangles which make up the area of the given triangle and use tangent
formulas. 

$${y\over x}= {\sin \alpha \cos (\alpha - \theta)\over \cos \alpha \cos (\alpha = \theta)}= \tan \alpha.$$
and so $P$ lies on the straight line $y = x\tan \alpha$.

The position $(x,y)$ depends only on $cos(\alpha - \theta)$, $\alpha$ being a 
constant, and $\theta$ a variable. The distance of the point $P$ from $O$ is 
given by 
$$OP^2 = x^2 + y^2 = \cos^2(\alpha - \theta)(\cos^2\alpha + \sin^2\alpha) =
\cos^2(\alpha - \theta).$$
Hence
$OP = \cos(\alpha - \theta)$ which is a maximum when $\cos(\alpha - \theta) = 1$, that is when $\alpha = \theta$. This occurs when $OPQR$ is a rectangle as 
shown in the diagram.

\vskip 5 cm

We get an even simpler method of solution by using the fact that the angles 
$QOR$ and $QPR$ are both 90 degrees so that $OQPR$ is a cyclic quadrilateral 
with $PR$ as a diameter. We have $\angle POR = \angle PQR = \alpha$ because 
these two angles are subtended by the same chord of the circle. This shows that 
$\angle POR$ is constant and hence that the locus of $P$ is the straight line 
$y = x \tan \alpha.$

What can you say about the locus of $P$ if the triangle $PQR$ is not a right 
angled triangle? 
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