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A Shade Crossed
Age 14 to 16
Challenge Level 
There were two solutions from Madras College, one from Thomas, James, Mike and Euan and the other from Sue Liu which is reproduced below.
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This triangle is a right angled isosceles triangle, the
hypotenuse being $\sqrt{2}b$. We draw a line through $A$ and point
$M$, the midpoint of the line $BC$. We draw the line $B`C`$ giving
another right angles isosceles triangle $AB`C`$, similar to
triangle $ABC$ but with sides a and hypotenuse $\sqrt{2}a$. Now $N$
is the point where the line $B`C`$ meets the line $AM$, and $P$ is
the point where $BC`$ meets $B`C$ (also on $AM$).
It is clear that triangle $PBC$ is similar to triangle $P
B`C`$ and the enlargement factor from $B`C`$ to $BC$ is $b/a$. So
the line $PM$ is $b/a$ times as long as the line $PN$. Also the
line $AN$ is half the length of $B`C`$, so it is $\sqrt{2}a/2$. The
line $AM$ is half the length of $BC$ so it is $\sqrt{2}b/2$.
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If we let $PN$ be $x$ then we have an equation:
$$x + \frac{bx}{a} + \frac{\sqrt{2}}{2}a =
\frac{\sqrt{2}}{2}b$$
Solving this equation gives $$x =
{\frac{\sqrt{2}}{2}}{\frac{(b - a)a}{(a + b)}}$$
$PM$ is the height of the shaded triangle and $PM =
xb/a$.
Area of the shaded triangle is:
$$\frac{xb}{a} \cdot \frac{\sqrt{2}}{2} \cdot b =
\frac{b^{2}(b - a)}{2(b + a)}$$
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