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##### 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. 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 $BC$ giving another right angles isosceles triangle $ABC$, similar to triangle $ABC$ but with sides a and hypotenuse $\sqrt{2}a$. Now $N$ is the point where the line $BC$ meets the line $AM$, and $P$ is the point where $BC$ meets $BC$ (also on $AM$). It is clear that triangle $PBC$ is similar to triangle $P BC$ and the enlargement factor from $BC$ 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 $BC$, so it is $\sqrt{2}a/2$. The line $AM$ is half the length of $BC$ so it is $\sqrt{2}b/2$.

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:

$$\frac{xb}{a} \cdot \frac{\sqrt{2}}{2} \cdot b = \frac{b^{2}(b - a)}{2(b + a)}$$