### Three Way Split

Take any point P inside an equilateral triangle. Draw PA, PB and PC from P perpendicular to the sides of the triangle where A, B and C are points on the sides. Prove that PA + PB + PC is a constant.

### Pareq Calc

Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel lines are 1 unit and 2 units.

### Pareq Exists

Prove that, given any three parallel lines, an equilateral triangle always exists with one vertex on each of the three lines.

##### Stage: 4 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:

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