A primer for mathematics competitions - Section 1.1 - Appetizer Problem 3

Problem:

Solution:

We know the method for computing the radius for the inscribed circle. Now let's think about the radius of the circumscribed circle.

The key observation is that it is a right angled triangle inside a circle, so the hypotenuse must be the diameter, and therefore $R = \frac{c}{2}$, in our previous case, that would be $\frac{5}{2}$.

To avoid the algebra, we can simply test the options and see which one gives $1 : 2.5$. That gives option (A) right away.

Knowing the answer is (A), working backwards, we know the radius of the inscribed circle is $\frac{ab}{a + b + c}$.

This is a particularly good looking formula. Let's see why.

Using exactly the same computation we had in the last problem, we have

$\frac{d}{a} = \frac{1 - \frac{a}{c}}{\frac{b}{c}} = \frac{c - a}{b}$.

The equation of the angle bisector is

$\frac{x}{a} + \frac{y}{\frac{a(c-a)}{b}} = 1$.

Solving for $x$ by subsituting $y = x$, we get

$x = \frac{a}{1 + \frac{b}{c-a}} = \frac{a (c - a)}{b + c - a}$ and that is our representation for the radius of the inscribed circle.

I claim that the two representations are equal, to see that, we write

$\begin{eqnarray*} \frac{a (c - a)}{b + c - a} &=& \frac{ab}{a+b+c} \\ a (c - a)(a + b + c) &=& (ab)(b + c - a) \\ a^2c + abc + ac^2 - a^3 - a^2b - a^2c &=& ab^2 + abc - a^2b \\ abc + ac^2 - a^3 - a^2b &=& ab^2 + abc - a^2b \\ ac^2 - a^3 &=& ab^2 \\ ac^2 - a^3 &=& a(c^2 - a^2) \\ ac^2 - a^3 &=& ac^2 - a^3 \end{eqnarray*}$

Reading backwards, now we have shown the both representations are equal (of course, the given one look much better)