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Saturday, March 19, 2016

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)

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