Two High School Math Problems

Problem 1

What is the radius of the incircle of a triangle with side lengths $a, \ b$ and $c$?

Solution: Let the radius be $r$, and let

$$ s = \frac{a+b+c}{2}. $$

Then,

$$ \hbox{Area of triangle} = sr = \sqrt{s(s-a)(s-b)(s-c)}. $$

It follows from the above equation that

$$ r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}. $$

NB: I got inspiration from this Quora post.

Problem 2

Suppose $a,\ b$ and $c$ are three positive numbers. Show that the following three inequalities

$$ \left|\frac{a^2 + b^2 - c^2}{2ab}\right| < 1, $$

and

$$ \left|\frac{a^2 + c^2 - b^2}{2ac}\right| < 1, $$

and

$$ \left|\frac{b^2 + c^2 - a^2}{2bc}\right| < 1, $$

are all equivalent to

$$ \{a+b > c,\ a+c > b,\ \hbox{and}\ b+c > a\}. $$

That is, the $a,\ b$ and $c$ in the inequalities can form a triangle.

Proof:

\begin{align} & \left|\frac{a^2 + b^2 - c^2}{2ab}\right| < 1 \\ \Leftrightarrow & |a^2 + b^2 - c^2| < 2ab \\ \Leftrightarrow & -2ab < a^2 + b^2 - c^2 < 2ab \\ \Leftrightarrow & (a+b)^2 > c^2\ \hbox{and}\ (a-b)^2 < c^2 \\ \Leftrightarrow & a + b > c\ \hbox{and}\ |a - b| < c \\ \Leftrightarrow & \{a+b > c,\ a+c > b,\ \hbox{and}\ b+c > a\}. \end{align}

Similarly, we can prove the other equivalence.

Since the necessary and sufficient condition that positive numbers $a,\ b$ and $c$ are lengths of three sides of a triangle is that

$$ \{a+b > c,\ a+c > b,\ \hbox{and}\ b+c > a\}, $$

the $a,\ b$ and $c$ in the inequalities can form a triangle.

NB: See this nice StackExchange post, which gives an answer to the question: “lengths of the sides of a triangle: sufficient and necessary condition?”