Given a triangle with side lengths $a$, $b$, and $c$, and the length of the median to side $a$ denoted by $t_a$, prove that $$ \frac{b+c-a}{2} < t_a < \frac{b+c}{2} $$

1. Problem Description

Given a triangle with side lengths $a$, $b$, and $c$, and the length of the median to side $a$ denoted by $t_a$, prove that

$$\frac{b+c-a}{2} < t_a < \frac{b+c}{2}$$

2. Solution Steps

We will use Apollonius's theorem (also known as the median theorem), which relates the length of a median of a triangle to the lengths of its sides. The theorem states that if $m$ is the median to side $a$, then

$$b^2 + c^2 = 2(m^2 + (a/2)^2)$$

In our case, $m = t_a$, so we have

$$b^2 + c^2 = 2(t_a^2 + (a/2)^2) = 2t_a^2 + \frac{a^2}{2}$$

Solving for $t_a^2$:

$$2t_a^2 = b^2 + c^2 - \frac{a^2}{2}$$

$$t_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$$

$$t_a = \sqrt{\frac{2b^2 + 2c^2 - a^2}{4}} = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}$$

We want to prove $\frac{b+c-a}{2} < t_a < \frac{b+c}{2}$.

First, let's prove $t_a < \frac{b+c}{2}$:

We need to show that $\frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2} < \frac{b+c}{2}$, which is equivalent to showing that $\sqrt{2b^2 + 2c^2 - a^2} < b+c$.

Squaring both sides, we get $2b^2 + 2c^2 - a^2 < (b+c)^2 = b^2 + 2bc + c^2$.

Thus, $b^2 + c^2 - a^2 < 2bc$, or $b^2 + c^2 - 2bc < a^2$, so $(b-c)^2 < a^2$.

Taking the square root of both sides gives $|b-c| < a$, which is equivalent to $-a < b-c < a$, or $c-a < b < a+c$. This is one of the triangle inequality conditions, so $|b-c| < a$ holds.

Now, let's prove $\frac{b+c-a}{2} < t_a$:

We need to show that $\frac{b+c-a}{2} < \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$, which is equivalent to $b+c-a < \sqrt{2b^2 + 2c^2 - a^2}$.

Squaring both sides, we get $(b+c-a)^2 < 2b^2 + 2c^2 - a^2$.

Expanding the left side, we have $b^2 + c^2 + a^2 + 2bc - 2ab - 2ac < 2b^2 + 2c^2 - a^2$.

Thus, $2a^2 + 2bc - 2ab - 2ac < b^2 + c^2$, or $2a^2 < b^2 + c^2 - 2bc + 2ab + 2ac$, so $2a^2 < (b-c)^2 + 2a(b+c)$.

Since $a < b+c$, it means that $a^2 < a(b+c)$.

Also, since $b < a+c$ and $c < a+b$, we have that $b-c < a$. Therefore, $(b-c)^2 < a^2$. Thus

$2a^2 - 2a(b+c) < (b-c)^2 < a^2$.

Then $2a^2 < (a+b+c)(a+b-c)$.

This implies that the inequalities hold.

3. Final Answer

$\frac{b+c-a}{2} < t_a < \frac{b+c}{2}$