Prove that the altitudes of an acute triangle intersect inside the triangle. 
Prove that the altitudes of an acute triangle intersect inside the triangle.

I can pretty easily see that this is true by a pythagorean theorem argument. Given any two sides, the smaller length to the side will be closer on the base so thus the orthocenter must lie inside the triangle. What would be a more mathematical way of proving this?

Here is a proof
by analytic geometry.
Assume that the vertices are
$A=(0, 0), B=(1, 0)$,
and
$C=(a, 1)$.
It turns out that
the condition for
the intersection of the altitudes
to be inside the triangle
is $0 < a < 1$.
Looking at the triangle,
it is intuitively obvious
that this makes the triangle acute,
and it is nice that
this falls out of the proof.
The vertical altitude
through $C$ is
$x = a$.
The line $BC$
has slope
$\frac{1-0}{a-1}
=\frac{1}{a-1}
$.
Therefore the altitude through $A$
has slope
$1-a$,
so its equation is
$y = (1-a)x$.
Therefore, its intersection with
the altitude through $C$
has
$x=a, y=a(1-a)$,
so the intersection of the altitudes
is it
$V=(a, a(1-a))$.
A point inside the triangle is
$D
=(A+B+C)/3
=(a+1, 1)/3
=((a+1)/3, 1/3)
$.
Therefore $V$ is inside the triangle
if and only if it is
on the same side of each side
as $D$.
The equation of side
$AB$ is
$y=0$.
Evaluating this at $D$
is
$1/3$
and at $V$ is
$a(1-a)$.
Since $1/3 > 0$,
we must have
$a(1-a) > 0$
or
$0 < a < 1$.
The equation of side
$AC$ is
$y - x/a=0$.
Evaluating this at $D$
is
$\frac13-\frac{a+1}{3a}
=\frac{a-(a+1)}{3a}
=\frac{-1}{3a}
$
and at $V$ is
$a(1-a)-a/a
=a-a^2-1
$.
Since $\frac{-1}{3a}$
has the opposite sign of $a$,
and we now know that
$0 < a < 1$,
we must have
$a-a^2-1 < 0$
or
$a(1-a) < 1$.
Since we know that
$0 < a < 1$,
this is true.
The equation of side
$BC$ is
$\frac{1}{a-1}
=\frac{y}{x-1}$
or
$x-1-y(a-1) = 0$.
Evaluating this at $D$
is
$0
=\frac{a+1}{3}-1-\frac13(a-1)
=\frac{a+1-3-(a-1)}{3}
=\frac{-1}{3}
$
and at $V$ is
$x-1-y(a-1)
=a-1-a(a-1)(a-1)
=(a-1)(1-a(a-1))
$.
Since
$\frac{-1}{3} < 0$,
we must have
$(a-1)(1-a(a-1))
< 0$.
Since $0 < a < 1$,
$a-1 < 0$
and
$a(a-1) < 0$
so 
$1-a(a-1) > 1 > 0$,
so we do have
$(a-1)(1-a(a-1))
< 0$.
