**Problem**: (2001 USAMO - #4) Let $P$ be a point in the plane of triangle $ABC$ such that the segments $PA$, $PB$, and $PC$ are the sides of an obtuse triangle. Assume that in this triangle the obtuse angle opposes the side congruent to $PA$. Prove that $\angle BAC$ is acute.

**Solution**: Personally, I like to kill geometry problems by putting them on the coordinate plane, which can sometimes be an extremely effective way, as shown here.

WLOG, assume $A$ is the point $(0,0)$ and $B$ be $(1,0)$.

Let $C$ be $(x,y)$.

We want to show that $x > 0$ (then $\angle BAC < 90^{\circ}$).

Let $P$ be $(a,b)$.

By the given condition, we have $PA^2 > PB^2+PC^2$, giving

$a^2+b^2 > (a-1)^2+b^2+(a-x)^2+(b-y)^2$.

$0 > (a-1)^2+x(x-2a)+(b-y)^2$.

Now suppose $x \le 0$. The $(b-y)^2$ term can be arbitrarily small, so we have

$(a-1)^2+x(x-2a)+(b-y)^2 \ge (a-1)^2+x(x-2a) = a^2-(2+2x)a+(1+x^2)$.

The discriminant of this (as a quadratic in $a$) is $(2+2x)^2-4(1+x^2) = 8x \le 0$, so

$0 > (a-1)^2+x(x-2a)+(b-y)^2 \ge a^2-(2+2x)a+(1+x^2) \ge 0$

since it can have at most one root for $a$.

But that gives a contradiction, so $x > 0 \Rightarrow \angle BAC$ is acute.

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Practice Problem: Do this problem geometrically.

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