2001 USAMO Problems/Problem 4

Problem

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

We know that $PB^2+PC^2 < PA^2$ and we wish to prove that $AB^2 + AC^2 > BC^2$ . It would be sufficient to prove that $PB^2+PC^2+AB^2+AC^2 \geq PA^2 + BC^2.$ Set $A(0,0)$ , $B(1,0)$ , $C(x,y)$ , $P(p,q)$ . Then, we wish to show

$\begin{align*} (p-1)^2 + q^2 + (p-x)^2 + (q-y)^2 + 1 + x^2 + y^2 &\geq p^2 + q^2 + (x-1)^2 + y^2 \\ 2p^2 + 2q^2 + 2x^2 + 2y^2 - 2p - 2px - 2qy + 2 &\geq p^2 + q^2 + x^2 + y^2 - 2x + 1 \\ p^2 + q^2 + x^2 + y^2 + 2x - 2p - 2px - 2qy + 1 &\geq 0 \\ (x-p)^2 + (q-y)^2 + 2(x-p) + 1 &\geq 0 \\ (x-p+1)^2 + (q-y)^2 &\geq 0, \end{align*}$

which is true by the trivial inequality.

2001 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
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2001 USAMO Problems/Problem 4

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