USA IMO

By the Cauchy-Schwarz Inequality, $$\sqrt{P{B}^2+P{C}^2}\sqrt{A{C}^2+A{B}^2}\ge PB\cdot AC+PC\cdot AB.$$ Applying the (Generalized) Ptolemy's Inequality to quadrilateral $ABPC$ yields $$PB\cdot AC+PC\cdot AB\ge PA\cdot BC.$$ Because $PA$ is the longest side of an obtuse triangle with side lengths $PA$, $PB$, $PC$, we have $PA>\sqrt{P{B}^2+P{C}^2}$ and hence $$PA\cdot BC\ge \sqrt{P{B}^2+P{C}^2}\cdot BC.$$ Combining these three inequalities yields $\sqrt{A{B}^2+A{C}^2}>BC$, implying that angle $BAC$ is acute.

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Alternative solution.

Let $D$ and $Q$ be the feet of the perpendiculars from $B$ and $P$ to line $AC$, respectively. Then $DQ\le BP$. Furthermore, the given conditions imply that $A{P}^2>B{P}^2+P{C}^2$, which can be written as $A{P}^2-P{C}^2>B{P}^2$. Hence, $$\begin{array}{rl}A{Q}^2& \ge A{Q}^2-Q{C}^2\\ & =(A{P}^2-P{Q}^2)-(C{P}^2-P{Q}^2)\\ & =A{P}^2-P{C}^2\\ & >B{P}^2\\ & \ge D{Q}^2.\end{array}$$ Let $\ell$ be the ray $AC$ minus the point $A$. Note that, since $PA>PC$, $Q$ lies on ray $\ell$. If $D$ did not lie on $\ell$, then $AQ$ would be less than or equal to $DQ$, a contradiction. Thus, $D$ lies on $\ell$, and angle $BAC$ is acute.

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Alternative solution.

Set up a coordinate system on the plane with $A=(0,0)$, $B=(a,0)$, $C=(b,c)$, and $P=(x,y)$. Without loss of generality, we may assume that $a>0$ and that $c>0$. Proving that angle $BAC$ is acute is equivalent to proving that $b>0$. Since $P{A}^2>P{B}^2+P{C}^2$, $$x^2+y^2>(x-a{)}^2+y^2+(x-b{)}^2+(y-c{)}^2.$$ Hence, $$0>(x-a{)}^2-2bx+b^2+(y-c{)}^2\ge -2bx.$$ Since $PA>PB$, we have $x>\frac{a}{2}>0$. It follows that $b>0$, as desired.

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Alternative solution.

We first prove the following Lemma.

Lemma. For any four points $W$, $X$, $Y$, and $Z$ in the plane, $$W{Y}^2+X{Z}^2\le W{X}^2+X{Y}^2+Y{Z}^2+Z{W}^2.$$ Proof. Pick an arbitrary origin $O$ and let $w$, $x$, $y$, $z$ denote the vectors from $O$ to $W$, $X$, $Y$, $Z$, respectively. Then $$\begin{array}{rl}W{X}^2+X{Y}^2+Y{Z}^2+Z{W}^2-W{Y}^2-X{Z}^2& \\ & =|w-x{|}^2+|x-y{|}^2+|y-z{|}^2+|z-w{|}^2-|w-y{|}^2-|x-z{|}^2\\ & =w\cdot w+x\cdot x+y\cdot y+z\cdot z\\ & -2(w\cdot x+x\cdot y+y\cdot z+z\cdot w-w\cdot y-x\cdot z)\\ & =|w+y-x-z{|}^2,\end{array}$$ which is always nonnegative. Equality holds if and only if $w+y=x+z$, which is true if and only if $WXYZ$ is a (possibly degenerate) parallelogram.

Applying the Lemma to points $A$, $B$, $C$, and $P$ gives $$\begin{array}{rl}0\le A{B}^2+B{P}^2+P{C}^2+C{A}^2-A{P}^2-B{C}^2& \\ & =(P{B}^2+P{C}^2-P{A}^2)+(A{B}^2+A{C}^2-B{C}^2)\\ & <0+(A{B}^2+A{C}^2-B{C}^2)\\ & =A{B}^2+A{C}^2-B{C}^2.\end{array}$$ Therefore, angle $BAC$ is acute.

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Alternative solution.

In this solution, ${\sin }^{-1}$ takes on values between $0^{\circ }$ and $90^{\circ }$. Note that $\angle PAB<90^{\circ }$, since $PB<PA$. Applying the Law of Sines to triangle $PAB$ yields $$\sin \angle PAB=\frac{PB}{PA}\sin \angle ABP\le \frac{PB}{PA}.$$ It follows that $$\angle PAB\le {\sin }^{-1}\frac{PB}{PA}.$$ Since $P{A}^2>P{B}^2+P{C}^2$, we have similarly $$\angle PAC\le {\sin }^{-1}\frac{PC}{PA}<{\sin }^{-1}\frac{\sqrt{P{A}^2-P{B}^2}}{PA}.$$ Thus, $$\begin{array}{rl}\angle BAC& \le \angle BAP+\angle PAC\\ & <{\sin }^{-1}\frac{PB}{PA}+{\sin }^{-1}\frac{\sqrt{P{A}^2-P{B}^2}}{PA}.\end{array}$$ If $\theta ={\sin }^{-1}\frac{PB}{PA}$, then $\sin (90^{\circ }-\theta )=\cos \theta =\sqrt{1-{\sin }^2\theta }=\frac{\sqrt{P{A}^2-P{B}^2}}{PA}$. Hence, $$\angle BAC<{\sin }^{-1}\frac{PB}{PA}+{\sin }^{-1}\frac{\sqrt{P{A}^2-P{B}^2}}{PA}=90^{\circ },$$ and angle $BAC$ is acute.

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Alternative solution.

Note that $P{A}^2>P{B}^2+P{C}^2$. Regard $P$ as fixed and $A,B$, and $C$ as free to rotate on circles of radii $PA,PB$, and $PC$ about $P$, respectively. As $A,B,C$ vary, $\angle BAC$ will be maximized when $B$ and $C$ are on opposite sides of line $PA$ and $\angle ABP$ and $\angle ACP$ are right angles, i.e., lines $AB$ and $AC$ are tangent to the circles passing through $B$ and $C$. Without loss of generality, we assume that $PA>PB\ge PC$. In this case, $ABPC$ is cyclic and $A{B}^2=P{A}^2-P{B}^2>P{C}^2$, and similarly $A{C}^2>P{B}^2$. Hence, on the circumcircle of $ABPC$, arcs $AB$ and $AC$ are bigger than arcs $PC$ and $PB$, respectively. Thus, $\angle BPC>\angle BAC$. Since these two angles are supplementary, angle $BAC$ is acute.