When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

Actually, the hypothesis that $PA,PB,PC$ are sides of a triangle is not really needed. We can prove the following: if $\angle BAC\ge\frac\pi 2$, then $PA^2\le PB^2+PC^2\ (*)$ for all points $P$ in the plane. Suppose the contrary. Divide the plane into four regions by drawing the perpendicular bisectors of $AB,AC$. If $(*)$ doesn't hold, then, since $PA>PB,PC$, and we're assuming that $\angle BAC\ge\frac\pi 2,\ P$ must lie in the region completely contained in the semiplane bounded by $BC$ which does not contain $A$. Now keep $P,B,C$ fixed, and push $A$ along $(PA$ until $\angle BAC=\frac\pi 2$. $(*)$ still doesn't hold, so the problem is reduced to the case when $\angle BAC=\frac\pi 2$. Let the parallel through $P$ to $AB$ and the parallel through $A$ to $PB$ intersect in $D$, and let $T=AC\cap PD$. We have $PA^2=AT^2+PT^2$ and $PB^2+PC^2=AD^2+PC^2=(AT^2+DT^2)+(PT^2+TC^2)$, which is $\ge AT^2+PT^2$, meaning that $(*)$ does hold, and contradicting our assumption.

Place $ABC$ on the coordinate plane. WLOG, assume $A$ be the point $(0,0)$ and $B$ be $(1,0)$. Let $C$ be $(x,y)$. We want to show that $x > 0$. 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.

this is one of my first solutions using vectors, and i think it's nice Let $A$ be the origin. then, we have that $PA^{2}>PB^{2}+PC^{2}$ is equivalent to $\overrightarrow{P}^{2}>(\overrightarrow{P}-\overrightarrow{B})^{2}+(\overrightarrow{P}-\overrightarrow{C})^{2}$ which is equivalent to $2\overrightarrow{B}\cdot\overrightarrow{C}>\overrightarrow{P}^{2}+\overrightarrow{B}^{2}+\overrightarrow{C}^{2}-2\overrightarrow{P}\cdot\overrightarrow{B}-2\overrightarrow{P}\cdot\overrightarrow{C}+2\overrightarrow{B}\cdot\overrightarrow{C}=(\overrightarrow{P}-\overrightarrow{B}-\overrightarrow{C})^{2}\geq 0$ which implies that angle $BAC$ is acute.

Let $ PA,PB,PC$ be $ a,b,c$ respectively, and $ AB,BC,CA$ be $ z,x,y$ respectively. Assume for contradiction that $ \angle BAC$ is not acute, that is, that $ x^2 \ge y^2 + z^2$. By the condition in the problem, $ a^2 > b^2 + c^2$. Therefore, by Cauchy-Schwarz, $ a^2x^2 > (b^2 + c^2)(y^2 + z^2) \ge (by + cz)^2 \implies ax > by + cz$. But by Ptolemy's Inequality on quadrilateral $ ABPC$, $ ax \le by + cz$; contradiction.

We will show that $ AP^{2}+BC^{2}\le AB^{2}+AC^{2}+BP^{2}+CP^{2}$. Note that this will immediately imply that $ \angle{BAC}$ is acute, since we have $ PA^{2}>PB^{2}+PC^{2}$, which means that $ BC^{2}<AB^{2}+AC^{2}$, or $ \angle{A}$ is acute. The claim is simple with coordinates . Apply the appropriate transformations preserving the sign of $ AB^{2}+AC^{2}+BP^{2}+CP^{2}-AP^{2}-BC^{2}$ remains the same, so that $ A=(0,0),B=(0,1),C=(a,b),P=(c,d)$. We compute this to be $ 1+(a^{2}+b^{2})+(c^{2}+(d-1)^{2})+((a-c)^{2}+(b-d)^{2})-(c^{2}+d^{2})-(a^{2}+(b-1)^{2})$ $ =(a-c)^{2}+(b-d-1)^{2}\ge0$, so we're done.

Much like paladin8's solution, let $A$ be $(0,0)$ and $B$ $(1,0)$. Let $C$ be $(m,n)$ with $m,n$ constants, WLOG $n>0$. It suffices to show $m>0$. The locus of possible points $P$ is the locus of all points such that $PA^2 > PB^2+PC^2$ or $x^2+y^2>(x-1)^2+y^2+(x-m)^2+(y-n)^2$ Rearranging, $0>-2x+1+x^2-2mx+m^2+(y-n)^2=m^2-2mx+[(y-n)^2+(x-1)^2]$. Note that this inequality necessarily has solutions, as $P$ is one of them. By the quadratic formula, $\frac{2x-\sqrt{4x^2-4[(y-n)^2+(x-1)^2]}}{2}<m<\frac{2x+\sqrt{4x^2-4[(y-n)^2+(x-1)^2]}}{2}$ Clearly, the upper bound is positive (and real, as our inequality has solutions). The lower bound is also real, but we must prove it's positive. The lower bound simplifies to: $x-\sqrt{x^2-[(y-n)^2+(x-1)^2]}$ which we must prove is greater than 0. $x-\sqrt{x^2-[(y-n)^2+(x-1)^2]} > 0 \Longleftrightarrow x \ge \sqrt{x^2-[(y-n)^2+(x-1)^2]} \Longleftrightarrow (y-n)^2+(x-1)^2>0$ which is always true except when $y=n$ and $x=1$. However, if that is the case, we have $m=0$, so $C$ is located at $(0,n)$. That means that $PB^2+PC^2=n^2+1=PA^2$, meaning the triangle with side lengths $PA,PB,PC$ is not obtuse, contradiction. Thus, $m>0$ and $ABC$ is acute.

WLOG let $B$ be at $(-1,0)$ and $C$ be at $(1,0)$. Let $P$ be at $(x,y)$ and let $A$ be at $(a,b)$. Now the condition that $PA$ is equal to the side opposite the obtuse angle is equivalent to $PA^2>PB^2+PC^2$ so \begin{align*}(x-a)^2+(y-b)^2&> (x+1)^2+y^2+(x-1)^2+y^2 \\ a^2+b^2&>2xa+2yb+x^2+y^2+2 \\ 2(a^2+b^2)&>(x+a)^2+(y+b)^2+2\geq2 \\ a^2+b^2&>1 \\ \end{align*} Thus $(a,b)$'s distance from the origin is greater than $1$, so $A$ is outside the circle with diameter $BC$ therefore $\angle BAC$ is acute.

^very nice solution JSGandora!

There are many fairly simple solutions to this problem, however, I would like to point out that it can be trivialized if one knows a certain (quite useful, in my opinion) lemma.

We show the contrapositive. Assume that $ABC$ is obtuse at $A$, we will show that, for any point $P$, $PA^2\le PB^2+PC^2$. Note that \[ PC^2=AC^2+PA^2-2(PA)(AC)\cos{PAC}\\ PB^2=AB^2+PA^2-2(PA)(AB)\cos{PAB} \] Then, \[ PC^2+PB^2=AB^2+AC^2+2PA^2-2(PA)(AC)\cos{PAC}-2(PA)(AB)\cos{PAB}. \] We wish, then, to show that \[ AB^2+AC^2+PA^2\ge2(PA)(AC)\cos{PAC}+2(PA)(AB)\cos{PAB}. \] Note that, because $\angle PAB+\angle PAC=\angle BAC$ is obtuse, \begin{align*} && \cos{(PAB+PAC)} &< 0\\ \Rightarrow\;\; && \cos{PAB}\cos{PAC} &< \sin{PAB}\cos{PAC}\\ \Rightarrow\;\; && \cos^2{PAB}\cos^2{PAC} &< \sin^2{PAB}\cos^2{PAC}\\ && &=(1-\cos^2{PAB})(1-\cos^2{PAC})\\ && &=1-(\cos^2{PAB}+\cos^2{PAC})+\cos^2{PAB}\cos^2{PAC}\\ \Rightarrow\;\; && \cos^2{PAB}+\cos^2{PAC} &<1. \end{align*} Thus, we can find an $x$ such that \[ \cos^2{PAB}\le x\\ \cos^2{PAC}\le 1-x \] whence we find that \[ \cos{PAB}\le \sqrt{x}\\ \cos{PAC}\le \sqrt{1-x} \] Finally, we use AM-GM to get \begin{align*} AB^2+xPA^2+AC^2+(1-x)PA^2 &\ge 2(AB)(AP)\sqrt{x}+2(AC)(AP)\sqrt{1-x}\\ &\ge 2(AB)(AP)\cos{PAB}+2(AC)(AP)\cos{PAC} \end{align*} and we conclude.

One more solution (I think this works?): Let $x, y, z$ be the lengths of $PA, PB, PC$ respectively. Fix $P, x, y, z$. Then note that $A, B, C$ lie on the three circles centered at $P$ with radii $x, y, z$ respectively. In order to maximize $\angle BAC$, we choose A arbitrarily and then B and C to maximize $\angle BAP$ and $\angle CAP$ by making $AB$ and $AC$ tangent to the circles of radius $y$ and $z$ respectively. Then it suffices to show that $\angle BAP + \angle CAP < 90$, which is equivalent to $\sin^-1(y/x)+\sin^-1(z/x) < 90$. This can be seen by considering the triangle $XYZ$ with sidelengths $x, y, z$ (opposite the corresponding vertices per the standard notation); because $\angle YXZ$ is obtuse, we have $\sin^-1(y/x)+\sin^-1(z/x) < \angle XYZ + \angle XZY < 90$, and we are done. $\blacksquare$

someone told me a solution that seems too easy, If $\angle{BAC}>90$, and $PA,PB,PC$ forms a obtuse triangle with $PA$ being the opposite side of the obtuse angle, $PA>PB,PA>PC$ so $$\angle{PCA}>\angle{PAC},\angle{PBA}>\angle{PAB}$$$$\angle{PCA}+\angle{PBA}>\angle{PAC}+\angle{PBC}=\angle{BAC}$$, => $\angle{BCA}+\angle{CBA}>\angle{BAC}$ which is a contradiction.

Using Ptolemy's inequality and Cauchy-Schwarz, \begin{align*} PA \cdot BC &\le PB \cdot AC + PC \cdot AB \\ &\le \sqrt{(PB^2+PC^2)(AB^2+AC^2)} \\ &< \sqrt{PA^2 \cdot (AB^2+AC)^2} = PA \cdot \sqrt{AB^2+AC^2} \end{align*}meaning $BC^2 < AB^2+AC^2$, so $\angle BAC$ is acute.

Is this solution correct? Because of the obtuse angle we know that $PA>PB$ and $PA>PC$, thus we know that $P$ lies on the side of the perpendicular bisectors of $AB$ and $AC$ which don't contain $A$ and at the same time it lies in the plane of the triangle $ABC$, which obviously is impossible when $\angle BAC\ge 90^{\circ}$(by seeing where is the circumcenter) .

Let $\angle BPC=\alpha,\angle CPA=\beta,\angle APB=\gamma$. Then we need to show \[PB^2+PC^2-2PB\cdot PC\cos\alpha=BC^2<AB^2+AC^2=2PA^2+PB^2+PC^2-2PB\cdot PA\cos\gamma-2PC\cdot PA\cos\beta\iff\]\[0<4PA^2-4PC\cdot PA\cos\beta-4PB\cdot PA\cos\gamma+4PB\cdot PC\cos\alpha = \]\[(2PA-PC\cos\beta-PB\cos\gamma)^2+4PB\cdot PC\cos\alpha-(PC\cos\beta+PB\cos\gamma)^2.\]It is clear that $PA>(PB+PC)/2$, so we can instead show an inequality with $PA=\sqrt{PB^2+PC^2}$. By homogenizing we can assume $PA=1$ and let $PB=y,PC=z$ with $y^2+z^2=1$. Then the desired inequality is \[1\ge z\cos\beta+y\cos\gamma-yz\cos\alpha=z\cos\beta+y\cos\gamma-yz\cos(\beta+\gamma).\]With respect to $\beta$, the derivative of this expression is $-z\sin\beta+yz\sin(\beta+\gamma)$. Similarly, with respect to $\gamma$, the derivative of this expression is $-y\sin\gamma+yz\sin(\beta+\gamma)$. If one of $y,z$ is equal to zero the inequality is obvious, so suppose neither is. Then we must have $\sin \beta=y\sin(\beta+\gamma),\sin\gamma=z\sin(\beta+\gamma)$. Let $p=e^{i\beta},q=e^{i\gamma}$ so we have the equations \[p-1/p=y(pq-1/pq)\implies p^2q-q=z(p^2q^2-1),q-1/q=z(pq-1/pq)\implies pq^2-p=z(p^2q^2-1).\]Since this is a system of two quadratic equations, it has at most four solutions: as $\sin$ is cyclic mod $\pi$, we can restrict our search to choices of $\beta,\gamma\in [-\pi/2,\pi/2]$. Moreover, if $\beta=0$ or $\gamma=0$ the equation is not quadratic (its only solution is actually $\gamma=0$) so we only expect three solutions. If $y=\sin\theta,z=\cos\theta$ with $\theta\in (0,\pi/2)$ we expect the solutions to be $(\beta,\gamma)=(0,0),(\theta,\pi/2-\theta),(-\theta,\theta-\pi/2)$. Since the last one is just the second with the sign flipped and $\cos$ does not care about signs, we have two cases. We need to show $1\ge y+z-yz$ and $1\ge y^2+z^2$, the latter of which is trivially true. To see the first, write it suffices to show \[1\ge y+z-yz\iff 0\ge -(1-y)(1-z),\]which is trivial.

Assume for the sake of contradiction that $\angle BAC$ is not acute, so $BC^2 \ge AB^2+AC^2$. Multiplying this with $PA^2 > PB^2+PC^2$, we have \[(PA \cdot BC)^2 > (PB \cdot AB)^2+(PB \cdot AC)^2+(PC \cdot AB)^2+(PC \cdot AC)^2. \qquad \qquad (1)\]By Ptolemy's Inequality, we have $PA \cdot BC \le PC\cdot AB+PB \cdot AC$, so \[(PA\cdot BC)^2 \le (PC\cdot AB)^2+(PB\cdot AC)^2 +2(PC\cdot PC \cdot AB \cdot AC).\]Combined with $(1)$, \[(PC\cdot AB)^2+(PB\cdot AC)^2 +2(PC\cdot PC \cdot AB \cdot AC) >(PB\cdot AB)^2+(PB\cdot AC)^2+(PC \cdot AB)^2+(PC \cdot AC)^2,\]which means that \[2 PC \cdot PC \cdot AB \cdot AC > (PB \cdot AB)^2+(PC \cdot AC)^2.\] However this inequality is false by AM-GM so we have a contradiction. Hence $\angle BAC$ must be acute.

QM-AM and Median Length Formula when $AP$ passes through the midpoint $M$ of $BC$, $2(AM^2+PM^2)>PB^2+PC^2$ implies $AB^2+AC^2>BC^2$

Suppose that $\angle BAC$ is obtuse. Let $A=(0,0), B=(x_b,y_b), C=(x_c,y_c)$ and $P=(x_p,y_p)$, with $x_b,y_b>0$ and $x_c,y_c<0$. We have that: $$BC^2>AB^2+AC^2\implies x_a^2+y_a^2>(x_p-x_b-x_c)^2+(y_p-y_b-y_c)^2-2(x_bx_c+y_by_c) (i).$$The statement gives us that $PA^2>PB^2+PC^2\implies-(x_bx_c+y_by_c)>x_a^2+y_a^2 (ii)$. With $(i)$ and $(ii)$: $$x_bx_c+y_by_c>(x_p-x_b-x_c)^2+(y_p-y_b-y_c)^2\geq0$$which is a contradiction. $\blacksquare$

Notice $AP^2>BP^2+CP^2$ and $AB\cdot CP+AC\cdot BP\ge AP\cdot BC.$ Suppose FTSOC that $\angle BAC\ge 90.$ Then, $BC^2\ge AB^2+AC^2.$ Hence, $$(AB\cdot CP+AC\cdot BP)^2\ge (AP\cdot BC)^2>(BP^2+CP^2)(AB^2+AC^2),$$a contradiction by Cauchy-Schwarz. $\square$

Let $M, N$ be midpoints of $AP$ and $BC$, respectively. For the points $A, B, P, C$; let's apply Euler's quadrilateral formula, $$ AB^2 + BP^2 + PC^2 + CA^2 = AP^2 + BC^2 + 4MN^2 \geq AP^2 + BC^2 .$$Given that $AP^2 > BP^2 + PC^2$. Thus, $$ AB^2 + AC^2 > BC^2 .$$and we get $\angle BAC$ is acute. (Lokman GÖKÇE)

Mogmog8 wrote: Notice $AP^2>BP^2+CP^2$ and $AB\cdot CP+AC\cdot BP\ge AP\cdot BC.$ Suppose FTSOC that $\angle BAC\ge 90.$ Then, $BC^2\ge AB^2+AC^2.$ Hence, $$(AB\cdot CP+AC\cdot BP)^2\ge (AP\cdot BC)^2>(BP^2+CP^2)(AB^2+AC^2),$$a contradiction by Cauchy-Schwarz. $\square$ Regarding this: Based on my diagram, it seems like it should be quadrilateral APCB, so $AB \cdot CP + AP \cdot BC \geq AC \cdot BP$. Am I being dumb? (I have P to the right of triangle ABC, and it does seem to satisfy PA > PB > PC).

By the condition, $AP^2>BP^2+CP^2.$ By Ptolemy, $AP\cdot BC \le AC\cdot PB + AB\cdot PC.$ By C-S inequality, \[(AC^2+AB^2)(PB^2+PC^2)\ge (AC\cdot PB+AB\cdot PC)^2 \ge AP^2 \cdot BC^2.\]Since $BP^2+CP^2< AP^2$ we must have $BP^2+CP^2> BC^2$ which is what we wanted to prove.

#3amsolves We will use coordinates. Let $A=(a,b),B=(-1,0),C=(1,0).$ Let $$P=(x,y).$$We wish to show that $a^2+b^2>1.$ From the obtuse condition, we have $$(x-a)^2+(y-b)^2>(x+1)^2+y^2+(x-1)^2+y^2$$$$a^2+b^2>x^2+y^2+2ax+2by+2.$$Rewrite this as $$a^2+b^2>(x+a)^2+(y+b)^2+2-a^2-b^2,$$so we have $$a^2+b^2>(x+a)^2+(y+b)^2+2-a^2-b^2\geq 2-a^2-b^2\rightarrow a^2+b^2>1,$$as desired.

Assume by way of contradiction that $PA^2 > PB^2 + PC^2$ and simultaneously $A$ is obtuse, i.e., $BC^2 > AB^2 + AC^2$. Multiplying these two inequalities provides us with \[PA^2 \cdot BC^2 > AC^2(PB^2 + PC^2) + AB^2(PB^2 + PC^2).\]However by Ptolemy's theorem on any four arbitrary points, we have that \begin{align*} PA \cdot BC &\leq AC \cdot BP + AB \cdot CP \\ \implies PA^2 \cdot BC^2 &\leq (AC \cdot BP + AB \cdot CP)^2 \\ \end{align*}However combining our two results gives us that \[AC^2(PB^2 + PC^2) + AB^2(PB^2 + PC^2) < (AC \cdot BP + AB \cdot CP)^2 \iff AC^2 \cdot PC^2 + AB^2 \cdot PB^2 < 2AC \cdot BP \cdot AB \cdot CP.\]However by AM-GM on the LHS of the inequality, we obtain that $2AC \cdot PC \cdot AB \cdot PB < 2AC \cdot PC \cdot AB \cdot PB$, which is a contradiction. Therefore we cannot both have $PA^2 > PB^2 + PC^2$ and $BC^2 > AB^2 + AC^2$. $\blacksquare$

Suppose otherwise. Then $AB^2+AC^2 > BC^2$ and $PA^2 > PB^2+PC^2$. But by Ptolemy's inequality, $$(PB^2+PC^2)(AB^2+AC^2) < (PA)^2 (PC)^2 \leq (AB \cdot PC + AC \cdot PB)^2.$$This is a contradiction by C-S.

Let $a$, $b$ and $c$ be positive reals such that $a^2 > b^2 + c^2$ – then, we assign $PA = a$, $PB = b$ and $PC = c$. Imagine fixing $A$ and $P$ while varying $B$ and $C$. In the worst case situation, both $\angle PAC$ and $\angle PAB$ are maximized. This happens when $\angle PBA = \angle PCA = 90^{\circ}$ (consider drawing circles of radius $b$ and radius $c$ centered at $P$.) Even in this worst case situation, we have \[ AB^2 + AC^2 = (a^2 - b^2) + (a^2 - c^2) > b^2 + c^2 = PB^2 + PC^2,\]and since $ABPC$ is cyclic, this implies that $\angle BAC$ is acute.

We proceed with vectors. Let $\mathbf{p} = \overrightarrow{PA}$, $\mathbf{b} = \overrightarrow{BA}$, and $\mathbf{c} = \overrightarrow{CA}$. We wish to prove $\mathbf{b} \cdot \mathbf{c} > 0$. The obtuse triangle condition can be written as \[ \| \mathbf{p} - \mathbf{b} \|^2 + \| \mathbf{p} - \mathbf{c} \|^2 < \| \mathbf{p} \| ^2. \]Using the identity $\| \mathbf{v} \| = \mathbf{v} \cdot \mathbf{v}$, we have \begin{align*} (\mathbf{p} - \mathbf{b}) \cdot (\mathbf{p} - \mathbf{b}) + (\mathbf{p} - \mathbf{c}) \cdot (\mathbf{p} - \mathbf{c}) &= 2(\mathbf{p} \cdot \mathbf{p}) - 2\mathbf{p} \cdot (\mathbf{b} + \mathbf{c}) + \mathbf{b} \cdot \mathbf{b} + \mathbf{c} \cdot \mathbf{c} \\ &< \mathbf{p} \cdot \mathbf{p}. \end{align*}Simplifying and adding $2(\mathbf{b} \cdot \mathbf{c})$ to both sides, we get \[ \mathbf{p} \cdot \mathbf{p} - 2\mathbf{p} \cdot (\mathbf{b} + \mathbf{c}) + (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) < 2(\mathbf{b} \cdot \mathbf{c}). \]The left side factors as $(\mathbf{p} - (\mathbf{b} + \mathbf{c})) \cdot (\mathbf{p} - (\mathbf{b} + \mathbf{c}))$. That is nonnegative, so we conclude that $\mathbf{b} \cdot \mathbf{c} > 0$, which is what we wanted to prove.