USAMO 2001 Problem 4

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USAMO 2001 Problem 4

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MithsApprentice

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MithsApprentice

#1 h Sep 30, 2005, 8:11 PM • 3 Y

Y by Adventure10, HWenslawski, Mango247

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.

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MithsApprentice

2390 posts

MithsApprentice

#2 h Sep 30, 2005, 8:12 PM • 3 Y

Y by Adventure10, HWenslawski, Mango247

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.

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7849 posts

#3 h Oct 13, 2005, 11:31 PM • 2 Y

Y by Adventure10, jrsbr

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.

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3237 posts

#4 h Dec 14, 2005, 6:53 AM • 2 Y

Y by Adventure10, Mango247

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.

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campos

#5 h Nov 10, 2006, 3:05 PM • 5 Y

Y by vsathiam, Adventure10, jrsbr, Mango247, and 1 other user

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.

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aspiringchesslete

29 posts

aspiringchesslete

#6 h Apr 5, 2008, 2:54 AM • 4 Y

Y by vsathiam, Adventure10, Mango247, and 1 other user

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.

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CatalystOfNostalgia

1479 posts

CatalystOfNostalgia

#7 h Apr 10, 2009, 10:51 PM • 2 Y

Y by Adventure10, Mango247

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.

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4210 posts

#8 h Apr 24, 2011, 10:26 PM • 2 Y

Y by vsathiam, Adventure10

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.

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#9 h Apr 12, 2013, 2:44 AM • 2 Y

Y by Adventure10, Mango247

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.

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#10 h Jul 14, 2013, 3:24 PM • 2 Y

Y by Adventure10, Mango247

^very nice solution JSGandora!

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#11 h Nov 10, 2013, 1:08 AM • 2 Y

Y by Adventure10, Mango247

Non-coord bash solution

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695 posts

#12 h Aug 29, 2014, 5:00 AM • 2 Y

Y by Adventure10, Mango247

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.
Solution

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491 posts

#13 h Jan 26, 2015, 6:58 AM • 3 Y

Y by NuclearFusion, Adventure10, Mango247

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.

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1027 posts

#14 h Mar 10, 2016, 4:19 PM • 2 Y

Y by Adventure10, Mango247

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. $\blacksquare$

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#15 h Mar 24, 2016, 5:54 AM • 3 Y

Y by vsathiam, Adventure10, Mango247

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#16 h Apr 9, 2017, 7:08 AM • 2 Y

Y by Adventure10, Mango247

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yiwen

#17 h Feb 16, 2018, 2:12 PM • 1 Y

Y by Adventure10

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.

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6870 posts

#18 h Apr 5, 2019, 6:24 PM • 2 Y

Y by v4913, Adventure10

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.

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#19 h Dec 24, 2020, 4:55 PM • 1 Y

Y by Mango247

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) .

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GeronimoStilton

1521 posts

GeronimoStilton

#20 h Jan 16, 2021, 10:11 PM

Y by

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.

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#21 h Apr 29, 2021, 11:23 PM • 2 Y

Y by Mango247, Mango247

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.

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7586 posts

#22 h Dec 12, 2021, 12:32 AM

Y by

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 $AB^2+AC^2>BC^2$

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jrsbr

#23 h Jan 12, 2022, 3:13 AM • 1 Y

Y by megarnie

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. $\blacksquare$

This post has been edited 3 times. Last edited by jrsbr, Jan 12, 2022, 3:17 AM

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#24 h Jan 12, 2022, 4:05 AM • 2 Y

Y by centslordm, megarnie

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. $\square$

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#25 h Feb 24, 2022, 2:32 PM • 2 Y

Y by SerdarBozdag, Tafi_ak

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)

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17019 posts

#26 h Sep 12, 2022, 4:09 AM • 1 Y

Y by Mango247

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).

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awesomeming327.

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awesomeming327.

#27 h Dec 25, 2022, 2:19 PM

Y by

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.

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#28 h Mar 4, 2023, 8:12 AM

Y by

#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.

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MagicalToaster53

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MagicalToaster53

#30 h Oct 22, 2023, 1:10 AM

Y by

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 $BC^2 > AB^2 + AC^2$ . $\blacksquare$

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#31 h Feb 27, 2024, 2:25 AM

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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.

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#32 h Feb 27, 2024, 4:29 AM

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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.

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#33 h Jun 16, 2024, 9:12 PM

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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.

This post has been edited 1 time. Last edited by BestAOPS, Jun 16, 2024, 9:13 PM

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#34 h Mar 14, 2025, 9:53 PM

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HamstPan38825 wrote:

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.

Past me was a lot smarter.

Assume for the sake of contradiction that $\angle BAC$ is obtuse, i.e. $A$ lies inside the circle with diameter $(BC)$ . Let $M$ be the midpoint of $\overline{BC}$ . Then the length $PA^2$ is upper bounded by $PA^2 \leq \left(PM+\frac{BC}2\right)^2 = \left(\frac{\sqrt{2b^2+2c^2-a^2}}2 + \frac a2\right)^2$ by setting $BC = a$ , $PB = b$ , $PC = c$ . The equation $PA^2 > PB^2+PC^2$ expands to $a^2\left(2b^2+2c^2-a^2\right) > 4b^4+4c^4 + 8b^2c^2 \iff a^4+4\left(b^2+c^2\right)^2 < 2a^2\left(b^2+c^2\right)$ which violates the AM-GM inequality, contradiction.

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#35 h Mar 17, 2025, 3:48 PM

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We proceed with Cartesian coordinates. Let $B = (-1,0)$ and $C = (1,0),$ and let $A = (x,y)$ and $P = (p,q).$ Then
$\begin{align*} &PA^2 > PB^2 + PC^2 \\ &\implies (p-x)^2 + (q-y)^2 > (p+1)^2 + q^2 + (p-1)^2 + q^2 \\ &\implies p^2 + q^2 - 2px - 2qy + x^2 + y^2 \ge 2p^2 + 2q^2 + 2 \\ &\implies x^2 + y^2 > p^2 + 2px + q^2 + 2qy + 2 \\ &\implies 2(x^2 + y^2) > 2 + (p+x)^2 + (q+y)^2 \ge 2 \\ &\implies x^2 + y^2 > 1. \end{align*}$ Thus $A$ lies outside of the circle with diameter $BC,$ so $\angle BAC$ is acute.

This post has been edited 1 time. Last edited by EpicBird08, Mar 17, 2025, 3:48 PM

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