it's enough to notice that $P$ is the isogonal conjugate of the fermat point. the rest is an easy angle chasing.

By Apollonius circle theorem, AP/PC = AB1/B1C = c/a. So AP*a=CP*c, and analogously we get AP*a=BP*b. Therefore AP sin A = BP sin B = CP sin C. So the pedal triangle DEF of P wrt ABC is equilateral. Now it's just angle chasing: BPC = BPD + CPD = BFD + CED = 120 - AFE + 120 - AEF = 60 + A = 90.

We can also use complex numbers after the Apollonius circles... if $|(p-a)(b-c)|=|(p-b)(c-a)|=|(p-c)(a-b)|$ then some simple computation yields that \[\frac{1}{p-a},\frac{1}{p-b},\frac{1}{p-c}\]form an equilateral triangle, so either \[p=-\frac{bc+ca\omega^2+ab\omega^4}{a+b\omega^2+c\omega^4}=\frac{c^2\omega}{c(1+\omega)-a}\]or \[p=-\frac{bc+ca\omega^4+ab\omega^2}{a+b\omega^4+c\omega^2}=\frac{c[a(1+\omega)-c\omega]}{a},\]where $\omega=e^{\pi i/3}$ and WLOG $b=c\omega$. The latter is the reflection of $A$ over $BC$, so because $P$ is inside $\triangle{ABC}$ we must have \[p=\frac{c^2\omega}{c(1+\omega)-a}=\frac{bc}{b+c-a}\implies \frac{p-b}{p-c}=\frac{b(a-b)}{c(a-c)}=\frac{\omega(a-c\omega)}{a-c}\in\mathbb{R}i,\]as desired.

Let $w$ be the circumcircle of $ABC$. Let $B_3$ be the intersection of tangent from $B$ with line $AC$ and $C_3$ be the intersection of tangent from $C$ with line $AB$. It is not hard to prove that $B_3$ is the midpoint of $B_1B_2$ and $C_3$ is the midpoint of $C_1C_2$. Let $AP,BP,CP$ meets $w$ again at $A_4,B_4,C_4$ respectively. Using $B_3P=B_3B$ and $C_3P=C_3C$ and IMO#4 2010, we conclude that $A_4B_4C_4$ is equilateral. Finding $\angle{BPC}=90^0$ is just an easy angle chasing now.

let $X,Y$ be on rays $PB,PC$ such that $PX*PB=PC*PY=PA^2$ the Apollonius circles give $PC/PB=AC/AB$ $PC/PA=BC/CA$ and $PA/PB=AC/BC$ now using $\triangle PCA\sim \triangle PYA$ and $\triangle PBA\sim \triangle PXA$ by very little length chase we get $AX=AY$ now combining with $\triangle PXY\sim \triangle PBC$ with also easy length chase we get $XY=AY$ so $AXY$ is equilateral. now finally $\angle BPC=30+\angle PCA+\angle PBA=30+\angle YAP+\angle XAP=30+\angle XAY=30+60=90$

Well-known... $P$ is the first isodynamic point of triangle $ABC$ so $\angle BPC=\angle A+60=90$. $\blacksquare$

Hello - I am studying for this contest, but I find it difficult to clearly visualize the diagram. Can someone please kindly upload the diagram for this question? Thanks

$P$ is the intersection of $B$-Apollonius Circle and $C$-Apollonius Circle. This gives $\frac{AP}{PC}=\frac{AB}{BC}$ and $\frac{AP}{PB}=\frac{AC}{CB}$, so $PA \sin A = PB \sin B = PC \sin C$ by Sine Law. This gives that the Pedal Triangle of $P$ is equilateral, then angle chasing gives $\angle BPC = 60 + \angle A = 90$ as required.

How did you come up with this amazing and short proof rkm0959? Thank you very much for all your help!

Notice that $P$ is the first isodynamic point of $\triangle ABC$, that is, an inversion around $P$ maps $ABC$ to an equilateral triangle. Thus, $\angle BPC=\angle BAC+60=90$ and we are done. This was sort of trivial for anyone who has seen the isodynamic point being used. As far as I remember, here are some basic properties of the isodynamic points, just for reference. 1.) The three Appolonian circles in a triangle are coaxial and concur at two points, about which an inversion sends the triangle to an equilateral one. 2.) The angles sub tended by the sides are $60$ more than the sub tended ones at vertices, in a directed sense. 3.) The two isodynamic points are collinear with the circumcenter and the Lemoine centre. There are many, but these are the basic ones, that I remember. (Since this is all from memory, please tell me if there is a typo or something is incorrect )

Why didn't any of you use the Pythagorean Theorem to prove it? And what's the Apollonius Circle Theorem?

al_wang wrote: Why didn't any of you use the Pythagorean Theorem to prove it? And what's the Apollonius Circle Theorem? You are welcome to show a proof using Pythagorean Theorem.

This P is $X_{15}$ in disguise, Its properties is $\angle BX_{15}C=\angle BAC+60^{\circ}$, so $\angle BPC=90^{\circ}$. If a point $P$ inside $\triangle BAC$ has $\frac{\sin{\angle B}}{\sin{\angle C}}=\frac{AB}{AC}=\frac{PB}{PC}$ then pedal triangle $\triangle XYZ$ of $P$ has $XY=XZ$. This is because $$XY=BP\sin{\angle B}=CP\sin{\angle C}=XZ$$So by this, the pedal triangle of $X_{15}$ is equilateral, so $\angle BX_{15}C=\angle BZX+\angle CYX=\angle A+\angle YXZ=\angle A+60^{\circ}$.

Hmm so this can be barybashed... Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$, so $B_1=(a:0:c),B_2=(-a:0:c),C_1=(a:b:0), C_2=(-a:b:0)$. Now let $X$ be the point such that $BB_1XB_2$ is a rectangle, and let $Y$ be the point such that $CC_1YC_2$ is a rectangle. Now it's easy to compute $X=B_1+B_2-B=(2a^2:c^2-a^2:-2c^2)$. Similarly, $Y=(2a^2:-2b^2:b^2-a^2)$. Now let $P'=BY\cap CX$. We can instantly get $P'=(2a^2:c^2-a^2:b^2-a^2)$. Now all we have to show is that $(BB_1B_2P)$ is cyclic. By plugging in $B,B_1,B_2$, we can easily find the equation of $(BB_1B_2)$: \[a^2yz+b^2zx+c^2xy=(x+y+z)(\tfrac{-b^2c^2}{a^2-c^2}x+\tfrac{a^2b^2}{a^2-c^2}z).\]At this point I want to find our condition; we have $a^2=b^2+c^2-bc\sqrt{3}$ by Law of Cosines. This gives $bc\sqrt{3}=b^2+c^2-a^2$, so \[3b^2c^2=(b^2+c^2-a^2)^2\implies a^4+b^4+c^4-b^2c^2-2a^2b^2-2a^2c^2=0.\]Great, so now let's plug $P'$ into the circle equation and hope it all works out. We want \[a^2(c^2-a^2)(b^2-a^2)+b^2(2a^2)(b^2-a^2)+c^2(2a^2)(c^2-a^2)=(b^2+c^2)(\tfrac{-2a^2b^2c^2+a^2b^2(b^2-a^2)}{a^2-c^2}).\]Factoring out $a^2$ and clearing the denominator, this is \[(a^2-c^2)(b^2c^2-(b^2+c^2)a^2+a^4+2b^4-2a^2b^2+2c^4-2a^2c^2)=(b^2+c^2)(b^2)(b^2-a^2-2c^2).\]Now we can use our condition once on the left side to reduce it to \[(a^2-c^2)(2b^2c^2-a^2(b^2+c^2)+b^4+c^4)=(b^2+c^2)(b^2)(b^2-a^2-2c^2).\]Now we can pull out a factor of $(b^2+c^2)$ on the left side, so dividing it out we get \[(a^2-c^2)(b^2+c^2-a^2)=b^2(b^2-a^2-2c^2).\]And fortunately for us, this expands to be exactly our condition, so we are done.

Note that $P$ is an isodynamic point, so $\triangle ABC$ inverts to an equilateral triangle at $P$, giving $$\angle PBA + \angle PCA = 60^\circ\implies \angle BPC = 90^\circ$$.