Claim: Both $TPYC$ and $TPXA$ are cyclic. Proof: In fact, $\triangle TXC$ ~ $\triangle TAY$ since $\angle XTC = \angle ATY = 150^\circ$ and $AT \cdot TC = BT^2 = TX \cdot TY \Rightarrow AT:TY = TX:TC$. Hence, $\angle PYT = \angle PCT$ and $\angle PXT = \angle PAT$, as we wanted. Now, we have $CP \cdot CX = CT \cdot CA = CB^2 \Rightarrow CB$ is tangent to $(BPX) \Rightarrow \angle CBP = \angle BXC$. By the Cevian Ratio Lemma in $\triangle CBX$, $$ \dfrac{PX}{PC} = \dfrac{BX}{BC} \cdot \dfrac{sin \angle XBP}{sin \angle BXP} = \dfrac{BX \cdot PX}{BC \cdot BP} \Rightarrow \dfrac{BP}{PC} = \dfrac{BX}{BC} = \dfrac{BT}{BC} = \dfrac{AB}{AC}$$. The other ratios are completely analogous and we're done.

Construct $A_1$ outwards from $ABC$ such that $BCA_1$ is equilateral, and notice that quadrilaterals $ATYB$ and $ABA_1C$ are similar (because $ABT \sim ACB$ and $BTY \sim CBA_1$). Another way to think of this is that they can be transformed into each other by a homothety with center $A$ and a reflection through the bisector of $\angle BAC$. This similarity gives us $\angle CAY = \angle BAA_1$. Doing an analogous process on cevian $CX$, we notice that $P$ is the isogonal conjugate of the Fermat Point of $ABC$ (The point $F$ for which $\angle AFB = \angle AFC = \angle BFC = 120^{\circ}$). The isogonal conjugate of the Fermat Point is the first isodynamic point of $ABC$ which indeed satisfies the equations. But, if you didn't know this, you could also extend $PA$, $PB$ and $PC$ until they intersected $(ABC)$ at $A'$, $B'$ and $C'$. Then: $\angle A'B'C' = \angle A'B'B + \angle BB'C = \angle BAA' + \angle BCC' = \angle FAC + \angle FBC = 60^{\circ}$ We repeat the same process on the other two angles of $A'B'C'$ to conclude it's equilateral. Lastly, from $BPC \sim C'PB'$ and $APB \sim B'PA'$, we get: $\dfrac{BC}{CP} = \dfrac{B'C'}{PB'} = \dfrac{A'B'}{PB'} = \dfrac{AB}{AP} \implies BC \cdot AP = AB\cdot CP$ Repeating this for another pair of sides yields the conclusion.

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Analytic geometry bash: Let $T = (0;0), A = (a; 0), B = (0; b), C = (c; 0)$. WLOG $a < 0, b > 0, c > 0$. Now, $\Delta ABT$ ~ $\Delta BCT \Leftrightarrow \frac{|AT|}{|BT|} = \frac{|BT|}{|CT|} \Rightarrow b^2 = -ac$. Now, $X = \left(x; \frac{b}{2}\right)$ and $|XT| = |BT| \Rightarrow \sqrt{x^2 + \frac{b^2}{4}} = \sqrt{b^2} \Rightarrow $ $ x^2 = \frac{3b^2}{4}$. Given that $b > 0$ and $X$ is on the same halfplane of A wrt the $y$ axis, $X = \left(-\frac{b\sqrt{3}}{2}; \frac{b}{2}\right)$. Similarly, $Y = \left(\frac{b\sqrt{3}}{2}; \frac{b}{2}\right)$. The equation of the line through $A$ and $Y$ is $y = \frac{b}{b\sqrt{3} - 2a}x - \frac{ab}{b\sqrt{3} - 2a}$. Similarly, the equation of the line through $C$ and $X$ is $y = -\frac{b}{b\sqrt{3} + 2c}x + \frac{bc}{b\sqrt{3} + 2c}$. Therefore, $P = \left(\frac{(a+c)b\sqrt{3}}{2b\sqrt{3} + 2c - 2a};\frac{b(c - a)}{2b\sqrt{3} + 2c - 2a}\right)$. Now, it is enough to prove that $|PA|\cdot|BC| = |PB|\cdot|AC|$, then the other equality is similar. $|PA|\cdot|BC| = |PB|\cdot|AC| \Leftrightarrow $ $ \sqrt{\left(\frac{(a+c)b\sqrt{3}}{2b\sqrt{3} + 2c - 2a} - a\right)^2 + \left(\frac{b(c - a)}{2b\sqrt{3} + 2c - 2a}\right)^2} \cdot \sqrt{b^2 + c^2} = $ $ \sqrt{\left(\frac{(a+c)b\sqrt{3}}{2b\sqrt{3} + 2c - 2a}\right)^2 + \left(\frac{b(c - a)}{2b\sqrt{3} + 2c - 2a} - b\right)^2} \cdot \sqrt{(c - a)^2} \Leftrightarrow $ $ (((c - a)(b\sqrt{3} - 2a))^2 + (b(c - a))^2)(b^2 + c^2) = $ $ (((a + c)b\sqrt{3})^2 + (b(a - c - 2b\sqrt{3}))^2)(c - a)^2 \Leftrightarrow $ $ (3b^2 - 4ab\sqrt{3} + 4a^2 + b^2)(b^2 + c^2) = $ $b^2(3a^2 + 6ac + 3c^2 + a^2 + c^2 + 12b^2 - 2ac - 4ab\sqrt{3} + 4bc\sqrt{3}) \Leftrightarrow $ $(a^2 + b^2 - ab\sqrt{3})(b^2 + c^2) = b^2(a^2 + c^2 + ac + 3b^2 - ab\sqrt{3} + bc\sqrt{3}) \Leftrightarrow $ $ a^2b^2 + a^2c^2 + b^4 + b^2c^2 - ab^3\sqrt{3} - abc^2\sqrt{3} = $ $ a^2b^2 + b^2c^2 + ab^2c + 3b^4 - ab^3\sqrt{3} + b^3c\sqrt{3} \Leftrightarrow $ $ a^2c^2 - abc^2\sqrt{3} = ab^2c + 2b^4 + b^3c\sqrt{3}$. But $b^2 = -ac \Rightarrow a^2c^2 = b^4, ab^2c = -b^4, -abc^2\sqrt{3} = b^3c\sqrt{3}$, as desired.

Genius132 wrote: voce pode mandar o link da prova por favor https://www.obm.org.br/content/uploads/2023/10/Nivel_3_OBM_2023.pdf

Let $T = (0;0), A = (a; 0), B = (0; 2), C = (-c; 0), Y = (\sqrt{3}, 1), X = (-\sqrt{3}, 1)$. Then, $ac = 4$. Let $P(x; y)$. Then ${P \in AY} \Rightarrow \frac{y}{x+a} = \frac{1}{\sqrt{3}+a}$ and ${P \in CX} \Rightarrow \frac{y}{x-c} = \frac{-1}{\sqrt{3}+c}$, yielding: $$(x+a)(\sqrt{3}+c) = (c-x)(\sqrt{3}+a)$$$$x = \frac{\sqrt{3}(c-a)}{2\sqrt{3}+a+c}$$Writing $d$ for this denominator, we get: $$x+a = \frac{\sqrt{3}(c+a)+a(a+c)}{d} = \frac{(a+c)(\sqrt{3}+a)}{d}$$So $c - x = \frac{(a+c)(\sqrt{3}+c)}{d}$, $y = \frac{a+c}{d}$, and $2-y = \frac{a+c+4\sqrt{3}}{d}$. Now, using the distance formula, we get: $$PA^{2} \cdot BC^{2} \cdot d^{2} = d^{2}\cdot((x+a)^{2}+y^2)\cdot(c^2+4)$$Since $c^{2}+4 = c^{2}+ac = c(c+a)$ and we have a $(c+a)^{2}$ common term in both $(x+a)^{2}$ and $y^{2}$, we can write: $$PA^{2} \cdot BC^{2} \cdot d^{2} = (a+c)^{3} \cdot ((a+\sqrt{3})^{2}+1) \cdot c = (a+c)^3 \cdot (4a + 4c + 8\sqrt{3}) $$As this expression is symmetric in $a$ and $c$, it will also be the same for $PC^{2} \cdot BA^{2} \cdot d^{2}$. Finally, we have: $$ PB^{2} \cdot AC^{2} \cdot d^{2} = d^{2} \cdot (x^{2} + (2-y)^{2} ) \cdot (c+a)^{2} = $$$$ (a+c)^{2} \cdot ( 3(c-a)^{2} + (a+c+4\sqrt{3})^{2}) = $$$$ (a+c)^{2} \cdot ( 4a^{2} + 4c^{2} - 4ac + 8\sqrt{3}(a+c) + 48 )$$ Writing $48 = 12ac$, enables the final factoring $$ PB^{2} \cdot AC^{2} \cdot d^{2} = d^{2} \cdot (4(a+c)^{2} + 8\sqrt{3}(a+c)) = (a+c)^{3}\cdot(4a+4c+8\sqrt{3}) $$

Let $Y'$ in the exterior of side $BC$ such that $\Delta BY'C$ is equilateral. Because $\Delta ABC$ is rightangle, we get $\Delta BAT\sim\Delta CAB\Rightarrow BATY\sim CABY'\Rightarrow\angle{BAY'}=\angle{CAY}.$ Defining $X'$ analogously and $F=AY'\cap CX'$ we conclude that $P$ is the isogonal conjugate of $F$, but $F$ is the $fermat$ point of $\Delta ABC\Rightarrow P$ is an $isodynamic$ point, and that gives us the desired equality.