parmenides51 wrote: Let $ABC$ be a triangle with $\angle BAC \ne 45^o$ and $\angle ABC \ne 135^o$. Let $P$ be the point on the line $AB$ with $\angle CPB = 45^o$. Let $O_1$ and $O_2$ be the centers of the circumcircles of the triangles $ACP$ and $BCP$ respectively. Show that the area of the square $CO_1P O_2$ is equal to the area of the triangle $ABC$. How could that be a square??

I think (s)he meant 'quadrilateral'. Solution: $\angle AO_1C=2\angle APC=90^{\circ}$ Similarly $\angle BO_2C=90^{\circ}$ Since $O_1A=O_1C$ and $O_2B=O_2C$, $\triangle AO_1C\sim \triangle BO_2C$. Hence $\triangle ABC\sim \triangle O_1O_2C$, and their ratio is $\sqrt{2}:1$. $\therefore [ABC]=2[O_1O_2C]=[CO_1PO_2]$. Here, $[X]$ means the area of $X$.

parmenides51 wrote: Let $ABC$ be a triangle with $\angle BAC \ne 45^o$ and $\angle ABC \ne 135^o$. Let $P$ be the point on the line $AB$ with $\angle CPB = 45^o$. Let $O_1$ and $O_2$ be the centers of the circumcircles of the triangles $ACP$ and $BCP$ respectively. Show that the area of the square $CO_1P O_2$ is equal to the area of the triangle $ABC$. This configuration allows us to obtain the center of the triangle X(20303) as the center of a conic (more details in HG221023): Let $ABC$ be a triangle and $A'B'C'$ be the Kiepert triangle corresponding to $\theta=\pi/4$. The circle with center at $A'$ and passing through $B$ and $C$ again cuts $AC$ at $A_b$, and $AB$ at $A_c$. Let $O_{ab}$ and $O_{ac}$ be the centers of the circles ($ABA_b$) and ($ACA_c$), respectively. The points $A_2=BA_b \cap A'O_{ab}$ and $A_3=CA_c \cap A'O_{ac}$ are considered. Points $B_3, B_1$ and $C_1, C_2$ are defined cyclically. The six points $A_2, A_3, B_3, B_1, C_1, C_2$ lie on a conic, whose center is X(20303). Baricentric equation of this conic: $$ \Sigma_{abc,xyz}(a^8-2 a^4 (3 b^4+2 b^2 c^2+3 c^4)+8 a^2 (b^6+c^6)-(b^2-c^2)^2 (3 b^4+2 b^2 c^2+3 c^4)) x^2+2 (a^8-(b^2-c^2)^4-2 a^6 (b^2+c^2)+2 a^2 (b^2-c^2)^2 (b^2+c^2)) y z] = 0. $$ Note: $CO_1PO_2$ = $CO_{ac}A_cA'$.

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