Denote by $ X$, $ Y$, $ Z$ the intersections $ BB^{\prime} \cap CC^{\prime}$, $ CC^{\prime} \cap AA^{\prime}$ and $ AA^{\prime} \cap BB^{\prime}$, respectively. According to Pascal's theorem applied to the hexagon $ ABB^{\prime}PC^{\prime}C$ (or for example, from Pappus' theorem) we get that the points $ B_{1}$, $ X$, $ C_{1}$ are collinear. Similarly, the points $ C_{1}$, $ Y$, $ A_{1}$ are collinear and so are $ A_{1}$, $ Z$, $ B_{1}$. In this case, since the lines $ AY$, $ XB$, $ B_{1}A_{1}$ are concurrent at $ Z$, the triangles $ AXB_{1}$ and $ YBA_{1}$ are perspective, and therefore, by Desargues' theorem, the intersection points $ AX \cap YB$, $ XB_{1} \cap BA_{1} = \infty$, $ AB_{1} \cap YA = \infty$ are collinear. Thus, $ AX \| BY$. Similarly, we will get that $ BY \| CZ$ and so $ AX \| BY \| CZ$. Denote by $ \left[\mathcal{P}\right]$ the area of the polygon $ \mathcal{P}$. In this case $ \left[XYZ\right]=\left[ABX\right]+\left[ACX\right]$, and since $ X$, $ B_{1}$, $ C_{1}$ are collinear it follows that $ \left[XYZ\right]=2\left[AC_{1}X\right]+2\left[AB_{1}X\right]=2\left[AB_{1}C_{1}\right]=\left[ABC\right]/2$, which is independent of the position of $ P$.

Here are another two (more) interesting results related to this configuration: Extension. Let $ I_{a}$, $ I_{b}$, $ I_{c}$ be the corresponding excenters of triangle $ ABC$. Denote by $ \alpha$, $ \beta$, $ \gamma$ the reflections of $ AX$, $ BY$, $ CZ$ into the lines $ I_{b}I_{c}$, $ I_{c}I_{a}$ and $ I_{a}I_{b}$, respectively. Prove that $ \alpha$, $ \beta$, $ \gamma$ concur at $ P$. Extension 2. Let $ MNQ$ be the anticomplementary triangle of $ ABC$. Denote by $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ the reflections of $ AX$, $ BY$, $ CZ$ into the sideliens $ NQ$, $ QM$ and $ MN$, respectively. Prove that $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ concur at a point $ P^{\prime}$ lying on the circumcircle of triangle $ ABC$ such that $ PP^{\prime} \| AX \| BY \| CZ$.

Let $ AA' \cap BC = A_2, BB' \cap CA = B_2, CC' \cap AB = C_2.$ Suppose $ P$ and $ B$ are on the opposite side of $ AC,$ so that $ A_2$ and $ C_2$ lie on the sides $ BC$ and $ AB,$ but $ B_2$ lies on the extension of side $ CA.$ We have \[ \tfrac{|BA_2|}{|A_2C|} = \tfrac{|BA|}{|AC|} \tfrac{\sin \angle BAA'}{\sin \angle A'AC} = \tfrac{c}{b} \tfrac{\sin \angle BPA'}{ \sin \angle A'PC} = \tfrac{c}{b} \tfrac{|PC|}{|PB|} \tfrac{|BA_1|}{|A_1 C|} = \tfrac{c}{b} \tfrac{|PC|}{|PB|} = \tfrac{\mathcal C}{\mathcal B}\] where $ \mathcal A = a \cdot |PA|, \mathcal B = b \cdot |PB|, \mathcal C = c \cdot |PC|.$ Similarly, $ \tfrac{|CB_2|}{|B_2A|} = \tfrac{\mathcal A}{\mathcal C}$ and $ \tfrac{|AC_2|}{|C_2B|} = \tfrac{\mathcal B}{\mathcal A}.$ Further, by Ptolemy's theorem, we have $ \mathcal B = \mathcal A + \mathcal C.$ According to Routh's theorem and the fact that $ \lambda = \tfrac{\mathcal C}{\mathcal B}$ and $ \mu = - \tfrac{\mathcal A}{\mathcal C}$ and $ \nu = \tfrac{\mathcal B}{\mathcal A},$ we have that the area of the triangle is \[ |ABC| \left|\frac{4\mathcal A\mathcal B\mathcal C}{(\mathcal A + \mathcal B + \mathcal C) (-\mathcal A + \mathcal B + \mathcal C) (-\mathcal A - \mathcal B + \mathcal C)} \right| = |ABC| \left| \frac{4\mathcal A \mathcal B \mathcal C}{(2 \mathcal B) (2 \mathcal C) (-2 \mathcal A)} \right|\]\[ = \frac{|ABC|}{2}\]

Let $X = BB'\cap CC'$, $Y = CC'\cap AA'$, $Z = AA'\cap BB'$. By Pascal's theorem on $BB'PC'CA$, $X$ lies on line $B_1C_1$; similarly, $Y\in C_1A_1$ and $Z\in A_1B_1$. Since $B_1C_1\parallel BC$, $\measuredangle{AB_1X} = \measuredangle{ACB} = \measuredangle{AB'X}$, so $AB'B_1X$ is cyclic. Since $B' = BX\cap PB_1$ and $AB'PB$ is cyclic as well, we thus obtain the spiral similarity $\triangle{AB_1X}\sim\triangle{APB}$. Now we work in the complex plane with unit circle $(ABCP)$. Using $b_1 = \frac{a+c}{2}$, the aforementioned spiral similarity translates directly as \[ \frac{b_1-a}{x-a} = \frac{p-a}{b-a} \Longleftrightarrow x = a + \frac{(a-b)(a-c)}{2(p-a)}. \](In fact, it is only slightly more annoying to compute $x$ without noticing $X\in B_1C_1$ or the spiral similarity.) From here, one can directly compute $[XYZ]$ using a complex determinant. However, there is a nicer approach: in order to compute $\frac{x-b_1}{x-c_1}$, we use the analogous $\triangle{AC_1X}\sim\triangle{APC}$ to deduce \[ \frac{x-b_1}{x-c_1} = \frac{(a-b_1)\frac{b-p}{a-p}}{(a-c_1)\frac{c-p}{a-p}} = \frac{a-c}{a-b}\frac{p-b}{p-c}. \]Now define $\alpha = (b-c)(p-a)$, $\beta = (c-a)(p-b)$, and $\gamma = (a-b)(p-c)$, and observe that $\alpha+\beta+\gamma=0$ (Ptolemey's theorem with complex numbers). Then $\frac{x-b_1}{x-c_1} = -\frac\beta\gamma$, so $X$ has barycentric coordinates $(0,-\frac\gamma\alpha,-\frac\beta\alpha)$ with respect to $\triangle{A_1B_1C_1}$. Hence the ratio between the signed areas of triangles $XYZ$ and $A_1B_1C_1$ is \[ \det\begin{pmatrix} 0 & -\frac\gamma\alpha & -\frac\beta\alpha \\ -\frac\gamma\beta & 0 & -\frac\alpha\beta \\ -\frac\beta\gamma & -\frac\alpha\gamma & 0 \\ \end{pmatrix} = -\frac1{\alpha\beta\gamma} \det\begin{pmatrix} 0 & \gamma & \beta \\ \gamma & 0 & \alpha \\ \beta & \alpha & 0 \\ \end{pmatrix} = -\frac1{\alpha\beta\gamma} (-\gamma(-\alpha\beta) + \beta(\gamma\alpha)). \](This is the matrix, in barycentric coordinates, corresponding to the affine transformation sending $\triangle{A_1B_1C_1}$ to $\triangle{XYZ}$.) Hence $[XYZ] = 2[A_1B_1C_1] = \frac12[ABC]$ is independent of $P$, as desired.

can't we solve it without using strong theorems like pascal's theorem?…

Could any one explain me this ??? area(XYZ) = area(ABX) +area(ACX) ??? Is true??? Sir

Yes pohatza is wrong in that step but still the equality is valid.Because $[XYZ]=[XYB]+[BYZ]=[AYB]+[BYC]=[BA_1C_1]+[AC_1Y]+[CA_1Y]=[BA_1C_1]+[C_1B_1Y]+[A_1B_1Y]=[BA_1C_1]+[B_1A_1C_1]=\frac{[ABC]}{2}$ I think the proof is really nice.Well done pohatza!!

The vertices lie on the medial triangle of ABC by Pascal and by Thales' theorem we easily see the area is (ABC)/2

My solution: Let $ X=BB' \cap CC', Y=CC' \cap AA', Z=AA' \cap BB' $ . From Pascal theorem ( for $ BB'PC'CA $ ) we get $ X \in B_1C_1 $ . Similarly, we can prove $ Y \in C_1A_1 $ and $ Z \in A_1B_1 $ . From Pappus theorem ( for $ B-A_1-C $ and $ Z-A-Y $ ) we get $ BY \parallel CZ $ . Similarly, we can prove $ CZ \parallel AX $ and $ AX \parallel BY $ . Now we get $ [XYZ]=[XYA]+[XAZ]=[XBA]+[XAC]=\tfrac{1}{2} [ABC]=\text{Const} $ . Q.E.D ____________________________________________________________ pohoatza wrote: Extension. Let $ I_{a}$, $ I_{b}$, $ I_{c}$ be the corresponding excenters of triangle $ ABC$. Denote by $ \alpha$, $ \beta$, $ \gamma$ the reflections of $ AX$, $ BY$, $ CZ$ into the lines $ I_{b}I_{c}$, $ I_{c}I_{a}$ and $ I_{a}I_{b}$, respectively. Prove that $ \alpha$, $ \beta$, $ \gamma$ concur at $ P$. Let $ D=YZ \cap BC, E=ZX \cap CA, F=XY \cap XY $ . Let $ P^* $ be the isogonal conjugate of $ P $ WRT $ \triangle ABC $ (at infinity) . Since $ AX \parallel BY \parallel CZ $ , so from Desargue theorem ( for $ \triangle ABC $ and $ \triangle XYZ $ ) we get $ D, E, F $ are collinear , hence $ P(AX \cap \odot (ABC), A_1; B, C)=A(X, A'; B, C)=A(X,D;B,C)=-1 $ , so combine with $ A_1B=A_1C \Longrightarrow P \cup AX \cap \odot (ABC) $ is parallel to $ BC \Longrightarrow P^* \in AX $ . Similarly, we get $ P^* \in BY, P^* \in CZ $ (or follows from $ AX \parallel BY \parallel CZ $ ) $ \Longrightarrow \alpha, \beta, \gamma $ are concurrent at $ P $ . ____________________________________________________________ pohoatza wrote: Extension 2. Let $ MNQ$ be the anticomplementary triangle of $ ABC$. Denote by $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ the reflections of $ AX$, $ BY$, $ CZ$ into the sideliens $ NQ$, $ QM$ and $ MN$, respectively. Prove that $ \alpha^{\prime}$, $ \beta^{\prime}$, $ \gamma^{\prime}$ concur at a point $ P^{\prime}$ lying on the circumcircle of triangle $ ABC$ such that $ PP^{\prime} \| AX \| BY \| CZ$. Seems not true

CantonMathGuy wrote: \begin{eqnarray*} [XYZ] & = & [XYA] + [XAZ]\\ & = & ([XCA] + [CYA]) + ([XAB] + [BAZ])\\ & = & [C_1CA]\cdot\frac{B_1X}{B_1C_1} + [CC_1A] + [B_1AB]\cdot\frac{XC_1}{B_1C_1} + [BAB_1]\\ & = & \frac{[BCA]}{2}\cdot\frac{B_1X}{B_1C_1} + \frac{[CBA]}{2} + \frac{[CAB]}{2}\cdot\frac{XC_1}{B_1C_1} + \frac{[BAC]}{2}\\ & = & \frac{[ABC]}{2}\cdot\left(\frac{B_1X}{B_1C_1}-1+\frac{XC_1}{B_1C_1}-1\right)\\ & = & -\frac{[ABC]}{2}, \end{eqnarray*}where areas are signed. Thus, $[XYZ]$ is independent of $P$. Please explain what's the motivation for the steps in this manipulation. Thanks.

April wrote: Let $ ABC$ be a fixed triangle, and let $ A_1$, $ B_1$, $ C_1$ be the midpoints of sides $ BC$, $ CA$, $ AB$, respectively. Let $ P$ be a variable point on the circumcircle. Let lines $ PA_1$, $ PB_1$, $ PC_1$ meet the circumcircle again at $ A'$, $ B'$, $ C'$, respectively. Assume that the points $ A$, $ B$, $ C$, $ A'$, $ B'$, $ C'$ are distinct, and lines $ AA'$, $ BB'$, $ CC'$ form a triangle. Prove that the area of this triangle does not depend on $ P$. Author: Christopher Bradley, United Kingdom By Pascal's Theorem, we see that $X=\overline{BB'} \cap \overline{CC'}$ lies on line $\overline{B_1C_1}$. Define $Y, Z$ similarly. Observe the Lemma. Let $A, B, C, D$ lie on a circle $\omega$ and $E$ be a point inside it. Lines $\overline{EA}, \overline{EB}, \overline{EC}, \overline{ED}$ meet $\omega$ again at $A', B', C', D'$ respectively. Then $(A,B,C,D)=(A',B',C',D')$. (Proof) Consider a projective transformation which fixes $\omega$ and takes $E$ to its center. The result is obvious now. $\blacksquare$ Let $\overline{BB_1}$ meet $(ABC)$ again at $B_0$. Apply the previous lemma, to get $$(B_1C_1, X\infty) \overset{B}{=} (B_0A, B'C) \overset{B_1}{=} (BC, PA).$$Let $P_1 \in (A_1B_1C_1)$ such that $A_1B_1C_1P_1$ and $ABCP$ are similar quadrilaterals. Introduce barycentrics, $A_1=(1,0,0), B_1=(0,1,0)$ and $C_1=(0,0,1)$. Set $P_1=(u, v, w)$ and $a'=\tfrac{a^2}{u}, b'=\tfrac{b^2}{v}, c'=\tfrac{c^2}{w}$; by equation of circumcircle, we have $$a'+b'+c'=0.$$By the cross-ratio condition, we have $X=(0:b':c')$. Likewise, $Y=(a':0:b')$ and $Z=(a':b':0)$. By area formula, $$\frac{[XYZ]}{[A_1B_1C_1]}=\frac{1}{(b'+c')(c'+a')(a'+b')}\det \begin{pmatrix} 0 & b' & c' \\ a' & 0 & c' \\ a' & b' & 0 \\ \end{pmatrix}=-2,$$where area is signed. Thus, $[XYZ]$ doesn't depend on $P$, as desired. $\blacksquare$

gold1 wrote: can't we solve it without using strong theorems like pascal's theorem?… Of course it can. The following thoughtless barybash was done by me without drawing any pictures, making zero synthetic observations, and not knowing Pascal's theorem. And it could be done in steady hands in <15 minutes (But for some reasons took me >5 hours because I was complex bashing first).

Konigsberg wrote: CantonMathGuy wrote: \begin{eqnarray*} [XYZ] & = & [XYA] + [XAZ]\\ & = & ([XCA] + [CYA]) + ([XAB] + [BAZ])\\ & = & [C_1CA]\cdot\frac{B_1X}{B_1C_1} + [CC_1A] + [B_1AB]\cdot\frac{XC_1}{B_1C_1} + [BAB_1]\\ & = & \frac{[BCA]}{2}\cdot\frac{B_1X}{B_1C_1} + \frac{[CBA]}{2} + \frac{[CAB]}{2}\cdot\frac{XC_1}{B_1C_1} + \frac{[BAC]}{2}\\ & = & \frac{[ABC]}{2}\cdot\left(\frac{B_1X}{B_1C_1}-1+\frac{XC_1}{B_1C_1}-1\right)\\ & = & -\frac{[ABC]}{2}, \end{eqnarray*}where areas are signed. Thus, $[XYZ]$ is independent of $P$. Please explain what's the motivation for the steps in this manipulation. Thanks. Surprisingly I have the same sol. as cantonmathguy but for the sake of the motivation I say - Note there are midpoints along with a preety diagram with area bisectors [medians] and so parallel lines for the collinearity by pascals th. So a good area chasing ! Still for a motivation you can call up the scenario of the proof[by trivial area chasing] of the th- The area of the parallelogram formed by joining the midpoints of a quadrilateral is half of the area of the original quadrilateral !

Define $X=BB'\cap CC'$, and similarly define $Y,Z$. Then, by Pascal on $PA'ABCC'$, we get that $Y$ lies on $A_1C_1$. Similarly, $X$ lies on $B_1C_1$ and $Z$ lies on $X_1Y_1$. Now, note that $\frac{XY}{XC}=\frac{XC_1}{a/2-XC_1}$, $\frac{XZ}{XB}=\frac{XB_1}{a/2-XB_1}=\frac{a/2-XC_1}{XC_1}$. So, $$\frac{[XYZ]}{[XBC]}=\frac{XY}{XC}\cdot\frac{XZ}{XB}\cdot\frac{\sin YXZ}{\sin BXC}=1\implies [XYZ]=[XBC]=\frac{1}{2}[ABC]$$which is a constant, as desired.

In what follows, \([\bullet]\) denotes area. Let the triangle formed by the three lines be \(XYZ\). I claim \([XYZ]=\frac12[ABC]\), which will suffice. In what follows, the two claims allow the lemma to solve the problem. [asy][asy] size(5cm); defaultpen(fontsize(10pt)); pair O,A,B,C,A1,B1,C1,P,Ap,Bp,Cp,X; O=(0,0); A=dir(110); B=dir(210); C=dir(330); P=dir(50); A1=(B+C)/2; B1=(C+A)/2; C1=(A+B)/2; Ap=2*foot(O,P,A1)-P; Bp=2*foot(O,P,B1)-P; Cp=2*foot(O,P,C1)-P; X=extension(B,Bp,C,Cp); draw(P--Bp,gray); draw(P--Cp,gray); draw(X--B1,linewidth(0.4)); draw(Bp--X--C,dashed); draw(circle(O,1)); draw(A--B--C--cycle); dot("\(A\)",A,A); dot("\(B\)",B,SW); dot("\(C\)",C,C); dot("\(B_1\)",B1,E); dot("\(C_1\)",C1,NW); dot("\(P\)",P,P); dot("\(B'\)",Bp,Bp); dot("\(C'\)",Cp,SW); dot("\(X\)",X,W); [/asy][/asy] Claim: \(X\) lies on \(\overline{B_1C_1}\), and similarly \(Y\in\overline{C_1A_1}\), \(Z\in\overline{A_1B_1}\). Proof. Apply Pascal theorem to \(BB'PC'CA\). \(\blacksquare\) Claim: \(\overline{AX}\parallel\overline{BY}\parallel\overline{CZ}\). Proof. Apply Pappus theorem to \(BACZA_1Y\) to show \(\overline{BY}\parallel\overline{CZ}\). \(\blacksquare\) Lemma: [Fixed area on medial triangle] Let \(ABC\) be a triangle with medial triangle \(A_1B_1C_1\). If \(X\), \(Y\), \(Z\) lie on \(\overline{B_1C_1}\), \(\overline{C_1A_1}\), \(\overline{A_1B_1}\) with \(\overline{AX}\parallel\overline{BY}\parallel\overline{CZ}\), then \([XYZ]=\frac12[ABC]\). Proof. Using directed areas, we have \[ [XYZ]=[XYA]+[XAZ]=[XBA]+[XAC]=\tfrac12[ABC],\]as needed. \(\blacksquare\) Remark: In general, for any triangle \(ABC\) with medial triangle \(A_1B_1C_1\) and points \(X\), \(Y\), \(Z\) on \(\overline{B_1C_1}\), \(\overline{C_1A_1}\), \(\overline{A_1B_1}\), the following two conditions are equivalent: \(\overline{AX}\parallel\overline{BY}\parallel\overline{CZ}\); \(A\in\overline{YZ}\), \(B\in\overline{ZX}\), \(C\in\overline{XY}\). After noting the first two claims, the problem statement is equivalent to the lemma.

[asy][asy] unitsize(1.25inch); pair A, B, C, P, D, E, F, G, H, I, X, Y, Z; A = dir(110); B = dir(200); C = dir(340); draw(A--B--C--cycle); P = dir(240); draw(circumcircle(A, B, C), dotted); D = .5B+.5C; E = .5C+.5A; F = .5A+.5B; G = 2*foot(0, P, D) - P; H = 2*foot(0, P, E) - P; I = 2*foot(0, P, F) - P; Z = extension(A, G, B, H); X = extension(B, H, C, I); Y = extension(C, I, A, G); draw(A--X, lightblue); draw(B--Y, lightblue); draw(C--Z, lightblue); draw(D--E--F--cycle, dashed); draw(F--Y, dashed); draw(E--Z, dashed); draw(Y--G, lightred); draw(B--H, lightred); draw(C--Y, lightred); draw(P--G, dotted); draw(P--H, dotted); draw(P--I, dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(P)); dot("$A'$", G, dir(G)); dot("$B'$", H, dir(H)); dot("$C'$", I, dir(90)); dot("$A_1$", D, dir(D)); dot("$B_1$", E, dir(0)); dot("$C_1$", F, dir(180)); dot("$X$", X, dir(265)); dot("$Y$", Y, dir(Y)); dot("$Z$", Z, dir(90)); [/asy][/asy] Let $BB'\cap CC' = X$, $CC'\cap AA' = Y$, and $AA'\cap BB' = Z$. By Pascal's on $PB'BACC'$, $B_1, X, C_1$ are collinear. Similarly, $Y$ lies on $C_1A_1$ and $Z$ lies on $A_1B_1$. By similar triangles from parallel lines, $\frac{XC}{XY} = \frac{XB_1}{XC_1} = \frac{XZ}{XB}$, so $XC\cdot XB = XY\cdot XZ$ and $[XYZ] = [XBC] = \frac{1}{2}[ABC]$.

[asy][asy] size(10cm); pair A=dir(110); pair B=dir(210); pair C=dir(330); pair P=dir(50); pair A_1=(B+C)/2; pair B_1=(A+C)/2; pair C_1=(A+B)/2; pair D=dir(256.1); pair E=dir(240.3); pair F=dir(177.3); pair X=extension(C,F,B,E); pair Y=extension(A,D,C,F); pair Z=extension(A,D,B,E); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$P$",P,dir(P)); dot("$A_1$",A_1,SE); dot("$B_1$",B_1,N*3); dot("$C_1$",C_1,NW); dot("$A'$",D,NE); dot("$B'$",E,dir(E)); dot("$C'$",F,NW); dot("$X$",X,dir(X)); dot("$Y$",Y,NE); dot("$Z$",Z,dir(Z)); draw(A--B--C--cycle); draw(unitcircle); draw(D--P);draw(E--P); draw(F--P); draw(X--Y--Z--cycle,blue+linewidth(3)); draw(X--B_1--Z,gray+dotted); draw(A_1--C_1,gray+dotted); draw(A--X,purple+linewidth(2)); draw(B--Y,purple+linewidth(2)); draw(C--Z,purple+linewidth(2)); [/asy][/asy] Three steps - First, Pascal on $PC'CABB'$ gives $B_1,C_1, X$ collinear (similarly, $Y\in C_1A_1$ and $Z\in A_1B_1$). Then, Pappus on $BACZA_1Y$ shows $\infty_{AB},\infty_{AC}$, and $BY\cap CZ$ collinear (similar applications show that $AX||BY||CZ$). Last we have \[[XYZ]=[XYA]+[XAZ]=[XYA]+[XAC]=[YAC]=\frac12[ABC]. \]

Define the intersection of $BC$ and $B'C'$ to be $D$. Let $E,F$ be defined similarly By Pappus on $\{F,A,E\}$ and $\{B,A_1,C\}$, we obtain $BE \parallel CF \parallel AD$. By Pappus on $\{E,A,F\}$ and $\{B,A_1,C\}$, we obtain $C_1,B_1,D$ are collinear. Hence we have : $$ [DEF]= [DEA]+[DAF]=[DBA]+[DAC]= 2[DC_1A]+2[DAB_1]= 2[AB_1C_1] = \underbrace {\tfrac {[ABC]}{2}}_{const}$$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9.415895739507954cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -2.946462440308137, xmax = 3.3308013860304992, ymin = -1.6367926738977168, ymax = 1.5473816612281053; /* image dimensions */ pen evffff = rgb(0.8980392156862745,1.,1.); pen qqffff = rgb(0.,1.,1.); pen ffevff = rgb(1.,0.8980392156862745,1.); pen ffqqff = rgb(1.,0.,1.); pair A = (-0.6,0.8), B = (-0.8607523200613243,-0.5090240107392259), C = (0.8944975633612999,-0.44707282308444696), A_1 = (0.016872621649987818,-0.4780484169118364), B_1 = (0.14724878168065003,0.1764635884577766), C_1 = (-0.7303761600306621,0.14548799463038714), P = (-0.3843472264507031,-0.923188609938214), X = (0.2352022062942822,0.6180057562088045), Y = (-0.20629981905345793,0.16398516249915795), Z = (-1.8380468069791136,1.069775858727651); filldraw(A--B--C--cycle, evffff, linewidth(0.8) + qqffff); filldraw(Z--C_1--B--cycle, ffevff, linewidth(0.8) + linetype("2 2") + ffqqff); filldraw(X--B_1--C--cycle, ffevff, linewidth(0.8) + linetype("2 2") + ffqqff); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.2) + linetype("0 3 4 3")); draw(A--B, linewidth(0.8) + qqffff); draw(B--C, linewidth(0.8) + qqffff); draw(C--A, linewidth(0.8) + qqffff); draw(P--(0.8784469471322907,0.47783989062651366), linewidth(0.8)); draw(P--(0.484764786029156,0.874644557649624), linewidth(0.8)); draw((-0.8525151849490672,0.522702457839312)--P, linewidth(0.8)); draw((0.8784469471322907,0.47783989062651366)--A, linewidth(0.8)); draw((0.484764786029156,0.874644557649624)--B, linewidth(0.8)); draw(C--(-0.8525151849490672,0.522702457839312), linewidth(0.8)); draw(A--Z, linewidth(0.8)); draw(Z--(-0.8525151849490672,0.522702457839312), linewidth(0.8)); draw(Z--A_1, linewidth(0.8)); draw(X--A_1, linewidth(0.8)); draw(C_1--B_1, linewidth(0.8)); draw(Z--C_1, linewidth(0.8) + linetype("2 2") + ffqqff); draw(C_1--B, linewidth(0.8) + linetype("2 2") + ffqqff); draw(B--Z, linewidth(0.8) + linetype("2 2") + red); draw(X--B_1, linewidth(0.8) + linetype("2 2") + ffqqff); draw(B_1--C, linewidth(0.8) + linetype("2 2") + ffqqff); draw(C--X, linewidth(0.8) + linetype("2 2") + red); draw(Y--A, linewidth(0.8) + linetype("2 2") + red); /* dots and labels */ dot(A,blue); label("$A$", (-0.5858469022291581,0.8376789190725288), NE * labelscalefactor,blue); dot(B,blue); label("$B$", (-0.9539814797109282,-0.623473785365423), NE * labelscalefactor,blue); dot(C,blue); label("$C$", (0.9512098388029746,-0.5172081341335719), NE * labelscalefactor,blue); dot(A_1,blue); label("$A_1$", (0.047951803332239855,-0.5893169688980422), NE * labelscalefactor,blue); dot(B_1,blue); label("$B_1$", (0.08210861979962059,0.29876025925385563), NE * labelscalefactor,blue); dot(C_1,blue); label("$C_1$", (-0.8742822412870398,0.11659057142782532), NE * labelscalefactor,blue); dot(P,blue); label("$P$", (-0.4492196363596352,-1.0371507848037003), NE * labelscalefactor,blue); dot((0.8784469471322907,0.47783989062651366),blue); label("$A'$", (0.8942818113573401,0.5150867635472667), NE * labelscalefactor,blue); dot((0.484764786029156,0.874644557649624),blue); label("$B'$", (0.49958082106760726,0.913582955666708), NE * labelscalefactor,blue); dot((-0.8525151849490672,0.522702457839312),blue); label("$C'$", (-0.9008486540950026,0.583400396482028), NE * labelscalefactor,blue); dot(X,blue); label("$X$", (0.20355507835030762,0.6744852403950432), NE * labelscalefactor,blue); dot(Y,blue); label("$Y$", (-0.20253151742855213,0.02930092934451911), NE * labelscalefactor,blue); dot(Z,blue); label("$Z$", (-1.8230826987142823,1.1071382489818653), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] ok here is my solution by pascal on $ABPB'A'C$ we observe that $X,B_1,A_1$ are collinear the same goes for $(Z,C_1,A_1)$ and $(C_1,Y,B_1)$.on the other hand $\triangle XB_1C$ and $\triangle ZC_1B$ are perspective so by desargues theorem we get that $BZ||XC$ and similarly $AY||BZ||CX$ from now on $S_{ABC}$ denotes the area of $\triangle ABC$ by the parallelity claim we have \begin{align*} S_{XYZ} &=S_{AXY}+S_{AYZ} \\ &=S_{ABY}+S_{ACY}\\ &=S_{ABC}-S_{BYC}=\frac{S_{ABC}}{2} \quad \text{which is independent of P} \end{align*}

Let $X=BB'\cap CC',Y=AA'\cap CC',Z=AA'\cap BB'.$ By the Pascal theorem for $ABB'PC'C$ we have $X\in B_1C_1,$ and analogously $Y\in A_1C_1,Z\in A_1B_1.$ Next triangles $AB_1X,YA_1B$ are homothetic, since $AY,XB,B_1A_1$ concur at $Z.$ Thus $AX\parallel BY,$ and also $AX\parallel CZ.$ Finally we see (directed areas) $$\text{area}(XYZ)=\text{area}(AXZ)-\text{area}(AXY)=\text{area}(AXZ)-\text{area}(AXB)=$$$$=\text{area}(ABZ)=\text{area}(ABA_1)=\frac{1}{2}\text{area}(ABC).$$

Let $\overline{AA'},\overline{BB'},$ and $\overline{CC'}$ form triangle $DEF.$ By Pascal on $ABB'PC'C,$ we see $D$ lies on $\overline{B_1C_1}.$ Similarly, $E$ and $F$ lie on $\overline{A_1C_1}$ and $\overline{A_1B_1},$ respectively. By Pappus on $A,E,F$ and $B,A_1,C,$ we see $\overline{BE}\parallel\overline{CF}.$ Similarly, $\overline{AD}\parallel\overline{BE}\parallel\overline{CF}.$ Notice $[DEF]=[CBD]$ and $$[DEF]=[ADE]+[ADF]=[ABD]+[ACD]=[ABC]-[CBD]$$so $[DEF]=\tfrac{1}{2}[ABC].$ $\square$

Let $AA', BB', CC'$ form a triangle $XYZ$. Using Pascal's Theorem, we have $D$ on $B_1C_1$, $E$ on $A_1C_1$, $F$ on $A_1B_1$. Now, Pappus' Theorem yields $AD \parallel CF \parallel BE$. Use barycentric coordinates. Note that we have $X=(0.5, m, 0.5-m), Y=(n, 0.5, 0.5-n), Z=(q,0.5-q,0.5)$. Thus the ratio area is $$\frac{-2mn+2mq-2nq+n}{2},$$which comes out as $\frac{1}{2}$ upon applying the parallel condition. $\blacksquare$

Let $P_A$ be the intersection of $BB'$ and $CC'$, and define $P_B$ and $P_C$ similarly. By Pascal on $PC'CABB'$, $C_1,B_1,P_A$ are collinear. Similarly, $A_1,B_1,P_C$ are collinear and $A_1,C_1,P_B$ are collinear. Then, by Pappus, $AP_A,BP_B,CP_C$ are pairwise parallel. Then, $$[P_AP_BP_C]=[AP_AP_C]-[AP_AP_B].$$Since $AP_A\parallel BP_B$, we have $[AP_AP_B]=[AP_AB]$, so $$[AP_AP_C]-[AP_AP_B]=[AP_AP_C]-[AP_AB]=[ABP_C].$$However, from $A_1B_1$ passes through $P_C$ as shown earlier, and $A_1B_1\parallel AB$, we have $$[ABP_C]=[ABA_1]=\frac{1}{2}[ABC],$$so we are done.

Let $AA'$ and $BB'$ intersect at $Z$ and similarly the other two points are $X$ and $Y$. By Pascal's on $B'PC'CAB$, $X$ is on $B_1C_1$, and similarly $Y$ is on $A_1C_1$, and $Z$ is on $A_1B_1$. Since $BA$ and $ZA_1$ are parallel, $AC$ and $A_1Y$ are parallel, by Pappus's theorem $BY$ and $ZC$ are parallel. Similarly, $AX$ is parallel to them both. Thus, we have \begin{align*} [XYZ] &= [XBY] + [ZBY] \\ &= [ABY] + [CBY] \\ &= [BAYC] \\ &= \frac12 [ABC] \end{align*}which is independent of $P$.

Pascal bash FTW wlog p is on arc AB not containing C, and let D, E, F be the cyclic permutations of intersections of AA’, BB’, and CC’, according by BB’ with CC’ first. Now, on degenerate hexagons we find DYZ is a line, EZX is one, and FXY is another. Since XE//AC and similarly we have by sim triangles FE/FA=FX/FY=FB/FD, and consequently [DEF]=FE*FDsinDEF/2=FA*FBsinAFB/2=[AFB]=[ABC]/2, where the last step is because trigially they have same base but half the height (altitude dropped from C to XY is half to AB by sim triangles), which is constant. $\blacksquare$

Wow everyone else has solutions that are actually good and easy. Rename $A’\to P_a$, $B’\to P_b$ and $C’\to P_c$. Let $X=\overline{BP_b}\cap\overline{CP_c}$, $Y=\overline{CP_c}\cap\overline{AP_a}$ and $Z=\overline{AP_a}\cap\overline{BP_b}$. Let $\omega=(ABC)$. Pascal on $ABP_bPP_cC$ then gives us that $B_1-X-C_1$. So far we haven’t derived any ratio-like or length-like condition to compute the area. Let’s do so right now. Notice that if $C_2=(ABC)\cap\overline{CC_1}\neq C$ then $(B,C;P,A)_\omega\stackrel{C_1}{=}(C_2,A;P_c,B)_\omega\stackrel{C}{=}(\infty_{BC},X;B_1,C_1)$. Hence expanding the cross-ratios: \begin{align*} \frac{\infty_{BC}B_1}{\infty_{BC}C_1}\frac{XC_1}{XB_1}&=\frac{BP}{BA}\frac{CA}{CP}\\ \frac{XC_1}{B_1X}&=\frac{BP}{PC}\frac{CA}{BA} \end{align*}where we’ve directed every single angle. So similarly: \begin{align*} \frac{ZA_1}{B_1Z}&=\frac{BP}{PA}\frac{AC}{BC}\\ \frac{YA_1}{C_1Y}&=\frac{CP}{PA}\frac{AB}{CB}\\ \end{align*}Now we’re almost done. What’s left now is to compute $\frac{[\triangle XYZ]}{[\triangle A_1B_1C_1]}$. However: \begin{align*} \frac{[\triangle XYZ]}{[\triangle A_1B_1C_1]}&=1-\frac{C_1X}{C_1B_1}\frac{C_1Y}{C_1A_1}-\frac{B_1X}{B_1C_1}\frac{B_1Z}{B_1A_1}-\frac{A_1Y}{A_1C_1}\frac{A_1Z}{A_1B_1}\\ &=1-\left(\frac{1}{1+\frac{CP}{PB}\frac{BA}{CA}}\frac{1}{1+\frac{CP}{PA}\frac{AB}{CB}}+\frac{1}{1+\frac{BP}{PC}\frac{CA}{BA}}\frac{1}{1+\frac{BP}{PA}\frac{AC}{BC}}+\frac{1}{1+\frac{AP}{PC}\frac{CB}{AB}}\frac{1}{1+\frac{AP}{PB}\frac{BC}{AC}}\right) \end{align*}so it suffices to show that the part in brackets is constant. \begin{align*} &\frac{1}{1+\frac{CP}{PB}\frac{BA}{CA}}\frac{1}{1+\frac{CP}{PA}\frac{AB}{CB}}+\frac{1}{1+\frac{BP}{PC}\frac{CA}{BA}}\frac{1}{1+\frac{BP}{PA}\frac{AC}{BC}}+\frac{1}{1+\frac{AP}{PC}\frac{CB}{AB}}\frac{1}{1+\frac{AP}{PB}\frac{BC}{AC}}\\ &=\frac{(PB\cdot CA)(PA\cdot CB)}{(PB\cdot CA+CP\cdot BA)(PA\cdot CB+CP\cdot AB)}+\frac{(PC\cdot BA)(PA\cdot BC)}{(PC\cdot BA+BP\cdot CA)(PA\cdot BC+BP\cdot AC)}+\frac{(PC\cdot AB)(PB\cdot AC)}{(PC\cdot AB+AP\cdot CB)(PB\cdot AC+AP\cdot BC)}\\ &=\frac{(PB\cdot CA)(PA\cdot CB)}{(PB\cdot CA)(PA\cdot CB)}+\frac{(PC\cdot BA)(PA\cdot BC)}{(PC\cdot BA)(PA\cdot BC)}+\frac{(PC\cdot AB)(PB\cdot AC)}{(PC\cdot AB)(PB\cdot AC)}\\ &=3 \end{align*}where we used a form of directed Ptolemy (since I don’t want to bash out every single little case so I’m cheating a bit here). Hence $\frac{[\triangle XYZ]}{[\triangle A_1B_1C_1]}=-2$, i.e. $[\triangle XYZ]=-\frac{1}{2}[\triangle ABC]$ and clearly we have a fixed area. We can thus stop right here.

Very nice problem, the trick is in bringing the area thing only once you you are sure of what you cooked in order to make everything add up nicely. Let $P_A, P_B, P_C$ points in $(ABC)$ such that $PP_A \parallel BC$ and cyclic permutations are true. Claim 1: $AX \parallel BY \parallel CZ$ Proof: Note that from Pascal we have $B_1, X, C_1$ colinear (and thus cyclic permutations are true as well), now from DDIT on $BC_1B_1C$ with perspector $P$ we have that $(BP, B_1P), (C_1P, CP), (AP, \infty_{BC}P)$ are in pairs of involution, so projecting on $(ABC)$ gives that $A,X,P_A$ are colinear (and cyclic permutations of this are true as well), so now note that from arcs on $(ABC)$ that $\widehat{P_AC}=\widehat{BP}=\widehat{AP_B}$ which implies adding up with the things above that $AX \parallel CZ$, same way you get $AX \parallel BY$ thus the claim is proven. The finish: From thales we have $YX \cdot XZ=BX \cdot XC$ so from sine area formula we have that $[XYZ]=[BXC]$ and as $X$ lies in $B_1C_1$ that area is just half altitude times $BC$ which is $0.5 \cdot [ABC]$ being constant, thus we are done .