croodinate ,let,B,C on x line,yA>0,only need to prove yB1+yC1+yA1>0 but yB1,yC1>yp>0,yA1=-yP

let $X,Y,Z$ be midpoints of $PA_{1},PB_{1},PC_{1}$ and $K,L,M$ be midpoints of $B_{1}C_{1},A_{1}C_{1},A_{1}B_{1}$ the centroid of $A_{1}B_{1}C_{1}$ is the centroid of $KLM$. now since $\angle KZP=180-\angle ZPY=\angle BAC<90=\angle AZP$ so $Y,K$ are on the same side of $AZ$ similarly $Z,K$ are on the same side of $AY$ and since $PYKZ$ is a parallelogram then $P,K$ are on opposite sides of $XY$ so $K,A$ are on the same side of $XY$ so $K$ is inside $\triangle AXY$ so it's inside $ABC$ similarly $L,M$ are inside $ABC$ and so $KLM$ is inside $ABC$ and since the centroid of $KLM$ is inside $KLM$ it is inside $ABC$.

leader wrote: let $X,Y,Z$ be midpoin of $PA,PB,PC$ and $K,L,M$ be midpoints of $B_{1}C_{1},A_{1}C_{1},A_{1}B_{1}$ the centroid of $A_{1}B_{1}C_{1}$ is the centroid of $KLM$. now since $\angle KZP=180-\angle ZPY=\angle BAC<90=\angle AZP$ so $Y,K$ are on the same side of $AZ$ similarly $Z,K$ are on the same side of $AY$ and since $PYKZ$ is a parallelogram then $P,K$ are on opposite sides of $XY$ so $K,A$ are on the same side of $XY$ so $K$ is inside $\triangle AXY$ so it's inside $ABC$ similarly $L,M$ are inside $ABC$ and so $KLM$ is inside $ABC$ and since the centroid of $KLM$ is inside $KLM$ it is inside $ABC$. Just e benign typo, and so $X,Y,Z$ are the midpoints of $PA_1, PB_1, PC_1$. On the other hand, congrats for the beautiful solution.

ArefS wrote: $P$ is an arbitrary point inside acute triangle $ABC$. Let $A_1,B_1,C_1$ be the reflections of point $P$ with respect to sides $BC,CA,AB$. Prove that the centroid of triangle $A_1B_1C_1$ lies inside triangle $ABC$. Please correct me if I'm wrong... Probably I don't understand the problem,but: Since $ABC$ is acute-angled and $P$ is inside of $ABC$, triangle $A_1B_1C_1$ lies inside of $ABC$. It is obvious that centroid of any triangle lies inside of them, so centroid of $A_1B_1C_1$ lies inside of $A_1B_1C_1$, which implies that it lies inside of $ABC$, we're done!

$\triangle A_1B_1C_1 $ is triangle which after reflect $P$ wrt $BC,CA,AB$ we get , so $\triangle A_1B_1C_1$ is not lie on $\triangle ABC$

War-Hammer wrote: $\triangle A_1B_1C_1 $ is triangle which after reflect $P$ wrt $BC,CA,AB$ we get , so $\triangle A_1B_1C_1$ is not lie on $\triangle ABC$ Aaah..)) I understand my mistake) I had some troubles with the translate... I thought that $A_1$, $B_1$ and $C_1$ are the feet of perpendiculars from P...

Let $PA_1$ and $PB_1$ intersect respectively $BC,CA$ in $X,Y$.Then $PX=XA_1$ and if the midpoint of $A_1B_1$ lies on $A_1X$ then we get that the point that cuts the perimeter of triangle $PXY$ considered from the point $P$ lies on $PX$-absurd.Analogously we get that the midpoint of $A_1B_1$ lies on $XY$.Then it is easy to see that point that cuts the median $2:1$ obviously lies incide $ABC$.

Suppose $C_1B_1$ intersect $AB$ and $BC$ at $L$ and $M$,similarly $C_1A_1$ at $N,D$ and $A_1B_1$ at $X,Y$. Now $[C_1NL]=[NLP],[PDX]=A_1DX],[PMY]=[MYB_1]$,so area of the hexagon greater then $[C_1NL]+[A_1DX]+[B_1MY]$,hence we are done.

First let us assume that the midpoint M of A1B1 lies outside the triangle ABC (wlg let M be opposite to A with respect to BC). Let PA1 cut BC at X and XM cut AC produced at D. XM||PB1, ie XD||PB1, <XDC=90(since PB1 is perpendicular to AC) then the exterior angle <XCA of triangle XCD is >90, but this is impossible given that ABC is an acute angled triangle. This implies that our assumption that M is outside ABC if false or otherwise M is inside ABC. Similarly it can be shown that the midpoints of all 3 sides of triangle A1B1C1 are inside ABC triangle. Then obviously the centroid of the medial triangle of A1B1C1 that coincides with the centroid of A1B1C1 lies inside ABC triangle.

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Outline: This problem is ideally suited to barycentric coordinates. We first prove a lemma, and then the coordinates kill the problem. Lemma: Let $ABC$ be an acute triangle and $P$ a point inside $\triangle ABC.$ Let $D, E, F$ be the feet of the perpendiculars from $P$ to sides $BC, CA, AB$, respectively. Then \[\frac{FB}{AB} + \frac{EC}{CA} > 2 \frac{[BPC]}{[ABC]}\] where $[XYZ]$ denotes the area of $\triangle XYZ.$ Cyclic permutations hold. Proof. We adopt the convention that $d(X, l)$ represents the distance from a point $X$ to line $l.$ The inequality is then equivalent to \[\frac{FB}{AB} + \frac{EC}{CA} > 2 \cdot \frac{d(P, BC)}{d(A, BC)}.\] Noting that $d(A, BC) = AB \cdot \sin B = AC \cdot \sin C$, this is then equivalent to \[FB \cdot \sin B + EC \cdot \sin C > 2 \cdot d(P, BC)\]\[\iff d(F, BC) + d(E, BC) > 2 \cdot d(P, BC)\] which is clearly true, since $d(F, BC) > d(P, BC)$ and $d(E, BC) > d(P, BC).$ Thus, the Lemma is proven. $\blacksquare$ Now, for the main proof, denote by $D, E, F$ the feet of the perpendiculars from $P$ to $\overline{BC}, \overline{CA}, \overline{AB}.$ We proceed with barycentric coordinates. Clearly we have $D = \left(0 : DC : DB\right) = \left(0, \frac{DC}{BC}, \frac{DB}{BC}\right).$ Similarly, we have that $E = \left(\frac{EC}{CA}, 0, \frac{EA}{CA}\right)$ and $F = \left(\frac{FB}{AB}, \frac{FA}{AB}, 0\right).$ In addition, since $D$ is the midpoint of $PA_1$, it follows that $P + A_1 = 2D \implies A_1 = 2D - P.$ Similarly, we have that $B_1 = 2E - P$ and $C_1 = 2F - P.$ Therefore, by our Lemma and some basic properties of barycentric coordinates, it follows that the $x$ coordinate of $G_1$, the centroid of $A{}_1B{}_1C{}_1$, satisfies \[G{}_1{}_x = \frac{1}{3}\left(A{}_1{}_x + B{}_1{}_x + C{}_1{}_x\right) = \frac{1}{3}\left(2D_x +2E_x + 2F_x - 3P_x\right)\]\[= \frac{1}{3}\left(2 \cdot \frac{EC}{CA} + 2 \cdot \frac{FB}{AB} - 3 \cdot \frac{[BPC]}{[ABC]}\right) > \frac{1}{3}\left(2 \cdot \frac{EC}{CA} + 2 \cdot \frac{FB}{AB} - 4 \cdot \frac{[BPC]}{[ABC]}\right) > 0.\] Similarly, we have $G{}_1{}_y > 0$ and $G{}_1{}_z > 0.$ Therefore, $G_1 \in \triangle ABC$, as desired. $\blacksquare$

My solution: First observe that since $\angle A<90^{\circ}\Longrightarrow \angle B_1PC_1=180^{\circ}-\angle A>90^{\circ}$ so $A$ the circumcenter of $\triangle B_1PC_1$ lies outside of $\triangle B_1PC_1$ and $B_1C_1$ cuts the segments $AB,AC$. we can say similar arguments about $B,C$.($\bigstar$) Let $Q$ be the isogonal conjugate of $P$. since $Q$ is circumcenter of $\triangle A_1B_1C_1$ we get that $AQ$ perpendicular bisector of $B_1C_1$ so $B_1C_1\cap AQ=A_0$ is midpoint of $B_1C_1$. combine this with ($\bigstar$) and the fact that $Q$ lies inside $\triangle ABC$ we get that $A_0$ is inside the $\triangle ABC$. similarly $B_0,C_0$ lie inside $\triangle ABC$ hence centroid of $\triangle A_0B_0C_0$ lies inside $\triangle ABC$ but it is also centroid of $\triangle A_1B_1C_1$. Q.E.D