Can you send me the photo of all problems for all grades. Doesn't matter in Russian or in English (as private message) please!

Isnt anybody to solve this question??!!!!

Let $D\equiv \omega \cap BC,E\equiv \omega \cap AB,F\equiv \omega \cap CA$ We want to prove that this way does work (Pay attention that 7 lines is eanough to find $A',B',C'$): 1- Draw $BC$. 2- Draw $EF$ and let $K \equiv BC \cap EF$. 3- Draw $ED$. 4- Draw $BF$ and let $P \equiv DE \cap BF$. 5- Draw $PK$ and let $A' \equiv KP \cap \overset{\frown }{FD}$. 6- Draw $CQ$ and suppose that $B'$ be the second intersection of $CA',\omega$. 7- Draw $AB'$ and suppose that $C'$ be the second intersection of $AB',\omega$. Proof: For each point on $\overset{\frown }{FD}$ like $X$ define the function $f(X)$ such that: Suppose that $X'$ be the second intersection of $CX,\omega$ and $X''$ be the second intersection of $AX',\omega$ and finally $f(X)$ be the second intersection of $BX'',\omega$. We call the point $X$ is a fix point if and only if $f(X)=X$. It's obvious that $f(X)$ fixed the inharmonic so if for three point like $X,Y,Z$ we have $f(X),f(Y),f(Z)$ we can get $f(T)$ for all point like $T$ on $\overset{\frown }{FD}$. For each point $T$ we want to find $f(T)$. Let $M\equiv Xf(Y) \cap Yf(X),K\equiv Yf(Z) \cap Zf(Y),N\equiv Yf(Y) \cap MK,P\equiv Tf(Y) \cap MK$ and $T'\equiv PY \cap \omega$. We have $(XYZT)=f(Y)(XYZT)=(MNKP)=Y(MNKP)=(f(X)f(Y)f(Z)T')$ And we know that $(XYZT)=(f(X)f(Y)f(Z)f(T))$ so $T'\equiv f(T)$ and by this fact we can conclude that the fix point is the intersection of $MK$ and $\omega$. Now for finding $A'$ (A fix point) let $X\equiv D$ and suppose that $U$ be the second intersection of $AD,\omega$ and $Y$ be the second intersection of $CU,\omega$ and $Z$ be the second intersection of $CE,\omega$. Now it's obvious that $f(Y)=X,f(Z)=E,f(X)=$ the second intersection of $BU,\omega$. So if we let $K\equiv Xf(Y)\cap Yf(X),M\equiv Zf(X)\cap Xf(Z)$ we have $A'\in MK$. Now by these obvious lemmas the result follows:

Use a projective transformation sending the Gergonne point of ABC to it's incenter and preserving the incircle. Then it's easy to see the symmedian due to harmonic quadrilaterals, and the line A'X is parallel to BC where X is the midpoint of A-intouch chord (angle chase and similarity). So we get a projective fact if we convert this to intersection of lines. And using the fact that projective transformations preserve separation, we get the construction above. Note: This was a projective transformation I was using on AoPS for the last some days, but brushed this thought aside thrice in the paper, so couldn't solve it in the time.

MNZ2000 wrote: Given triangle $ABC$ and its incircle $\omega$ prove you can use just a ruler and drawing at most 8 lines to construct points$A',B',C'$ on $\omega$ such that $A,B',C'$ and $B,C',A'$ and $C,A',B'$ are collinear.

Take a projective transformation that sends the incircle to itself and the Gergonne Point of $\triangle ABC$ to it's center. Notice that all constructions by a ruler are projective, so a construction here has an analog (explicitly stated in post #5) for a general triangle. Refer to diagram of post #5 again. Let's characterize $A', B', C'$. First, by symmetry, we note that $\triangle A'B'C'$ must be regular. Let $A_0B_0C_0$ be the intouch triangle and $K=AA_0 \cap B_0C_0$. Claim: $KB' \parallel AC$, i.e., lines $KB', AC$ and $A_0C_0$ are concurrent. (Proof) WLOG, let $B_0C_0=2$ and let $B_0B'=x$ and $C_0B'=y$; since $C'B_0B'C_0$ is an isosceles trapezoid, we have $C'C_0=x$ and $C'B_0=\frac{y^2}{x}=z$. Apply cosine law to triangles $C_0C'B'$ and $\triangle B_0B'C'$ to obtain $$x^2+y^2+xy=4=x^2+z^2-xz,$$hence $x^2+xy=y^2 \implies 2y^2=4$. Hence, $\angle B_0KB'=60^{\circ}$ as desired. $\blacksquare$ Note. This is basically the generalized form of Sharygin 2017, Grade 9, #1, as can be seen by the arguments for proving the claim.