"It's like someone made configuration issues into a problem" --- during USA training test development

Call a point $X$ $k$-bad if at most $k$ lines $\ell$ through $X$ satisfy $f(\ell)\ne X$. Claim: [due to Jeffrey Kwan] We can't have three collinear $k$-bad points. Proof: Suppose $X,Y,Z$ collinear and $k$-bad. Then vary $P$ not collinear with them until $f(PX)=X$, $f(PY)=Y$, $f(PZ)$, at which point we lose as $P,X,Y,Z$ can't be cyclic. $\blacksquare$ Claim: There are at most $2k+1$ $k$-bad points. Proof: Suppose we have $k$-bad points $X_1,\ldots,X_n$. Double count the number of pairs $(i,j)$ such that $f(X_iX_j)=X_i$. Fixing $i$, we see that at least $n-1-k$ $j$ work, so the number of pairs is at least $n(n-1-k)$. Fixing $j$, we see that at most $k$ $i$ work, so the number of pairs is at most $nk$, so we have \[n(n-1-k)\le nk,\]or $n\le 2k+1$, as desired. $\blacksquare$ For each point $X$, let $\omega(X)$ be a circle of nonzero radius such that $f(\ell)\in\omega(X)$ for each $X\in \ell$ (exists by the problem condition). Pick an arbitrary line $\ell$, and pick $X$ and $Y$ on $\ell$ not equal to $f(\ell)$, and pick $X$ and $Y$ to be $100$-good (we can do this as there are at most $201$ $100$-bad points in the first place). Since $X$ is $100$-good, we may pick a line $\ell'\ne\ell$ through $X$ such that $f(\ell')\ne X$. Draw $99$ more lines through $Y$ called $m_1,\ldots,m_{99}$ such that $f(m_i)\ne Y$, and let $Z_i=m_i\cap\ell'$. \We have at most two of the $Z_i$ such that $f(YZ_i)=Z_i$ since $Z_i\in\omega(Y)$, and $\omega(Y)$ has at most two intersections with $\ell'$. So WLOG, we have $Z_1,\ldots,Z_{97}\in\ell'$ such that $f(YZ_i)\ne Z_i,Y$. There are at most $81$ $40$-bad points, so some $Z_i$ is $40$-good, let this be $Z$. Thus, we have $X$, $Y$, $Z$, all $40$-good such that $f(XY)\ne X,Y$, $f(XZ)\ne X,Z$, and $f(YZ)\ne Y,Z$. Letting $A=f(YZ)$, $B=f(XZ)$, $C=f(XY)$, we see that $\omega(A)$, $\omega(B)$, $\omega(C)$ concur at some $P$ by Miquel's theorem. Claim: $P$ is the desired point. Proof: Now, pick some $Y'\in YZ$ such that $Y'\ne Y,A,Z$, $Y'\not\in(XBC)$, and $f(XY')\ne X$ (this exists since $X$ is $40$-good). Similarly, pick $Z'\in ZX$, $X'\in XY$. Now, we see that the similarly defined Miquel point of $\triangle XY'Z$ is also $P$, since it is again $\omega(X)\cap\omega(Z)\cap\{B\}$. This is similarly true for $\triangle XYZ'$ and $\triangle X'YZ$. Now pick a point $Q$ far enough away from these $6$ points such that it is on none of $\omega(X),\ldots,\omega(Z')$, and on none of the possible $\binom{6}{3}$ circumcircles. We'll try to show that $P\in\omega(Q)$. We see that $f(QX)\ne Q$ as $Q\not\in\omega(X)$, and similarly for the other $5$. Now, if $f(QX)\ne X$ and $f(QY)\ne Y$, then we'd be done by Miquel on $\triangle QXY$, so WLOG $f(QX)=X$ or $f(QY)=Y$, so $X\in\omega(Q)$ or $Y\in\omega(Q)$. Writing out all similar statements, we actually have \begin{align*} X\in\omega(Q) &\quad\text{or}\quad Y\in\omega(Q) \\ X\in\omega(Q) &\quad\text{or}\quad Z\in\omega(Q) \\ Y\in\omega(Q) &\quad\text{or}\quad Z\in\omega(Q) \\ X\in\omega(Q) &\quad\text{or}\quad Y'\in\omega(Q) \\ Y'\in\omega(Q) &\quad\text{or}\quad Z\in\omega(Q) \\ Y\in\omega(Q) &\quad\text{or}\quad Z'\in\omega(Q) \\ X\in\omega(Q) &\quad\text{or}\quad Z'\in\omega(Q) \\ X'\in\omega(Q) &\quad\text{or}\quad Y\in\omega(Q) \\ X'\in\omega(Q) &\quad\text{or}\quad Z\in\omega(Q). \\ \end{align*}It is easy to manually verify that these Boolean conditions actually imply that three of $\{X,Y,Z,X',Y',Z'\}$ lie on $\omega(Q)$, which is a contradiction, as this implies $Q$ is on one of the $\binom{6}{3}$ circumcircles, which is false for far enough out $Q$. Thus, for far enough out $Q$, we have $P\in\omega(Q)$. Now, fix any line $t$ through $P$. We see that for far enough $Q$ on $t$, we have $f(t)\in\omega(Q)$, so $f(t)\in \omega(Q)\cap\ell=\{P,Q\}$, so by varying $Q$, we see that $f(t)=P$. So for all lines $t$ passing through $P$, we have $f(t)=P$. $\blacksquare$ This shows the existence of at least one such point. If there was another, say $P'$, then $f(PP')$ cannot be both $P$ and $P'$, so we're done. This completes the proof.

During ISL guessing, I guessed that this was G4.

The_Turtle wrote: Let $Z$ be a variable point on $\ell$. The given condition implies $f(\overline{AZ}) \in (AEF)$ along with cyclic variants, and \[Z, f(\ell), f(\overline{AZ}), f(\overline{BZ}), f(\overline{CZ})\]lie on a common circle $\Gamma_Z$. Since $Z$ is variable, \[\lvert\{A, B, C\} \cap \{f(\overline{AZ}), f(\overline{BZ}), f(\overline{CZ})\}\rvert \leq 1;\]otherwise, $\Gamma_Z$ would be fixed. There is a little issue. What happens if $f(l)\in \{A,B,C\}$? For example, if $l$ passes through $C$, $f(AZ)=A$, $f(BZ)=B_Z$, and $f(l)=f(CZ)=C$. $\Gamma_Z$ won't be fixed, but the inequality would be false.

By the way, it's supposed to say "to each line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$" not "$f(\ell)$ on $f(\ell)$" (which makes no sense).

Let $g(X)=\odot (Xf(\ell_1)f(\ell_2)f(\ell_3)).$ Note that for every noncollinear points $A,B,C$ such that $\left\{ A,B,C\right\} \cap\left\{ f(AB),f(BC),f(CA)\right\}=\varnothing$ circles $g(A),g(B),g(C)$ concur by Miquel theorem. Consider arbitrary line $\ell;\mathcal{A} =\bigcup_{i=1}^6 X_i\in \ell\backslash f(\ell),Y\notin \ell \cup \bigcup g(X_i)$. Since $|\mathcal{A} \backslash g(Y)|\geq 4$ we may suppose $\mathcal{B} =\mathcal{A} \backslash \left\{ X_5,X_6\right\} \notin g(Y),$ therefore $g(X_1),g(X_2),g(Y)$ has common point $Q.$ Consider arbitrary point $Z\notin \ell \cup \bigcup g(X_i).$ By the same reason we may suppose that $\mathcal{C} =\mathcal{B} \backslash \left\{ X_3,X_4\right\} \notin g(Z),$ thus $Q\in g(Z).$ Now we claim that $P=Q$ satisfies the condition, so take arbitrary line $\ell '$ passing through $Y.$ If $\ell '=\ell$ we obtain $Y=\ell' \cap g(X_1)\backslash X_1=f(\ell ').$ In the other case take two different cases $Z_1,Z_2\in \ell '$ of $Z.$ We get $$f(\ell ')=g(Z_1)\cap g(Z_2) \cap \ell '=\left\{ Q,Z_1\right\}\cap \left\{ Q,Z_2\right\}=Q.$$ If there exist point $R\neq Q$ satisfying the condition we conclude $R=f(QR)=Q,$ contradiction.

Can anyone proof read my solution?

PS: How do I hide attachments?

First, note that really the problem is assigning a circle to every point such that the circle goes through the point and if $X \in \ell$, then $f(\ell)$ is one of the intersections of $\ell$ with the circle corresponding to $X$. The main point of the problem is to look at a miquel-ish configuration. Suppose there exists a triangle $ABC$ such that $D = f(BC) \neq B,C$ and define $E,F$ similarly. Then by miquel, the circles $(AEF), (BDF), (CDE)$ all concur at a point. First, I'll show that we can find such a triangle. Suppose not, then pick a line $\ell$ and pick $100$ points on it. Pick a point $A$ not on any of the $100$ circles created, so that $f(AP_i) \neq A$. Call a point $P$ among the hundred bad if $f(AP) = P$. But note that this can happen for at most two points, if this happened for $P_1, P_2, P_3$, then the points $(AP_1P_2P_3)$ must be cyclic, impossible since $P_1, P_2, P_3$ are collinear. Further, ignore the case when one of the points equals $f(\ell)$. Among the remaining, call two of them to be $B,C$, they work. Now, say this miquel point is $M$. Pick a line $L$ passing through $B$ (say), that is not one of the sides, I claim $f(L) \neq B$ unless $L$ is tangent to $(BFD)$. Suppose not, and that $f(L) = B$. Pick a point $P$ on $L$. Let $X,Y,Z$ be the second intersections of $PA, PB, PC$ with the circles. Consider a couple of cases: If $f(PA) = A$ and $f(PC) = C$ as well, then $P$ must lie on $(ABC)$, but we can choose $P$ far enough so it doesn't, so this is impossible. Now say $f(PA) = X$ and $f(PC) = C$. Then $P$ must lie on $(BCX)$. But since $\angle PBC$ is fixed, we must have $\angle PXC$ also to be fixed, or that $\angle CXA$ to be fixed. But this happens for finitely many $X$, and so for finitely many $P$ only, contradiction. Now consider the final case, when $f(PA) = X$ and $f(PC) = Z$. Then $P$ lies on $(XZB)$. But by miquel again, $P$ already lies on $(XYZ)$. So the five points $X,Y,Z,B,P$ must be concyclic, impossible as $P,B,Y$ are collinear. So all in all, we get that $f(L)$ for any line passing through a vertex is the second intersection with the circles. In particular, this actually implies that the circle corresponding to any point not on one of the three sides of the three tangents, passes through $M$. Now, pick a line $L$ through $M$ and suppose $f(L) \neq M$. But then pick a point $P \in L$ not lying on one of the six lines mentioned above, and also not equal to $f(L), M$. Then we get a contradiction since $f(L)$ must lie on the circle which intersects $L$ at exactly $P,M$. So we must actually have $f(L) = M$ for all lines $L$ passing through $M$. Finally, to prove uniqueness of this point, say $N \neq M$ also satisfies this property. But then $f(MN)$ equals both $M,N$ contradiction. So we must actually have that $M$ is the unique point such that $f(\ell) = M$ for any line $\ell$ going through $M$.$\blacksquare$

If there exists two points $P$ and $P'$ satisfying the condition, then $f(PQ)=P=Q$, contradiction. Therefore, it suffices to find one such $P$. For any $X$, consider the locus of $f(\ell)$ where $\ell$ passes through $X$. Any three distinct points on this locus lie on a circle through $X$, so if the locus contains at least two points other than $X$, fixing two of those points and varying the third implies that $f(\ell)$ lies on a circle. Let $g(X)$ be the reflection of $X$ over the center of this circle. Then, for all $A\in\ell$, $f(\ell)$ must be the foot from $g(A)$ onto $\ell$. This implies $AB\perp g(A)g(B)$. It suffices to find a fixed point of $g$. If $g$ is a constant $C$, then $g(C)=C$ Otherwise, pick $A$ and $B$ such that $g(A)$ and $g(B)$ are distinct. Let $h$ be the result of $g$ after a spiral similarity mapping $g(A)g(B)$ to $AB$, so $h(A)=A$, $h(B)=B$, and $h(X)h(Y)\parallel XY$. For any $C$ not on $AB$, we get $h(C)=C$ by substituting $Y=A$ and $Y=B$, so we get $h$ is the identity by considering $Y=A$ and $Y=C$ for the points on $AB$. Therefore, the spiral center is a fixed point.

Very beautifull my most fav problem from the shortlist . My solution:- Now say the statement in the above mentioned problem is not true. First observe that the problem can be restated as for any fixed point $P$ if some line $l$ pivots around $P$ then $f(l)$ always lies on some circle. with non zero radius. For any point $P$ call its this circle $C_P$ [if there are multiple pick any $1$] . Claim: $f$ is surjective. Proof: For any point $P$ say the tangent to $C_P$ at $P$ is $t$ then $f(t) \in C_P$ and $f(t) \in t$ this forces $f(t)=P$ . Claim: $f$ is not injective. Proof: Say it is , for any point $P$ pick any point $X \neq P \in C_P$ . and such that $X$ is not the antipode of $P$ in $C_P$, observe $XP$ is tangent to $C_X$ this implies $C_X$ and $C_P$ intersect again at some point $T$ and such that $ T,P,X$ are not collinear and then $f(\overline{TP})=T$ and $f(\overline{XT})=T$ contradiction ! . Now for any point $P$ we define its $f$ class as the set $S_P=\{l |f(l)=P,l \in \mathcal{L}\}$. Claim: There exists a point $P$ such that $S_P$ is infinite . Proof: Assume the contrary, Pick any random point $Q$ , say $t$ is the tangent to $C_Q$ at $Q$ and $l_1$ is another line through $Q$ such that $f(l_1)=Q$ say $l_1$ intersects $C_Q$ again at $P$ . pick any line $l_2 \in S_P$ obviously $l_2 \neq l_1$ . First we consider the points , $S=\{l \cup C_Q| \measuredangle{(t,l)}=\frac{2 \pi}{\sqrt{2}}k , k \in \mathbb{N}\}$ and pick any point $P_1$ in $S$ but not $S_Q$[if didnt exist just $Q$ would satisfy required condition contradiction!] similarly define $P_2$ for $\sqrt{3}$ and $P_3$ for $\sqrt{5}$ Now we describe an algorithm to generate more elemenets in $S_P$ using previous elements . We will call it by a sequence of lines $a_1,a_2,....$ where $a_1=l_1$ , Now we define $a_{n+1}$ using $a_n$ in the following algorithm, Say $C_P$ intersect $t$ at $K \neq A$ [if exists] , observe for a suitable choice of $P_i$ , $i \in \{1,2,3\}$ , $a_{n}$ will intersect $QP_i$ at some point $G \neq P$ such that $Q,G,P,K$ are not concyclic . Observe $C_G$ is basically $GPP_i$ , say the tangent to $GPP_i$ at $G$ is $t_1$ now say $t_1$ and $t$ intersects at some point $Q_1$ Claim: $f(\overline{PQ_1})=P$ Proof: By some basic angle chasing $PGQQ_1$ lie on circle. and then $C_{Q_1}=PGQ$ say F.T.S.O.C $f(\overline{PQ_1})=Q$ now play with $C_P$ observe $f(\overline{PQ_1})$ from that POV u easilty get contradiction. And now we define $a_{n+1}=\overline{PQ_1}$ Its easy to prove that $a_i$ is not recursive by basic NT and chasing down the angles between $a_{n+1}$ and $a_{n}$ . Claim: We can actually find two such points. as in previous claim. After we have located one such point $X$ , pick a point $P$ such that $X$ is not in $C_P$[easy to prove existence otherwise obviously $C_X$ as $0$ radius] and repeat similar process as what we did before. Say after all this huge work the two points we got are $X_1,X_2$ , pick two lines $l_1,l_2$ from $S_{X_1}$ and $l_3,l_4$ from $S_{X_2}$ , such that if $l_1,l_3$ intersects at $Y_1$ and $l_2,l_4$ and $Y_2$ then $Y_i$ never lies on $\overline{X_1X_2}$ and $Y_2,Y_1,X_2,X_1$ do not lie on a circle but now observe the line $\overline{Y_2Y_1}=l$ observe $Y_1,X,Y,f(l)$ are concyclic and $Y_2,X,Y,f(l)$ are as well but that implies $Y_2,Y_1,X_2,X_1$ are concylic contradiction!