Subfield: Geometry


Question: Let $\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$. Suppose that for any point $X$, and for any three lines $\ell_{1}, \ell_{2}, \ell_{3}$ passing through $X$, the points $f\left(\ell_{1}\right), f\left(\ell_{2}\right), f\left(\ell_{3}\right)$ and $X$ lie on a circle.

Prove that there is a unique point $P$ such that $f(\ell)=P$ for any line $\ell$ passing through $P$.

Solution:
We provide a complete characterisation of the functions satisfying the given condition.

Write $\angle\left(\ell_{1}, \ell_{2}\right)$ for the directed angle modulo $180^{\circ}$ between the lines $\ell_{1}$ and $\ell_{2}$. Given a point $P$ and an angle $\alpha \in\left(0,180^{\circ}\right)$, for each line $\ell$, let $\ell^{\prime}$ be the line through $P$ satisfying $\angle\left(\ell^{\prime}, \ell\right)=\alpha$, and let $h_{P, \alpha}(\ell)$ be the intersection point of $\ell$ and $\ell^{\prime}$. We will prove that there is some pair $(P, \alpha)$ such that $f$ and $h_{P, \alpha}$ are the same function. Then $P$ is the unique point in the problem statement.

Given an angle $\alpha$ and a point $P$, let a line $\ell$ be called $(P, \alpha)$-good if $f(\ell)=h_{P, \alpha}(\ell)$. Let a point $X \neq P$ be called $(P, \alpha)$-good if the circle $g(X)$ passes through $P$ and some point $Y \neq P, X$ on $g(X)$ satisfies $\angle(P Y, Y X)=\alpha$. It follows from this definition that if $X$ is $(P, \alpha)$ good then every point $Y \neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \alpha}(X Y)=Y$ for every $Y \in g(X)$. Equivalently, $f(\ell) \in\left\{X, h_{P, \alpha}(\ell)\right\}$ for each line $\ell$ passing through $X$. This shows the following lemma.

Lemma 1. If $X$ is $(P, \alpha)$-good and $\ell$ is a line passing through $X$ then either $f(\ell)=X$ or $\ell$ is $(P, \alpha)$-good.

Lemma 2. If $X$ and $Y$ are different $(P, \alpha)$-good points, then line $X Y$ is $(P, \alpha)$-good.

Proof. If $X Y$ is not $(P, \alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \neq Y$.

Lemma 3. If $\ell_{1}$ and $\ell_{2}$ are different $(P, \alpha)$-good lines which intersect at $X \neq P$, then either $f\left(\ell_{1}\right)=X$ or $f\left(\ell_{2}\right)=X$ or $X$ is $(P, \alpha)$-good.

Proof. If $f\left(\ell_{1}\right), f\left(\ell_{2}\right) \neq X$, then $g(X)$ is the circumcircle of $X, f\left(\ell_{1}\right)$ and $f\left(\ell_{2}\right)$. Since $\ell_{1}$ and $\ell_{2}$ are $(P, \alpha)$-good lines, the angles

$$
\angle\left(\operatorname{Pf}\left(\ell_{1}\right), f\left(\ell_{1}\right) X\right)=\angle\left(\operatorname{Pf}\left(\ell_{2}\right), f\left(\ell_{2}\right) X\right)=\alpha
$$

so $P$ lies on $g(X)$. Hence, $X$ is $(P, \alpha)$-good.

Lemma 4. If $\ell_{1}, \ell_{2}$ and $\ell_{3}$ are different $(P, \alpha)$-good lines which intersect at $X \neq P$, then $X$ is $(P, \alpha)$-good.

Proof. This follows from the previous Lemma since at most one of the three lines $\ell_{i}$ can satisfy $f\left(\ell_{i}\right)=X$ as the three lines are all $(P, \alpha)$-good.

Lemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\alpha$ such that $A, B$ and $C$ are $(P, \alpha) \operatorname{good}$ points and $A B, B C$ and $C A$ are $(P, \alpha)$-good lines.



<img_3508>

Proof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles

$$
\angle(P D, D C)=\angle(P F, F B)=\angle(P E, E A)=\alpha
$$

for some angle $\alpha$. Hence, lines $A B, B C$ and $C A$ are all $(P, \alpha)$-good, so by Lemma $3, A, B$ and $C$ are $(P, \alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.)

Consider the set $\Omega$ of all points $(x, y)$ with integer coordinates $1 \leqslant x, y \leqslant 1000$, and consider the set $L_{\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\Omega}$. For each line $\ell$ in $L_{\Omega}$ we colour the point $f(\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5 , so these three vertices are all $(P, \alpha)$-good for some point $P$ and angle $\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \alpha)$-good for this particular pair $(P, \alpha)$. Now by Lemma 1 , the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that $(m+1, n+1)$ is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points $(m, n+2)$ and $(m+2, n)$ then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\Omega_{10}$ are good.

Now we will use this to prove that every point $S \neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\ell$ through $S$ intersects at most 10 points in $\Omega_{10}$. Hence, there are at least eight lines $\ell$ through $S$ which contain a point $Q$ in $\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\ell) \neq Q$. Hence, by Lemma 1, the line $\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4 , the point $S$ is good. Hence, every point $S \neq P$ is good, so by Lemma 2 , every line is good. In particular, every line $\ell$ passing through $P$ is good, and therefore satisfies $f(\ell)=P$, as required.
Note that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \in g(X) \cap g(Y) \cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$.

Lemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \notin\{X, Y\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$.

Proof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram below)


<img_3508>


Now pick any line $\ell$ and any six different points $Y_{1}, \ldots, Y_{6}$ on $\ell \backslash\{f(\ell)\}$. Pick a point $X$ not on $\ell$ or any of the circles $g\left(Y_{i}\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \ldots, Y_{4}$ do not lie on $g(X)$, so that $f\left(X Y_{i}\right) \notin\left\{X, Y_{i}\right\}$ for $1 \leqslant i \leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \leqslant i<j \leqslant 4$, we find that the points $s\left(Y_{i}, Y_{j}\right)$ and $s\left(X, Y_{i}\right)$ are all equal, to point $O$ say. Note that either $O$ does not lie on $\ell$, or $O=f(\ell)$, since $O \in g\left(Y_{i}\right)$.

Now consider an arbitrary point $X^{\prime}$ not on $\ell$ or any of the circles $g\left(Y_{i}\right)$ for $1 \leqslant i \leqslant 4$. As above, we see that there are two indices $1 \leqslant i<j \leqslant 4$ such that $Y_{i}$ and $Y_{j}$ do not lie on $g\left(X^{\prime}\right)$. By applying the above lemma to triangle $X^{\prime} Y_{i} Y_{j}$ we see that $s\left(X^{\prime}, Y_{i}\right)=O$, and in particular $g\left(X^{\prime}\right)$ passes through $O$.

We will now show that $f\left(\ell^{\prime}\right)=O$ for all lines $\ell^{\prime}$ through $O$. By the above note, we may assume that $\ell^{\prime} \neq \ell$. Consider a variable point $X^{\prime} \in \ell^{\prime} \backslash\{O\}$ not on $\ell$ or any of the circles $g\left(Y_{i}\right)$ for $1 \leqslant i \leqslant 4$. We know that $f\left(\ell^{\prime}\right) \in g\left(X^{\prime}\right) \cap \ell^{\prime}=\left\{X^{\prime}, O\right\}$. Since $X^{\prime}$ was suitably arbitrary, we have $f\left(\ell^{\prime}\right)=O$ as desired.
Notice that, for any two different points $X$ and $Y$, the point $f(X Y)$ lies on both $g(X)$ and $g(Y)$, so any two such circles meet in at least one point. We refer to two circles as cutting only in the case where they cross, and so meet at exactly two points, thus excluding the cases where they are tangent or are the same circle.

Lemma 1. Suppose there is a point $P$ such that all circles $g(X)$ pass through $P$. Then $P$ has the given property.

Proof. Consider some line $\ell$ passing through $P$, and suppose that $f(\ell) \neq P$. Consider some $X \in \ell$ with $X \neq P$ and $X \neq f(\ell)$. Then $g(X)$ passes through all of $P, f(\ell)$ and $X$, but those three points are collinear, a contradiction.

Lemma 2. Suppose that, for all $\epsilon>0$, there is a point $P_{\epsilon}$ with $g\left(P_{\epsilon}\right)$ of radius at most $\epsilon$. Then there is a point $P$ with the given property.

Proof. Consider a sequence $\epsilon_{i}=2^{-i}$ and corresponding points $P_{\epsilon_{i}}$. Because the two circles $g\left(P_{\epsilon_{i}}\right)$ and $g\left(P_{\epsilon_{j}}\right)$ meet, the distance between $P_{\epsilon_{i}}$ and $P_{\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\sum_{i} \epsilon_{i}$ converges, these points converge to some point $P$. For all $\epsilon>0$, the point $P$ has distance at most $2 \epsilon$ from $P_{\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \epsilon$ from $P_{\epsilon}$, so distance at most $4 \epsilon$ from $P$. A circle that passes distance at most $4 \epsilon$ from $P$ for all $\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property.

Lemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property.

Proof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2 .

Lemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\left(B_{i}\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property.

Proof. Consider some line $\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\left(B_{i}\right)$. Fix some direction along that line, and let $X_{\epsilon}$ be the point on $\ell$ that has distance $\epsilon$ from $A$ in that direction. In what follows we consider only those $\epsilon$ for which $X_{\epsilon}$ does not lie on any $g\left(B_{i}\right)$ (this restriction excludes only finitely many possible values of $\epsilon$ ).

Consider the circle $g\left(X_{\epsilon}\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\epsilon \rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\left(X_{\epsilon}\right)$ for any $\epsilon$ in that sequence.



Then $f\left(X_{\epsilon} B_{1}\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\epsilon} B_{1}$ with $g\left(B_{1}\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\epsilon}, f\left(X_{\epsilon} B_{1}\right)$ and $f\left(X_{\epsilon} B_{2}\right)$. As $\epsilon \rightarrow 0$, the angle at $X_{\epsilon}$ tends to $\angle B_{1} A B_{2}$ or $180^{\circ}-\angle B_{1} A B_{2}$, which is not 0 or $180^{\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\left(X_{\epsilon}\right)$, is also bounded above by a constant multiple of $\epsilon$, and so the result follows by Lemma 2 .

Lemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property.

Proof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.

Case 1: D does not lie on that line:

In this case, consider a sequence of $X_{\epsilon}$ at distance $\epsilon$ from $D$, tending to $D$ along some line that is not a tangent to either circle, but perturbed slightly (by at most $\epsilon^{2}$ ) to ensure that no three of the points $A, B$ and $X_{\epsilon}$ are collinear and no four are concyclic.

Consider the points $f\left(X_{\epsilon} A\right)$ and $f\left(X_{\epsilon} B\right)$, and the circles $g\left(X_{\epsilon}\right)$ on which they lie. The point $f\left(X_{\epsilon} A\right)$ might be either $A$ or the other intersection of $X_{\epsilon} A$ with the circle $g(A)$, and the same applies for $B$. If, for some sequence of $\epsilon \rightarrow 0$, both those points are the other point of intersection, the same argument as in the proof of Lemma 4 applies to find arbitrarily small circles. Otherwise, we have either infinitely many of those circles passing through $A$, or infinitely many passing through $B$; without loss of generality, suppose infinitely many through $A$.

We now show we can find five points $B_{i}$ satisfying the conditions of Lemma 4 (together with $A$ ). Let $B_{1}$ be any of the $X_{\epsilon}$ for which $g\left(X_{\epsilon}\right)$ passes through $A$. Then repeat the following four times, for $2 \leqslant i \leqslant 5$.

Consider some line $\ell=X_{\epsilon} A$ (different from those considered for previous $i$ ) that is not tangent to any of the $g\left(B_{j}\right)$ for $j<i$, and is such that $f(\ell)=A$, so $g(Y)$ passes through $A$ for all $Y$ on that line. If there are arbitrarily small circles $g(Y)$ we are done by Lemma 2, so the radii of such circles must be bounded below. But as $Y \rightarrow A$, along any line not tangent to $g\left(B_{j}\right)$, the radius of a circle through $Y$ and tangent to $g\left(B_{j}\right)$ at $A$ tends to 0 . So there must be some $Y$ such that $g(Y)$ cuts $g\left(B_{j}\right)$ at $A$ rather than being tangent to it there, for all of the previous $B_{j}$, and we may also pick it such that no three of the $B_{i}$ and $A$ are collinear and no four are concyclic. Let $B_{i}$ be this $Y$. Now the result follows by Lemma 4 .

Case 2: D does lie on that line:

In this case, we follow a similar argument, but the sequence of $X_{\epsilon}$ needs to be slightly different. $C$ and $D$ both lie on the line $A B$, so one must be $A$ and the other must be $B$. Consider a sequence of $X_{\epsilon}$ tending to $B$. Rather than tending to $B$ along a straight line (with small perturbations), let the sequence be such that all the points are inside the two circles, with the angle between $X_{\epsilon} B$ and the tangent to $g(B)$ at $B$ tending to 0 .

Again consider the points $f\left(X_{\epsilon} A\right)$ and $f\left(X_{\epsilon} B\right)$. If, for some sequence of $\epsilon \rightarrow 0$, both those points are the other point of intersection with the respective circles, we see that the angle at $X_{\epsilon}$ tends to the angle between $A B$ and the tangent to $g(B)$ at $B$, which is not 0 or $180^{\circ}$, while the distances tend to 0 (although possibly slower than any multiple of $\epsilon$ ), so we have arbitrarily small circumradii and the result follows by Lemma 2. Otherwise, we have either infinitely many of the circles $g\left(X_{\epsilon}\right)$ passing through $A$, or infinitely many passing through $B$, and the same argument as in the previous case enables us to reduce to Lemma 4.

Lemmas 3 and 5 together cover all cases, and so the required result is proved.
For any point $X$, denote by $t(X)$ the line tangent to $g(X)$ at $X$; notice that $f(t(X))=X$, so $f$ is surjective.

Step 1: We find a point $P$ for which there are at least two different lines $p_{1}$ and $p_{2}$ such that $f\left(p_{i}\right)=P$.

Choose any point $X$. If $X$ does not have this property, take any $Y \in g(X) \backslash\{X\}$; then $f(X Y)=Y$. If $Y$ does not have the property, $t(Y)=X Y$, and the circles $g(X)$ and $g(Y)$ meet again at some point $Z$. Then $f(X Z)=Z=f(Y Z)$, so $Z$ has the required property.

We will show that $P$ is the desired point. From now on, we fix two different lines $p_{1}$ and $p_{2}$ with $f\left(p_{1}\right)=f\left(p_{2}\right)=P$. Assume for contradiction that $f(\ell)=Q \neq P$ for some line $\ell$ through $P$. We fix $\ell$, and note that $Q \in g(P)$.

Step 2: We prove that $P \in g(Q)$.

Take an arbitrary point $X \in \ell \backslash\{P, Q\}$. Two cases are possible for the position of $t(X)$ in relation to the $p_{i}$; we will show that each case (and subcase) occurs for only finitely many positions of $X$, yielding a contradiction.

Case 2.1: $t(X)$ is parallel to one of the $p_{i}$; say, to $p_{1}$.

Let $t(X)$ cross $p_{2}$ at $R$. Then $g(R)$ is the circle $(P R X)$, as $f(R P)=P$ and $f(R X)=X$. Let $R Q$ cross $g(R)$ again at $S$. Then $f(R Q) \in\{R, S\} \cap g(Q)$, so $g(Q)$ contains one of the points $R$ and $S$.

If $R \in g(Q)$, then $R$ is one of finitely many points in the intersection $g(Q) \cap p_{2}$, and each of them corresponds to a unique position of $X$, since $R X$ is parallel to $p_{1}$.

If $S \in g(Q)$, then $\angle(Q S, S P)=\angle(R S, S P)=\angle(R X, X P)=\angle\left(p_{1}, \ell\right)$, so $\angle(Q S, S P)$ is constant for all such points $X$, and all points $S$ obtained in such a way lie on one circle $\gamma$ passing through $P$ and $Q$. Since $g(Q)$ does not contain $P$, it is different from $\gamma$, so there are only finitely many points $S$. Each of them uniquely determines $R$ and thus $X$.



<img_3606>

So, Case 2.1 can occur for only finitely many points $X$.

Case 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.

Clearly, $R_{1} \neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\ell$ only at $X$ and $Q$. Notice that $g\left(R_{i}\right)$ is the circle $\left(P X R_{i}\right)$. Let $R_{i} Q$ meet $g\left(R_{i}\right)$ again at $S_{i}$; then $S_{i} \neq Q$, as $g\left(R_{i}\right)$ meets $\ell$ only at $P$ and $X$. Then $f\left(R_{i} Q\right) \in\left\{R_{i}, S_{i}\right\}$, and we distinguish several subcases.

<img_3819>

Subcase 2.2.1: $f\left(R_{1} Q\right)=S_{1}, f\left(R_{2} Q\right)=S_{2}$; so $S_{1}, S_{2} \in g(Q)$.

In this case we have $0=\angle\left(R_{1} X, X P\right)+\angle\left(X P, R_{2} X\right)=\angle\left(R_{1} S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} R_{2}\right)=$ $\angle\left(Q S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} Q\right)$, which shows $P \in g(Q)$.

Subcase 2.2.2: $f\left(R_{1} Q\right)=R_{1}, f\left(R_{2} Q\right)=R_{2}$; so $R_{1}, R_{2} \in g(Q)$.

This can happen for at most four positions of $X$ - namely, at the intersections of $\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \in g(Q) \cap p_{i}$.



Subcase 2.2.3: $f\left(R_{1} Q\right)=S_{1}, f\left(R_{2} Q\right)=R_{2}$ (the case $f\left(R_{1} Q\right)=R_{1}, f\left(R_{2} Q\right)=S_{2}$ is similar).

In this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\angle\left(Q S_{1}, S_{1} X\right)=\angle\left(R_{1} S_{1}, S_{1} X\right)=$ $\angle\left(R_{1} P, P X\right)=\angle\left(p_{1}, \ell\right) \neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \rightarrow \ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction.

Thus Case 2.2 also occurs for only finitely many points $X$, as desired.

Step 3: We show that $f(P Q)=P$, as desired.

The argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \in \ell \backslash\{P, Q\}$, which is absurd.

Case 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$.

Let $t(Q) \operatorname{cross} p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X$ cross $g(R)$ again at $S$. Then $f(R X) \in\{R, S\} \cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$.

<img_4051>

Subcase 3.1.1: $S=f(R X) \in g(X)$.

We have $\angle(t(X), Q X)=\angle(S X, S Q)=\angle(S R, S Q)=\angle(P R, P Q)=\angle\left(p_{2}, \ell\right)$. Hence $t(X) \| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\prime}$, so $g\left(R^{\prime}\right)=\left(P R^{\prime} X\right)$, and let $R^{\prime} Q$ meet $g\left(R^{\prime}\right)$ again at $S^{\prime}$; notice that $S^{\prime} \neq Q$. Excluding one position of $X$, we may assume that $R^{\prime} \notin g(Q)$, so $R^{\prime} \neq f\left(R^{\prime} Q\right)$. Therefore, $S^{\prime}=f\left(R^{\prime} Q\right) \in g(Q)$. But then, as in Case 2.1, we get $\angle(t(Q), P Q)=\angle\left(Q S^{\prime}, S^{\prime} P\right)=\angle\left(R^{\prime} X, X P\right)=\angle\left(p_{2}, \ell\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible.

Subcase 3.1.2: $R=f(R X) \in g(X)$.

In this case, we have $\angle(t(X), \ell)=\angle(R X, R Q)=\angle\left(R X, p_{1}\right)$. Again, let $R^{\prime}=t(X) \cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\left(R^{\prime}\right)=\left(R^{\prime} X P\right)$; let $R^{\prime} Q$ meet $g\left(R^{\prime}\right)$ again at $S^{\prime}$. Due to $\angle\left(R^{\prime} X, X R\right)=\angle(Q X, Q R)=\angle\left(\ell, p_{1}\right), R^{\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\prime} \notin g(Q)$. Therefore, for those positions we have $S^{\prime}=f\left(R^{\prime} Q\right) \in g(Q)$. But then $\angle\left(R X, p_{1}\right)=\angle\left(R^{\prime} X, X P\right)=\angle\left(R^{\prime} S^{\prime}, S^{\prime} P\right)=$ $\angle\left(Q S^{\prime}, S^{\prime} P\right)=\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well.

Thus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction.



Case 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.

By Step $2, R_{1} \neq R_{2}$. Notice that $g\left(R_{i}\right)$ is the circle $\left(P Q R_{i}\right)$. Let $R_{i} X$ meet $g\left(R_{i}\right)$ at $S_{i}$; then $S_{i} \neq X$. Then $f\left(R_{i} X\right) \in\left\{R_{i}, S_{i}\right\}$, and we distinguish several subcases.

<img_3331>

Subcase 3.2.1: $f\left(R_{1} X\right)=S_{1}$ and $f\left(R_{2} X\right)=S_{2}$, so $S_{1}, S_{2} \in g(X)$.

As in Subcase 2.2.1, we have $0=\angle\left(R_{1} Q, Q P\right)+\angle\left(Q P, R_{2} Q\right)=\angle\left(X S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} X\right)$, which shows $P \in g(X)$. But $X, Q \in g(X)$ as well, so $g(X)$ meets $\ell$ at three distinct points, which is absurd.

Subcase 3.2.2: $f\left(R_{1} X\right)=R_{1}, f\left(R_{2} X\right)=R_{2}$, so $R_{1}, R_{2} \in g(X)$.

Now three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible.

Subcase 3.2.3: $f\left(R_{1} X\right)=S_{1}, f\left(R_{2} X\right)=R_{2}$ (the case $f\left(R_{1} X\right)=R_{1}, f\left(R_{2} X\right)=S_{2}$ is similar).

We have $\angle\left(X R_{2}, R_{2} Q\right)=\angle\left(X S_{1}, S_{1} Q\right)=\angle\left(R_{1} S_{1}, S_{1} Q\right)=\angle\left(R_{1} P, P Q\right)=\angle\left(p_{1}, \ell\right)$, so this case can occur for a unique position of $X$.

Thus, in Case 3.2, there is only a unique position of $X$, again yielding the required contradiction.


### 分析
The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\ell$ passing through $X$, the point $f(\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\ell)$ other than $X$; for such points, an arbitrary choice is made.)

If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.