nice tutorial for inversion

The trick is to invert the figure around a circle centered at $A$ of arbitrary radius. We let $\omega = \ell^\ast$ denote the image of $\ell$ under this inversion. Then, under the inversion, Evan's compass has the following behavior: Evan can draw a line through two points other than $A$; or Evan can draw a circle through three points other than $A$. In other words, the point $A$ is ``invisible'' to Evan, but Evan otherwise has a straightedge and the same compass. It is clear then that the answer to (ii) is no. Part (i) is equivalent to showing that Evan can construct the center of $\omega$; we give one construction here anyways. Take any cyclic quadrilateral $WXYZ$ inscribed in $\omega$, and let $P = \overline{WZ} \cap \overline{XY}$. Then the circumcircles of $\triangle PWX$ and $\triangle PYZ$ meet again at the Miquel point $M$, and the second intersection of $(MXZ)$ and $(MWY)$ is the center of $\omega$. Remark: The proof of (ii) implies that it's actually more or less impossible in this context to construct any point other than the reflection of $A$, as a function of $A$ and $\ell$. An alternative proof of (ii) is possible by inverting around a generic point $P$ on $\ell$ with radius $PA$; this necessarily preserves the entire construction, but the foot from $A$ to $\ell$ is not fixed by this inversion.

Reflection is very easy without inversion. Draw a circle centered at $A$ with radius greater than $d(A,l)$. You get two points of intersections. Now draw two circles with the same radius in achieved intersection points. They intersect in $A$ and reflection of $A$ over $l$. Since Evan can mark the intersections between two objects drawn the construction is correct.

WolfusA wrote: Reflection is very easy without inversion. Draw a circle centered at $A$ with radius greater than $d(A,l)$. You get two points of intersections. Now draw two circles with the same radius in achieved intersection points. They intersect in $A$ and reflection of $A$ over $l$. Since Evan can mark the intersections between two objects drawn the construction is correct. Are you kidding me? Have you ever read the problem yet.

Here is a solution I found for part (i) during the contest. We claim that the answer is $\boxed{YES}$. Here is an algorithm to construct $A^{*}$, the reflection of $A$ over $\ell$. 1) Mark arbitrary points $B, C$ on $\ell$. 2) Use a special compass to draw a circle $O$ through $A, B, C$. 3) Mark an arbitrary point $D$ not on $O$ and $\ell$ such that $D$ is on the different side of $A$ w.r.t. to $\ell$. 4) Use a special compass to draw circle $O_{1}$ through $A, B, D$ and $O_{2}$ through $A, C, D$. 5) Mark the intersection point of $O_{1}$ and $\ell$ other than $B$ and the intersection point of $O_{2}$ and $\ell$ other than $C$. Let the marked points be $E$ and $F$ respectively. 6) Use a special compass to draw a circle $O_{3}$ through $D, E, F$. 7) Mark an arbitrary point $G$ on $O_{3}$ such that $G \neq D, E, F$. 8) Use a special compass to draw a circle $O_{4}$ through $B, F, G$, and a circle $O_{5}$ through $C, E, G$. 9) Mark point $H$, the intersection point of $O_{4}$ and $O_{5}$ other than $G$. 10) Use a special compass to draw a circle $O_{6}$ through $B, C, H$. Claim: $A*$ is on $O_{6}$ Proof. We will prove by using directed angle. Therefore, it is sufficient to show that $\measuredangle BHC = \measuredangle BA^{*}C$ \begin{align*} \measuredangle BHC &= \measuredangle BHG + \measuredangle GHC &= \measuredangle BFG + \measuredangle GEC &= \measuredangle EFG + \measuredangle GEF &= -\measuredangle FGE &= -\measuredangle FDE &= \measuredangle DEF + \measuredangle EFD &= \measuredangle DEB + \measuredangle CFD &= \measuredangle DAB + \measuredangle CAD &= \measuredangle CAB &= \measuredangle BA^{*}C \end{align*} So, $A^{*}$ is on $O_{6}$ 11) Let $B_{1} = B$ and $C_{1}$ be a point different from $B, C$ on $\ell$ and construct $D_{1},...,H_{1}$ by applying step 3-10. 12) Use a special compass to draw a circle $O_{7}$ through $B, C_{1} H_{1}$. 13) Since $O_{6}$ and $O_{7}$ through $B$ and $A^{*}$, we can mark the intersection point of $O_{6}$ and $O_{7}$ different from $B$ and get the reflection of $A$ over $\ell$ as desired.

Consider an alternate universe where Evan instead has the statement of the problem inverted at $A$. Obviously, the answers to the questions (a) and (b) remain unchanged. If the original universe had the circle passing through $A$, then the new universe has a line not through $A$. If the original universe had a circle not through $A$, the new universe has another circle not through $A$. And $\ell$ in the original universe is now a circle $\omega$ through $A$. Let $M$ be the foot from $A$ to $\ell$. Then $M^*$ is a point on $\omega$, and by symmetry it must be the antipode of $A$. If $B$ is the reflection of $A$ over $\ell$, then since $AM/AB=1/2$, then $AM^*/AB^*=2$, i.e. $B^*$ is the center of $\omega$. The new problem reads: Restated wrote: We can draw lines and circles through three points not passing through a given point $A$. Given a circle $\omega$ passing through $A$, can we construct the antipode of $A$ and the center of $\omega$? Note that we cannot do anything with $A$; the only was to access it would be to draw a line or circle through it, which we cannot do. Hence, we cannot construct the antipode of $A$, since this is a function of $A$. Forget about $A$. We claim we can construct the center $O$ of $\omega$. Draw a quad $ABCD$ (a different $A$) inscribed in $A$. Let $E=AD\cap BC$, $F=AB\cap CD$, $G=AC\cap BD$. Let $M=(EAB)\cap (FBC)$ be the Miquel Point. Then it is well-known that $O\in MG$. Draw two different such quads to find $O$ by intersection.

Beautiful question. First, we invert the problem, so Evan is given a circle $\ell$ through some point $A$, he may draw a line through any two points that does not pass through $A$ and a circle through any three noncollinear points that does not pass through $A$, he may mark the intersection between any two objects, and an arbitrary point on any object or the plane. Part (i) asks if Evan can find the center of $\ell$ and part (ii) asks if Evan can find the antipode of $A$ wrt $\ell$. As $A$ is essentially a forbidden point, it is immediate that (ii) cannot be attained, since one cannot interact with $A$. We spend the remainder of this solution building a framework to show that the answer to (i) is yes. The first step is to use Reim's Theorem, as we are TRAINED OLYMPIAD GEOMETERS$^{\mbox{TM}}$. For noncollinear points $X,Y,Z$, we use Reim's Theorem to draw a line through $Z$ parallel to $XY$. Draw line $XZ$ and let $P\ne X, Z$ be arbitrary on line $XZ$. Draw $(PXY)$ and let $Q\ne P,X,Z$ be arbitrary on it, and draw $(PQZ)$. Let line $YQ$ intersect $(PQZ)$ for a second time at $K$, then $ZK\parallel XY$ by Reim's. Now, we draw the perpendicular bisector of an arbitrary line $KL$. Let $X,Y$ be arbitrary points in the plane with $Y$ not on $(XKL)$. Let $M$ be the intersection of the line through $X$ parallel to $KL$ and $(XKL)$, and $N$ be the intersection of the line through $Y$ parallel to $KL$ and $(YKL)$. Then, $KMN$ reflect to $LXY$ over the perpendicular bisector of $KL$. Now, note that for some choice of $P,Q,R$ with $P\in\{K,L\},Q\in\{M,X\},R\in\{N,Y\}$, the circle $(PQR)$ will intersect the perpendicular bisector of $KL$ on the Euclidean plane. Since Evan can try all 8 possibilities, we may assume that he gets it right on the first try and $P=K,Q=M,R=N$. Then, the circles $(KMN)$ and $(LXY)$ concur along their radical axis, which is the perpendicular bisector of $KL$ because they reflect to each other over it, so we can draw in the perpendicular bisector of $KL$. Finally, pick arbitrary but distinct $S,T$, $U,V$ on $\ell$ and note that the center of $\ell$ is the intersection of the perpendicular bisectors of $ST$ and $UV$. Remark: We need not worry about the possibility that one of the objects that is not $\ell$ passes through $A$; the probability thereof is zero.

Invert around $A$; in the inverted problem, $\ell$ becomes a circle through $A$ and we can choose arbitrary points on any object, mark the intersection of two objects, draw any circle through three points not including $A$ and draw any line through two points not including $A$. Part i then asks to find the center of the circle, and part ii asks for $A$'s antipode. The answer to part ii is \textbf{no}; indeed, we cannot gain any information on the point $A$ since it is unusable. The answer to part i is \textbf{yes}. We first plot points $A_1,A_2,A_3,A_4$ in clockwise order, plot $X=A_1A_3\cap A_2A_4,Y=(A_1A_2O)\cap (A_3A_4O)$. $Y$ is the center of the spiral similarity sending $A_1A_2$ to $A_3A_4$, and it's well-known that $\angle XYO=90^\circ$. Draw two more lines through $X$ that intersect $\ell$ at $A_5,A_6,A_7,A_8$ in clockwise order, and let $Z=(A_5A_6O)\cap (A_7A_8O)$; then, $(XYZ)$ contains $O$ because $\angle XYO=\angle XZO=90$. Draw another such circle $X_1Y_1Z_1$, and let $(XYZ),(X_1Y_1Z_1)$ intersect at $O_1,O_2$, one of which is $O$. To finish, choose a point on line $O_1O_2$ as $X_2$, draw four lines through $X_2$ and then construct a new circle through $O$ with the intersection points. It is obviously nondegenerate, and therefore cannot pass through $X_2,O_1,O_2$ and we will be able to identify the circumcenter.

Invert at $A$ so that we have a point $A$ on a circle $\omega$, we can draw circles and lines not through $A$, and we want to construct the center $O$ of $\omega$ and the $A$-antipode. Clearly the latter is impossible, as we cannot draw objects through $A$. Constructing the center is very much possible: take four points $W,X,Y,Z\in\omega$, take $P = WX\cap YZ$, $M = (PWZ)\cap (PXY)$, $R = WY\cap XZ$, and we have $O\in MR$ (in fact $M$ and $R$ are inverses about $\omega$). Now do this a second time and we know $O = MR\cap M'R'$.

Invert at $A$. Then we are given a circle $\omega$ missing one point $A$, and Evan doesn't know where the point $A$ is. Thus Evan cannot locate any point defined with reference to $A$. He has the same compass and also a straightedge. Part (ii) now asks if Evan can construct the $A$-antipode, which is obviously impossible under this scenario. Part (i) asks if Evan can construct the center $O$. This is straightforward actually by Miquel theorem for cyclic quadrilaterals: construct any complete quadrilateral $WXYZPQ$, and construct the Miquel point $M = (PWX) \cap (PYZ)$. Then $O = (MWY) \cap (MXZ)$.

Let $A'$ be the reflection of $A$ and $M$ be the foot. Invert around $A$ with radius $AM$. Then, for part 1, our goal is to construct the center of the circle, and in part 2 our goal is to construct the antipode of $A$ with respect to the circle. If Evan marks a circle through A and two other points, then in the inverted version he is marking a line between the two other points. A circle not going through A is still a circle not going through A. Thus, Evan still has the ability to construct circles and lines, but just not passing through $A$. For part 1, Evan marks 4 points on a circle, and then draws a $K_4$ and marks all intersections. By adding in circles as well, he can mark the Miquel point $M$. Finally, intersecting $(MAC)$ and $(MBD)$ would give $O$. For part 2, Evan cannot reach point $A$ at all after the inversion, so he cannot construct anything that depends on point $A$.

First, invert about $A.$ Then the compass can construct any cline not passing through $A.$ For part (i), we need to construct the center $O$ of the circle given by $\ell',$ the image of $\ell.$ First we choose a point $P$ and draw two lines through $P,$ with the first line intersecting $\ell'$ at $X,Y$ and the second intersecting at $Z,W$ such that $X,Y,Z,W$ appear on the circle in that order. Construct points $Q=XW \cap YZ$ and $R=XZ \cap YW,$ and $QR$ intersects $\ell'$ at $M,N.$ By Brokard's, $PM,PN$ are tangent to $\ell',$ so the circumcircle of $PMN$ is a circle with diameter $PO.$ We can repeat this for other points $P$ to find the point $O.$ For part (ii), we need to construct the antipode of $A.$ However, notice that it is impossible to use the point $A$ in a finite number of steps, so its antipode cannot be constructed. Therefore, the answer is yes for (i) and no for (ii).

We invert the problem about $A$. We can see that $\ell$ becomes a circle containing $A$. His compass can now draw any circle not containing $A$ and can draw any line not containing $A$. Since Evan can never do any operations with $A$, we can obviously see that he would never be able to satisfy $\text{(ii)}$. Now to satisfy $\text{(i)}$ We construct any cyclic quadrilateral $ABCD$ and we can let $AB$ and $CD$ intersect at $X$. We let $Y$ be the intersection of $(XBC)$ and $(XAD)$. We can see that the intersection of $(YAC)$ and $(YBD)$ is the center of the circle. $\blacksquare$

Invert at $A$ with radius 1. Then the compass has the ability to draw a line through 2 points, or an ordinary straightedge, and draw a circle through 3 points, or the original compass, other than $A$. Thus $A$ cannot be accessed with this tool, so (ii) is $\boxed{\text{no}}$. For (i), we need to construct the center of $\ell^*$. We can abuse Miquel's by taking two arbitrary cyclic quadrilaterals inscribed in $\ell^*$ with Miquel points $M_1$, $M_2$ and intersection of diagonals $R_1$, $R_2$. Our construction is just taking $M_1R_1 \cap M_2R_2$, so (i) is $\boxed{\text{yes}}$. $\blacksquare$

Saw this in the inversion unit of $OTIS$, Might not have solved it if I didn't know inversion would be involved. Anyways the solution:- We invert with center $A$ and arbitrary radius so now the new problem is- Quote: Given a circle passing through a fixed point $A$, one can draw a circle through any three points and a line through any two points, proivded they don't pass through $A$. Prove or disprove that one can mark the center and $A$-antipode of the given circle. For the first part, consider a quadrilateral inscribed in the circle. It is obvious that one can mark the intersection of diagonals and the miquel point of the complete quadrilateral and it is known that the line joining the miquel point and diagonal intersection passes through the center. Similarly draw another such line and the intersection is the center of the circle. For the second part, notice that since we can never draw any line or circle through $A$, we can draw an object which is defined with respect to $A$, so the answer is $NO$.

ELMO 2018 P3 Here is a sketch. Invert around $A$. (i) take two cyclic quadrilaterals in $\ell '$, then take $M,N$ to be their miquel points. And let $E,F$ be the intersection of their diagonals. then the needed point is $ME$ intersect $NF$. (ii) No, proof is easy.

Invert the diagram at point $A$ with radius $1.$ Then $\ell$ turns into a circle $\omega$ passing through $A.$ We can then construct any line not passing through $A$ and any circle not passing through $A.$ Thus we are excluding point $A$ from the plane. Part (ii) asks us to find the antipode of $A$ with respect to $\omega,$ which is impossible since we can't do anything with point $A.$ For part (i) we want to construct the center of $\omega.$ Take arbitrary points $B,C,D,E$ on $\omega.$ Construct $BC \cap DE = X$ and $BE \cap CD = Y.$ Then construct the circumcircles of $BEX$ and $CDX$ and let them intersect at point $K.$ Then by Miquel point properties, $(BDK)$ passes through the center of the circle, as does $(CEK).$ Therefore, we can just intersect these two circles, and we get the center of the circle, hence done. Remark: I found part (i) to be much easier to tackle than (ii) after inverting oops.

Tried the entire weekends to construct (ii) then it was impossible whoops