I'll give you some steps (it's a beautiful problem, but I seem to have some trouble with writing full solutions ): (1) Let L1 and L2 touch C at X and Y. Show that XY and OQ intersect in a fixed point S; (2) Show that If D and E are the intersection points of L1 and L2 with OQ then (S, Q; D, E)=-1 (harmonic cross-ratio); (3) Let R be the intersection point of AB and OQ for a fixed position of P; show that RD*RE=RA*RB=RQ²; (4) From (2) and (3) derive th fact that R is the midpoint of QS (meaning that it's fixed, because both Q and S are fixed; I think this is the only part which requires a bit of computation. The problem is very nice, by the way.

(2) Show that If D and E are the intersection points of L1 and L2 with OQ then (S, Q; D, E)=-1 (harmonic cross-ratio); Well, This is so cool！ I just need to know property(2) then can complete the proof. Thank you very much~ Chao

I have a different solution of the problem, using angle chasing and some properties of orthogonal circles. First I rewrite the problem with modified notations (I find the namings L, $L_1$, $L_2$ pretty useless): Theorem 1. Let k be a circle with center O, let Q be a point not lying on this circle k, and let P be a point on the perpendicular to the line OQ at the point Q. Assume that the point P lies outside the circle k. Let the two tangents from the point P to the circle k touch this circle k at the points X and Y. Denote by A and B the orthogonal projections of the point Q on the lines PX and PY. Let the line XY meet the line OQ at a point S. Let R be the midpoint of the segment QS. (a) The point S is the image of the point Q in the inversion with respect to the circle k. (b) The point R lies on the line AB. (c) The point R lies on the radical axis of the circle k and the circle with diameter PQ. (d) If the line OQ meets the circle k at two points D and E, then the points A, B, D and E are concyclic. Note that parts (a) and (b) of this theorem yield Chaogold's initial problem (in fact, since the point Q and the circle k are fixed, according to part (a) the point S must be fixed, too, so the midpoint R of the segment QS is fixed; but according to part (b), the point R lies on the line AB, so the line AB intersects the line QO at a fixed point - namely, at the point R). Part (d) yields the additional assertion that ABDE are cyclic. Proof of Theorem 1. (a) This is a trivial consequence of poles and polars, but here is an elementary proof: Since the tangents from the point P to the circle k (whose center is O) touch this circle k at the points X and Y, we have $PX\perp OX$ and $PY\perp OY$, so that < OXP = 90° and < OYP = 90°. Also, < OQP = 90° (since the point P lies on the perpendicular to the line OQ at the point Q). Hence, the points X, Y and Q lie on the circle with diameter OP. Thus, < OPX = < OQX. Now, since the points X and Y are the points where the tangents from the point P to the circle k (with center O) touch this circle k, these points X and Y are symmetric to each other with respect to the line OP. Thus, $XY\perp OP$, so that < (XY; OP) = 90°. On the other hand, $PX\perp OX$ yields < (OX; PX) = 90°. Thus, < SXO = < (XY; OX) = < (XY; OP) + < (OP; PX) - < (OX; PX) = 90° + < (OP; PX) - 90° = < (OP; PX) = < OPX = < OQX = - < XQO. Also, trivially, < SOX = - < XOQ. Thus, the triangles SXO and XQO are inversely similar, so that OS : OX = OX : OQ. In other words, $OS\cdot OQ=OX^2$. Since O is the center and OX is the radius of the circle k, this equation yields that the point S is the image of the point Q in the inversion with respect to the circle k. Thus, Theorem 1 (a) is proven. (b) Let W be the orthogonal projection of the point Q on the line XY. Then, < QWX = 90°. On the other hand, < QAX = 90°. Hence, the points W and A lie on the circle with diameter QX; consequently, < QWA = < QXA. But < QXA = < QXP, and since the points X, Y and Q lie on the circle with diameter OP, we have < QXP = < QOP. Thus, < QWA = < QOP. Since $XY\perp OP$ and $QW\perp XY$, we have QW || OP, and thus < (QO; OP) = < (QO; QW), what rewrites as < QOP = < RQW. Hence, < QWA = < QOP becomes < QWA = < RQW. On the other hand, since the triangle QWS is right-angled at its vertex W, the point R, being the midpoint of its hypotenuse QS, must be the circumcenter of this triangle QWS; hence, RQ = RW, so that the triangle WRQ is isosceles, and thus < RQW = < QWR. Hence, we get < QWA = < QWR. But this shows that the point A lies on the line WR. Similarly, the point B lies on the line WR. Thus, the four points A, B, W and R lie on one line, what yields that the point R lies on the line AB, and Theorem 1 (b) is proven. (c) The point R, being the midpoint of the segment QS, is the center of the circle with diameter QS. On the other hand, if we call U the midpoint of the segment PQ, then this point U is the center of the circle with diameter PQ. Since $UQ\perp RQ$ (this is just a rephrase of $PQ\perp OQ$), and the point Q is a common point of the circles with diameters QS and PQ, it follows that the circle with diameter QS is orthogonal to the circle with diameter PQ. On the other hand, since the point S is the image of the point Q in the inversion with respect to the circle k, the circle with diameter QS is orthogonal to the circle k (in fact, any circle through two distinct mutually inverse points with respect to a circle is orthogonal to this circle). Hence, the circle with diameter QS is orthogonal to both the circle k and the circle with diameter PQ. Thus, the center R of this circle with diameter QS lies on the radical axis of the circle k and the circle with diameter PQ. This proves Theorem 1 (c). (d) Since < PAQ = 90° and < PBQ = 90°, the points A and B lie on the circle with diameter PQ. Since the point R lies on the line AB, it follows that the power of the point R with respect to the circle with diameter PQ equals $RA\cdot RB$. On the other hand, the power of the point R with respect to the circle k equals $RD\cdot RE$. Now, since the point R lies on the radical axis of the circle k and the circle with diameter PQ, its powers with respect to the circle with diameter PQ and the circle k are equal; hence, $RA\cdot RB=RD\cdot RE$. By the converse of the intersecting chords theorem, this yields that the points A, B, D and E are concyclic, and Theorem 1 (d) is proven. $\blacksquare$ Darij

Approach by grobber: There are going to be quite a few definitions: Fix $ P$, and let $ M = PA\cap QB,N = PB\cap QA,\ell = MN$. Let $ S = AB\cap PQ,T = AB\cap\ell$, and let $ X,Y$ be the intersections between $ AB$ and the perpendiculars to $ PQ$ in $ Q$ and $ P$, respectively. If we move a point $ U$ on $ \ell$, then $ QU\cap AB\mapsto PU\cap AB$ is an involution on $ AB$. Through this involution, $ A$ corresponds to $ B$, $ X$ corresponds to $ Y$, while $ S,T$ are fixed points. This means that $ (S,T;A,B) = (S,T;X,Y) = - 1\ (*)$. From $ (PS,PT;PA,PB) = - 1$ we find that $ PT$ passes through the pole $ P'$ of $ PQ$ wrt the circle (the initial circle, centered at $ O$). Now, since $ PT = PP',PS = PQ$, the second equality in $ (*)$ can be written $ (PQ,PP';PX,PY) = - 1\ (**)$. Since $ PY\|QP',\ (**)$ shows that $ X$ is, in fact, the midpoint of $ QP'$. $ Q$ and $ P'$ do not depend on $ P$, so $ X = AB\cap OQ$ is fixed. A semicircle is drawn on one side of a straight line $ l.$ $ C$ and $ D$ are points on the semicircle. The tangents to the semicircle at $ C$ and $ D$ meet $ l$ at $ B$ and $ A$ respectively, with the centre of the semicircle between them. Let $ E$ be the point of intersection of $ AC$ and $ BD,$ and $ F$ be the point on $ l$ so $ EF$ is perpendicular to $ l.$: Let P be the intersection of the 2 tangents, R the radius of the semicircle and O its center. What we need first is to prove that the altitude from P in triangle ABP is concurrent with AC and BD. This is done by proving that BC/CP*PD/DA=tgA/tgB (we want to use the reciprocal of Ceva), where A and B are angles of the triangle ABP. This is done easily since PC=PD and BC=OC/tgB=R/tgB and AD=OD/tgA=R/tgA. By replacing these we get exactly what we want. Thisa means that line EF and line PF are, in fact, one and the same. Now to prove the actual result: We draw the reflection of the entire figure with respect to line l. Let the image of any pt X be named X'. It's obvious that P, F, P' are collinear (F is the middle of PP'). We know by Newton's thm that PP', AB, CD', DC' are concurrent (it states that for a circumscribed quadrilateral the diagonals and the 2 lines joining the 2 pts of tangency on 2 opposite sides are all concurrent). This means that the (C, F, D') and (D, F, C') are 2 triplets of collinear pts. angle PFD=angle P'FD' because the figure is symmetric with respect to l. Angle CFP=angle P'FD' because C, F, D' are collinear and so are P, F, F'. From these 2 equalities we get angle PFD=angle CFP, which means that EF=PF is the bisector of angle CFD Q.E.D. I love writing that (QED) Prove EF bisects angle CFD.

A line $ l$ does not meet a circle $ \omega$ with center $ O.$ $ E$ is the point on $ l$ such that $ OE$ is perpendicular to $ l.$ $ M$ is any point on $ l$ other than $ E.$ The tangents from $ M$ to $ \omega$ touch it at $ A$ and $ B.$ $ C$ is the point on $ MA$ such that $ EC$ is perpendicular to $ MA.$ $ D$ is the point on $ MB$ such that $ ED$ is perpendicular to $ MB.$ The line $ CD$ cuts $ OE$ at $ F.$ Prove that the location of $ F$ is independent of that of $ M.$ Approach by mr.danh: Four points M,E,A,B all lie on the circle diametered OM. Let I be the orthogonal projection of E on AB. E lies on the circumcircle (AMB), so I,C,D lie on Simson line. OE meets line (ICD) at F and line (IAB) at K. F is the midpoint of EK because of the righ-angle EIK. To prove that F is a fixed point, show that K is a fixed point. It is easy since $ OK.OE = ON.OM = R^2$ (where N is the intersection of AB and OM)

orl wrote: F is the midpoint of EK because of the righ-angle EIK. Whoops I failed to realize this even though I constructed $I$... It's easy to see that $CEMD$ and $OAEMB$ are both cyclic by right angles, so $\alpha=\angle{EDF}=\angle{EMA}=\angle{EOA}$. Since $FE\perp EM$ and $EM$ is the diameter of the circle passing through $CEMD$, $\angle{FEC}=\angle{EDF}=\alpha$ and \[\angle{FCE}=\pi-\angle{ECD}=\angle{EMD}=\angle{EDF}+2\angle{AMO}=\alpha+2\beta,\]where $\beta=\angle{AMO}=\angle{BMO}=\angle{AEO}$, using the fact that $MO$ is the angle bisector of $\angle{AMB}$. Thus \[\angle{EFC}=\pi-\angle{FEC}-\angle{FCE}=\pi-2\alpha-2\beta.\]Now the Law of Sines on $\triangle{EDF}$ gives us \begin{align*} EF=\frac{DE\sin\angle{EDF}}{\sin\angle{EFC}}=\frac{ME\sin\angle{EMD}\sin\alpha}{\sin2(\alpha+\beta)} &= \frac{ME\sin((\alpha+\beta)+\beta))\sin\alpha}{2\sin(\alpha+\beta)\cos(\alpha+\beta)}\\ &= \frac{ME(\sin(\alpha+\beta)\cos\beta+\cos(\alpha+\beta)\sin\beta)\sin\alpha}{2\sin(\alpha+\beta)\cos(\alpha+\beta)}\\ &= \frac{ME\left(\frac{OE}{OM}\cdot\frac{BM}{OM}+\frac{ME}{OM}\cdot\frac{BO}{OM}\right)\sin\alpha}{2\cdot\frac{OE}{OM}\cdot\frac{ME}{OM}}\\ &= \frac{ME(OE\cdot BM+ME\cdot BO)\sin\alpha}{2\cdot OE\cdot ME}\\ &= \frac{(OM\cdot BE)\sin\alpha}{2\cdot OE}, \end{align*}where we have used Ptolemy's Theorem on $OEMB$ in the last step. Now by the Law of Sines on $\triangle{AOE}$, we have \[\sin\alpha=\sin\angle{AOE}=\frac{AE\sin\angle{AEO}}{AO}=\frac{AE\sin\beta}{AO}=\frac{AE\cdot\frac{AO}{OM}}{AO}=\frac{AE}{OM}.\]Thus \[EF=\frac{(OM\cdot BE)\sin\alpha}{2\cdot OE}=\frac{AE\cdot BE}{2\cdot OE}.\]Finally, since $\angle{AEO}=\angle{BEO}=\beta$, we can reflect $B$ over $EO$ to $B'$ so that $E,A,B'$ are collinear with $B'$ on $\omega$. This tells us that $AE\cdot BE=AE\cdot B'E$ is the power of $E$ with respect to $\omega$, which is clearly constant as $M$ varies. But $OE$ is constant as well, so we're done.

Wait, I do not understand why F is the midpoint of KE. I do understand that angle KIE is a right angle, but why does that force F to be the midpoint?

At least one of the equalities $\angle{FEI}=\angle{FIE}$ and $\angle{FKI}=\angle{FIK}$ should be easy to angle chase directly.

Let $AB$ meet $OM$ at $G$ and $OE$ at $H$. Then $\angle MEH = \angle MGH = 90$, so $GHEM$ is cyclic, so $OH\cdot OE = OG\cdot OM$. Triangles $AOG$ and $MOA$ are similar, so $OG\cdot OM = OA^2$. Hence $OH = OA^2/OE$, which is constant. Let the lines $AB$ and $CD$ meet at $K$. Angles $OAM, OEM, OBM$ are all $90$, so $EAOBM$ is cyclic. Similarly $\angle ECM = \angle EDM = 90$, so $ECDM$ is cyclic. Hence $\angle EAK = \angle EMB = \angle EMD = \angle ECK$, so $EKAC$ is cylic. Hence $\angle EKA = 90$. So $EKBD$ is cyclic. So $\angle EKF = \angle EKD = \angle EBD = \angle EBM = \angle EOM$ ($EAOBM$ is cyclic)$ = \angle FEK (OM \parallel EK)$. Hence $EF = FK$. But $\angle EKH = 90$, so $EF = FH$. So $F$ is the midpoint of $EH$. But $E$ and $H$ are fixed, so $F$ is fixed also. $\Box$

Recall a few basic complex numbers things. First \[ \frac{a-b}{\overline{a-b}} = \frac{a-c}{\overline{a-c}} \implies A, B, C \] are collinear and \[ \frac{a-b}{\overline{a-b}} = -\frac{c-d}{\overline{c-d}} \implies AB \perp CD \] WLOG, let $\omega$ be the unit circle, and the line containing $E$ be parallel to the real axis. (We can assume this because we can rotate it). Let the points be represented by their lowercase equivalents. Let $a, b$ be points on the unit circle, and let $m$ be the intersection of the tangents. We can easily find that $m = \frac{2ab}{a+b}$. Note that since we assumed that $ME$ was parallel to the real axis, $E$ lies on the complex axis $\implies e = -\bar{e}$. Also $EM \perp OE$, so \[ \frac{e}{\overline{e}} = -\frac{e-m}{\bar{e}-\bar{m}} \] \[ e-\frac{2ab}{a+b} = -e-\frac{2}{a+b} \implies e = \frac{ab-1}{a+b} \] We can now get a system of equations for $c$. \[ \frac{c-a}{\bar{c} - \frac{1}{a}} = \frac{m-a}{\overline{m-a}} = -a^2, \frac{c-e}{\bar{c}-\bar{e}} = -\frac{m-a} {\overline{m-a}} = a^2 \] \[c = -a^2\bar{c}+2a, c = a^2\bar{c}-a^2\bar{e}+e\], so we can solve to get that \[c = \frac{2a+e-a^2\bar{e}}{2} = \frac{2a+e+a^2e}{2} = \frac{2a+\frac{ab-1}{a+b}+a^2\frac{ab-1}{a+b}}{2} = \frac{a^2+3ab+a^3b-1}{2(a+b)}\] Similarly, $d = \frac{2b+e+b^2e}{2}$. Now we proceed to find $f$. Note that since $f$ lies on the complex axis, $f = -\bar{f}$. Also, \[\frac{f-c}{\bar{f}-\bar{c}} = \frac{c-d}{\overline{c-d}} = \frac{\frac{2a+e+a^2e}{2}-\frac{2b+e+b^2e}{2}}{\overline{\frac{2a+e+a^2e}{2}-\frac{2b+e+b^2e}{2}}} = \frac{((a+b)e+2)(a-b)}{\overline{((a+b)e+2)(a-b)}} = \frac{(ab+1)(a-b)}{\frac{ab+1}{ab}\frac{b-a}{ab}} = -a^2b^2\] So \[f = \frac{c+a^2b^2\bar{c}}{1-a^2b^2} = \frac{\frac{a^2+3ab+a^3b-1}{2(a+b)}+a^2b^2\overline{\frac{a^2+3ab+a^3b-1}{2(a+b)}}}{1-a^2b^2} = \frac{a^3b^3-a^3b-ab^3-3a^2b^2-a^2-b^2-3ab+1}{2(a^2b^2-1)(a+b)} = \frac{(1+ab)(a^2b^2-a^2-b^2-4ab+1)}{2(1+ab)(ab-1)(a+b)} = \frac{a^2b^2-a^2-b^2-4ab+1}{2(ab-1)(a+b)}\]. We will show that $F$ is the midpoint of $E$ and the pole of $E$ wrt $\omega$. Note that the pole of $E$ would be the point $\frac{e}{\vert e \vert^2} = \frac{e}{e\bar{e}} = -\frac{1}{e}$. Therefore the midpoint is \[\frac{e - \frac{1}{e}}{2} = \frac{\frac{ab-1}{a+b} - \frac{a+b}{ab-1}}{2} = \frac{a^2b^2-a^2-b^2-4ab+1}{2(a+b)(ab-1)} = f\], as required.

What an EXCELLENT problem.

Let $PA$ touch $C$ at $D$, $PB$ at $E$. Let $G$ be the polar of $L$ wrt $C$. Let $C$ be the foot from $Q$ to $GD$. Then by Simson line on $\triangle PDE$, it follows that $C$ lies on $AB$. Note that $CDQA$ is cyclic. Claim: $\triangle ODG \sim \triangle OQD$. Proof. Follows by being a polar. $\blacksquare$ As such, it now follows that \[ \measuredangle GCA = \measuredangle DQA = \measuredangle DQA = \measuredangle DQG + \measuredangle GQA = \measuredangle ODG + GOD = \measuredangle OGD = \measuredangle QGC \]which gives the result. This took longer than I'd like to admit