It is also the Balkan Olympiad. I 'll give my solution later. Please sympathize me!

There was another tougher geometry problem in that TST, but I'll propose it when I know how to solve it .

If Oa and Ob are the antipodes of O wrt c(OCA) and c(OBD) respectively, then Oa, Ob and K are collinear. Indeed <ObKO + <OKOa = 180. Now if P=AC^BD it is well known that the polar of M wrt c(O) pass through P. As CA and BD are the polars of Oa and Ob wrt c(O) respectively and are concurrent with the polar of M, then Oa, M and Ob are collinear. From this the thesis.

Wow! I can't believe how many problems similar to this one have appeared on the forum . Let $N=AC\cap BD$. $K$ is the image of $O$ through the inversion of pole $N$ and power $NC\cdot NA=ND\cdot NB=$the power of $N$ wrt the circle $(ABDC)$. Let $NT_1,NT_2$ be the tangents from $N$ to the circle, and let $K'=T_1T_2\cap NO$. We get $NC\cdot NA=NT_1^2=NO\cdot NK'$, meaning that $K'=K$, so $K$ lies on the polar $T_1T_2$ of $N$ wrt the circle. Since $M$ also lies on the polar, $MK=T_1T_2$, and $T_1T_2\perp NO$.

I have a synthetic solution but it is longer than grobber's one , so I'll write only the idea. Let $L$ be the intersection of $AD$ and $CB$, the key is to show that $K, L, M$ are collinear. To prove this we must show that quadrilaterals $AKLB$ and $CKLD$ are cyclic, after this $M$ is the intersection point of the radical axes of the circumcircles of $AKLB, CKLD$ and $ABCD$. Other steps are simple angle chasing.

Actually, this is just a particular case of a more general result: Theorem 1. If A, B, C, D are four points on a circle with center O, if the line CD meets the line AB at a point M, and if the circumcircles of triangles AOC and DOB meet at a point K (apart from the point O), then < MKO = 90°. And indeed, Armo's proof perfectly works for this general result, too, and Grobber's proof also works if we allow the points $T_{1}$ and $T_{2}$ to be imaginary (in fact, the point N can happen to lie inside the circle, and then the points $T_{1}$ and $T_{2}$ are imaginary points). Here are two proofs of Theorem 1: First proof. This proof comes in two variants because I have posted it two times in different threads:

Second proof. We use directed angles modulo 180°. Since the point K lies on the circumcircle of triangle AOC, we have < CKO = < CAO, and since O is the center of the circle through the points A, B, C, D, the central angle theorem yields < CAO = 90° - < ABC. Thus, < CKO = 90° - < ABC. Similarly, < BKO = 90° - < DCB. Thus, < CKB = < CKO - < BKO = (90° - < ABC) - (90° - < DCB) = < DCB - < ABC = < (CD; BC) - < (AB; BC) = < (CD; AB) = < CMB. Hence, the point K lies on the circumcircle of triangle CMB, so that < MKC = < MBC. Thus, < MKO = < MKC + < CKO = < MBC + (90° - < ABC) = < ABC + (90° - < ABC) = 90°, and Theorem 1 is proven. Note that the second proof yields a remarkable corollary: Our point K lies on the circumcircle of triangle CMB. Similarly, we could show that our point K lies on the circumcircle of triangle AMD. Thus, the point K is the point of intersection of the circumcircles of triangles CMB and AMD (apart from M). Hence, we obtain the following extension of Theorem 1: Theorem 2. If A, B, C, D are four points on a circle with center O, and if the line CD meets the line AB at a point M, then the point of intersection of the circumcircles of triangles AOC and DOB (apart from O) coincides with the point of the intersection of the circumcircles of triangles CMB and AMD (apart from M), and if we denote this point of intersection by K, then < MKO = 90°. This theorem also yields the result of http://www.mathlinks.ro/Forum/viewtopic.php?t=18801 . Darij

I'll just solve the original problem. This solution uses the simpler nine-point circle in place of darij's "autopolar triangle" result, as I just solve the specific case where AOB is a line. Let AC ^ BD = L, and let the projection of L onto AB be N. Note that CODN lie on the nine-point circle of ABL, so if we invert in circle (O), we find that the image of N lies on both CD and AB, i.e. it is M. But we also have the image of K lies on AC and BD, i.e. it is L. Thus angle OKM is angle ONL (very basic fact of inversion), but we know that angle ONL is pi/2, so we are done.

Well I think I have seen this problem before. The idea is Inversion. Let’s invert centering at O and radius AO. Then Line CD becomes a circle that goes through CDO say it intersects AB at M’ which is image of M after the inversion. [Note: A, B, C, D will remain constant] Now look O is midpoint of AB, BC and AD are perpendicular on AK’ and BK’ [where K’ is image of K and K’ is got by extending AC and BD] hence COD is 9point circle of triangle AK’B. Hence K’M’ is perpendicular on AB. But M’K’O ~ KMO. So MKO=90.

AC x BD in N. ACDB is cyclic -> angle bisectors of M and N are orthogonal at Z. K is on the circumcircle of ZMN . W.

Hehe , I have a solution : Using Radical axix theorem , we show $ AC BD , OK $ intersect together at $ P $ Suppose $ AD $ intersects $ BC $ at $ E $ , $ PE $ intersects $ AB $ at $ Q $ . We have : $ PQ $ is perpendicular to $ AB $ It's easy to show : $ (CK , CO) = (PA , PO) $ So $ OC $ is the tangent of $ (PKC) $ and $ OC^2= OK.OP = R^2 $ We have : $ (PA,PQ,PB,PM) = -1 $ . So $ OQ.OM =R^2 $ This infers $ OQ.OM = OK.OP \iff \measuredangle{PKM} = 90 $

My approach: Let L be point of intersections of AD with BC.Easy to see:C,K,L,D lie on one circle.Anologously for A,K,L,B.Hence:K,L,M are colinear.Therefore,OKM=180-(CAB+CBA)=90.

Let $ X = AC \cap BD $ and $ Y $ be the foot of the perpendicular from $ X $ to $ AB $. As $ X $ lies on the pole of $ M $, and $ XY $ is perpendicular to $ OM $, $ XY $ must be the pole of $ M $, and as $ Y $ is the foot of the perpendicular, this implies that $ M $ and $ Y $ are inverse points. Now, we know that the perpendicular to $ AB $ through $ Y $, $ AC $, and $ BD $, are all collinear at $ X $. Invert this statement with respect to $ O $. $ AC $ and $ BD $ turn into circles $ AOC $ and $ BOD $ while the perpendicular to $ AB $ through $ Y $ inverts into the circle with diameter $ OM $. These circles are all still concurrent at the inverse of $ X $ - call this point $ K $. Then we see that circles $ AOC $ and $ BOD $ intersect at $ K $, and $ OK $ is perpendicular to $ MK $ (as $ K $ lies on the circle with diameter $ OM $), as desired.

I've just started playing with inversion and came up with the following with the help of a friend : Let’s invert with pole at $ O$ and radius $ AO$. Let's call $ P'$ the image of the point $ P$ through the inversion of pole $ O$. First note that $ K'=AC \cap BD$. Because of the inversion, we want to prove that $ \measuredangle K'M'O=90^{\circ}$. Due to Miquel's Theorem, $ CK'DK$ is cyclic. Hence $ \measuredangle CKK'=\measuredangle CDK'=\measuredangle CAB$. Since $ OD=OB$ we have that $ \measuredangle OKB=\measuredangle ODB=\measuredangle OBD=\measuredangle ABD$ hence $ \measuredangle K'KB=180^{\circ}-\measuredangle ABD=\measuredangle ACD$. So we have that $ \measuredangle CKB+\measuredangle CMA=180^{\circ}$, concluding that $ CKBM$ is cyclic. Note that $ CK'BM'$ is also cyclic because the circle $ CKBM$ doesn't go through O. Hence $ \measuredangle K'M'B=\measuredangle K'CB=90^{\circ}$ thus concluding that $ \measuredangle K'M'O=90^{\circ}$ $ QED$

Yes.Everybody is welcome here.This site is no doubt the most powerful and useful one for math students over the world!

keira_khtn wrote: My approach: Let L be point of intersections of AD with BC.Easy to see:C,K,L,D lie on one circle.Anologously for A,K,L,B.Hence:K,L,M are colinear.Therefore,OKM=180-(CAB+CBA)=90. Its not obvious to me that C,K,L,D lie on one circle, could someone please explain?

TheHeinrich wrote: keira_khtn wrote: My approach: Let L be point of intersections of AD with BC.Easy to see:C,K,L,D lie on one circle.Anologously for A,K,L,B.Hence:K,L,M are colinear.Therefore,OKM=180-(CAB+CBA)=90. Its not obvious to me that C,K,L,D lie on one circle, could someone please explain? Use inscribed angles in quadrilaterals AOKC and BOKD, and in ABCD, then it's easy to prove $ \angle CKD=\angle CLD$

I can't believe i didnt notice that. Thanks for your help!

grobber wrote: Wow! I can't believe how many problems similar to this one have appeared on the forum . Let $N=AC\cap BD$. $K$ is the image of $O$ through the inversion of pole $N$ and power $NC\cdot NA=ND\cdot NB=$the power of $N$ wrt the circle $(ABDC)$. Let $NT_1,NT_2$ be the tangents from $N$ to the circle, and let $K'=T_1T_2\cap NO$. We get $NC\cdot NA=NT_1^2=NO\cdot NK'$, meaning that $K'=K$, so $K$ lies on the polar $T_1T_2$ of $N$ wrt the circle. Since $M$ also lies on the polar, $MK=T_1T_2$, and $T_1T_2\perp NO$. I cannot understand "so $K$ lies on the polar $T_1T_2$ of $N$ wrt the circle", can someone explain it for me indetail? what means wrt?

nttu wrote: Hehe , I have a solution : Using Radical axix theorem , we show $ AC BD , OK $ intersect together at $ P $ Suppose $ AD $ intersects $ BC $ at $ E $ , $ PE $ intersects $ AB $ at $ Q $ . We have : $ PQ $ is perpendicular to $ AB $ It's easy to show : $ (CK , CO) = (PA , PO) $ So $ OC $ is the tangent of $ (PKC) $ and $ OC^2= OK.OP = R^2 $ We have : $ (PA,PQ,PB,PM) = -1 $ . So $ OQ.OM =R^2 $ This infers $ OQ.OM = OK.OP \iff \measuredangle{PKM} = 90 $ what $ (CK , CO) = (PA , PO) $ means?

If you invert about $O$ with the normal radius, then $K$ goes to the intersection of $AC \cap BD$. Thus, $K$ lies on the polar of $AC \cap BD$, but by brokards so does $M$, hence $MK$ is the polar of $AC \cap BD$.

gearss wrote: nttu wrote: Hehe , I have a solution : Using Radical axix theorem , we show $ AC BD , OK $ intersect together at $ P $ Suppose $ AD $ intersects $ BC $ at $ E $ , $ PE $ intersects $ AB $ at $ Q $ . We have : $ PQ $ is perpendicular to $ AB $ It's easy to show : $ (CK , CO) = (PA , PO) $ So $ OC $ is the tangent of $ (PKC) $ and $ OC^2= OK.OP = R^2 $ We have : $ (PA,PQ,PB,PM) = -1 $ . So $ OQ.OM =R^2 $ This infers $ OQ.OM = OK.OP \iff \measuredangle{PKM} = 90 $ what $ (CK , CO) = (PA , PO) $ means? I answered your previous in the above post, this one means the directed angle between lines $CK, CO$...

nttu wrote: Hehe , I have a solution : Using Radical axix theorem , we show $ AC BD , OK $ intersect together at $ P $ Suppose $ AD $ intersects $ BC $ at $ E $ , $ PE $ intersects $ AB $ at $ Q $ . We have : $ PQ $ is perpendicular to $ AB $ It's easy to show : $ (CK , CO) = (PA , PO) $ So $ OC $ is the tangent of $ (PKC) $ and $ OC^2= OK.OP = R^2 $ We have : $ (PA,PQ,PB,PM) = -1 $ . So $ OQ.OM =R^2 $ This infers $ OQ.OM = OK.OP \iff \measuredangle{PKM} = 90 $ It's easy to show : $ (CK , CO) = (PA , PO) $ I want to konw why $ \angle KCO = \angle APO $, it is not easy to see.

gearss wrote: nttu wrote: Hehe , I have a solution : Using Radical axix theorem , we show $ AC BD , OK $ intersect together at $ P $ Suppose $ AD $ intersects $ BC $ at $ E $ , $ PE $ intersects $ AB $ at $ Q $ . We have : $ PQ $ is perpendicular to $ AB $ It's easy to show : $ (CK , CO) = (PA , PO) $ So $ OC $ is the tangent of $ (PKC) $ and $ OC^2= OK.OP = R^2 $ We have : $ (PA,PQ,PB,PM) = -1 $ . So $ OQ.OM =R^2 $ This infers $ OQ.OM = OK.OP \iff \measuredangle{PKM} = 90 $ It's easy to show : $ (CK , CO) = (PA , PO) $ I want to konw why $ \angle KCO = \angle APO $, it is not easy to see. It is just an angle chase, note $OC = OA$ and that $OKP$ are collinear... so $\angle KPO = 180 - (\angle PCK + \angle PKC = 180 - (\angle PCK + \angle CAO) = 180 - (\angle PCK + \angle OCA) = \angle OCK$. Probably an even better way is to invert about $O$ with radius $OA$, and then note that $\odot PKO$ is invariant ($P \to K$ and $C \to C$), so they are orthogonal. Hence $OC^2 = OK \cdot OP$.

nttu wrote: Hehe , I have a solution : Using Radical axix theorem , we show $ AC BD , OK $ intersect together at $ P $ Suppose $ AD $ intersects $ BC $ at $ E $ , $ PE $ intersects $ AB $ at $ Q $ . We have : $ PQ $ is perpendicular to $ AB $ It's easy to show : $ (CK , CO) = (PA , PO) $ So $ OC $ is the tangent of $ (PKC) $ and $ OC^2= OK.OP = R^2 $ We have : $ (PA,PQ,PB,PM) = -1 $ . So $ OQ.OM =R^2 $ This infers $ OQ.OM = OK.OP \iff \measuredangle{PKM} = 90 $ what $ (PA,PQ,PB,PM) = -1 $ means？ why $ OQ.OM =R^2 $?

Sorry for reviving but here's a solution

Also posted here