Please, help me with difficult geometry problem?

Question

http://i24.tinypic.com/2dj9cti.jpg

C, D are points on the semicircle diameter AB, center O.
CD meets the line AB at M.
The circumcircles of AOC and DOB meet again at K.

Prove that angle MKO is right angle.

This problem is from russian mathematical olympiad '95.
I spent two days already reflecting on it, and still do not have any solution.

C and D are points on the semicircle, which has diameter AB and center O.

Sorry.

Answer 1

Beautiful problem, took me four hours to crack it. Here is the solution.

Apply an inversion with the center O and radius R=|OA|. Inversion is a transformation of the plain carrying every point P, except O, to the point P* such that |OP| |OP*| = R^2. See e.g. http://en.wikipedia.org/wiki/Inversive_geometry for details. We will use the following property of inversion: a straight line, not passing through O, is transformed into a circle, passing through O, and vice versa. The points lying on the circle, like A, C, D, B, are obviously transformed to themselves: A -> A etc.

The circumcircle AOC is transfomred into the line passing through points A and C (because we know that circles are transformed into lines, we know that A and C do not move, and it is possible to draw only one line through two points). Similarly, the circumcircle DOB is transformed into line passing through B and D. Since point K belongs to both circumcircles, it is transformed into the point K*, which is the intersection of lines (AC) and (BD).

Thus, we have a triangle AK*B. Since points C and D belongs to the circle ACDB, then |AD| and |BC| are altitudes in this triangle. Project point K* on the side |AB|, and get the third altitude |K*L|.

The angle ABD is intercept the arc ACD of the circle ACDB. The angle ACD is intercept the complimetary arc, and hence it is equal to 180-ABC. Then the angle K*CD is equal to 180-ACD, which is equal to ABD. Consequently, triangles K*CD and AK*B are similar. Applying the same arguments, we see that triangles LDB and LCA are also similar to AK*B.

From above similarity of triangles, the angle LDB is equal to the angle BAC. The triangle OBD is isosceles, so that the angle ODB is equal to the angle ABD.The angle ODL is equal to the difference ABD-BAC. Similarly, we get relations for angles ACL=ABD, ACO=BAC, and OCL=ABD-BAC.

The angles OCL and ODL are equal, it means that the points O, C, D, and L lie on a circle. Indeed, construct a circle through three points O,C,L. Extend the line OD until it intersects the circle at point E. The angle OEL is intercept the arc OL. Hence it is equal to OCL and ODL, so that the points D and E coincide.

Now go to the point M. When we apply the inversion, the line CM is transformed to a circle. This circle should pass through point CDO. Hence this is the circle OCDL. M*- the image of M-lies on the line AB and on the circle OCDL. The same is true for point L. There could be two points wich satisfy this property (circle and line intersect in two points). One of them is point O. Since neither L no M* are equal to O, then this two points should coinced, i.e. L=M*.

Recall, that K*L is the altitude, so that the angle OM*K* is right. By the property of inversion, |OK| |OK*| = |OM| |OM*|, so that |OK|/|OM*| = |OM|/|OK*|. Hence triangles OKM and OM*K* are similar and

OKM = OM*K* is the right angle.

Answer 2

Your given information is contradictory. If C and D are points on Diameter AB then line CD is coincident with line AB. That is, they meet everywhere, not at just a unique point of intersection.

You need to restate the problem so it begins:

C and D are points of the semicircle WITH diameter AB, center O.

Answer 3

The risk P that a factor chosen randomly from interior the circle could additionally be interior the sq. is: P = sq. section / circle section circle section = pi*R^2 sq. section = R^2/sixteen so P = a million / (sixteen*pi) (approx a million/50 or 2%)

Answer 4

Points A & B are the end points of the arc?

From what I have come up with, the line of OK is perpendicular to the line made by the two centerpoints of the circumcirles.

Answer 5

I need more information. What's with the two circles within the semicirclcle? Without knowing the circle, I can not identify point K.


XR

Answer 6

I agree with my above answer.