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This is the first serious stereometry problem I am solving, so I would be very glad to hear whether it is correct or not. I will use a lemma from plane geometry (a theory where I feel rather at home ): Lemma 1. If H is the orthocenter of a triangle ABC, and N is the foot of the altitude from the vertex A, then we have $NH\cdot NA = BN\cdot NC$, where the segments are directed. Proof. I won't use directed segments in the proof, so I will only show the equality $NH\cdot NA = BN\cdot NC$ for non-directed segments. However, showing that it also holds for directed segments will be then just a matter of looking at the sketch. If O is the foot of the altitude of triangle ABC from the vertex B, then the two lines AN and BO are altitudes of the triangle ABC and thus meet at its orthocenter H. Consequently, < HBN = < OBC = 90 - < OCB = 90 - < ACN = < CAN. On the other hand, clearly < HNB = 90 = < CNA. Thus, the triangles HBN and CAN are similar, and we can conclude that HN : BN = CN : AN. In other words, NH : BN = NC : NA, or, equivalently, $NH\cdot NA = BN\cdot NC$. This proves Lemma 1. Lemma 1 has a converse: Lemma 2. Let ABC be a triangle, let N be the foot of the altitude from the vertex A, and let H' be a point on the line AN such that the equation $NH^{\prime}\cdot NA = BN\cdot NC$ holds with directed segments. Then, this point H' is the orthocenter of triangle ABC. Proof. If H is the orthocenter of triangle ABC, then Lemma 1 yields $NH\cdot NA = BN\cdot NC$. Comparing this with $NH^{\prime}\cdot NA = BN\cdot NC$, we see that NH = NH'; since we are using directed segments and the points H and H' both lie on the line AN, we can conclude that H = H', so that the point H' is the orthocenter of triangle ABC. Lemma 2 is proven. Now let's come to the solution of the problem. For any two points U and V in space, I will denote the plane passing through the midpoint of the segment UV and being perpendicular to the line UV as the perbisplane of the segment UV. This is, of course, an abbreviation for "perpendicular bisector plane"; we will often use the fact that the perbisplane of a segment UV is the locus of all points W such that UW = VW. Since the tetrahedron ABCD is regular, the triangles ABC and ABD are equilateral. Let X and Y be the centers of these equilateral triangles. Then, of course, AX = BX and AY = BY; thus, the points X and Y lie on the perbisplane of the segment AB. Also, since the triangles ABC and ABD are equilateral, we have AC = BC and AD = BD, so that the points C and D lie on the perbisplane of the segment AB. The point M, lying on the line CD, also lies on the perbisplane of the segment AB. Hence, AM = BM. Trivially, another point lying on the perbisplane of the segment AB is the midpoint S of the segment AB. The definition of the point P that was given in the problem is just a complicated way to say that the point P is the orthocenter of the triangle ABM. Of course, since the triangle ABM is isosceles (we have AM = BM), we thus must have AP = BP, and consequently, the point P lies on the perbisplane of the segment AB. Altogether, the points X, Y, C, D, M, S and P all lie on the perbisplane of the segment AB. This will be very useful for us. Now, I am not really sure what the problem means when it says that the point M must lie on the edge CD: should it mean that the point M lies on the line CD, or should it mean that the point M lies on the segment CD ? In the following, I will care for both of these cases. In fact, I will show: (a) If the point M traces the whole line CD, the locus of the point P is the circumcircle of triangle SXY (except for the point S, which corresponds to an infinite position of the point M). (b) If the point M traces just the segment CD, the locus of the point P is the arc XY of the circumcircle of triangle SXY. At first, I will prove part (a): Since the point S is the midpoint of the segment AB, the line MS is a median of triangle ABM. But since this triangle ABM is isosceles (AM = BM), this median is also an altitude, so that the point S can be considered as the foot of the altitude of triangle ABM from the vertex M. Since the point P is the orthocenter of triangle ABM, it follows that the point P lies on the line MS, and by Lemma 1 we have $SP\cdot SM = AS\cdot SB$. Since the point S is the midpoint of the segment AB, we have $AS=SB=\frac{AB}{2}$; denoting $\frac{AB}{2} = s$, we get AS = SB = s, and thus $SP\cdot SM = s^2$. Now, if we consider the circle k which has center S and radius s and which lies in the perbisplane of the segment AB, then the fact that the point P lies on the line MS and satisfies $SP\cdot SM = s^2$ yields that the point P is the inverse of the point M in the circle k. Similarly, taking into account the points X and Y, which are the centers - in particular, the orthocenters - of the equilateral triangles ABC and ABD, we see that the points X and Y are the inverses of the points C and D in the circle k. Hence, the inverse of the line CD in the circle k is the circumcircle of triangle SXY (in fact, this inverse must be, like the inverse of any line which doesn't pass through the center of the inversion, a circle through the center of the inversion, and it must pass through the points X and Y since they are the inverses of the points C and D in the circle k). Now, since the point M lies on the line CD, the inverse P of the point M in the circle k must lie on the inverse of the line CD, i. e. on the circumcircle of triangle SXY. Conversely, if we take any point on the circumcircle of triangle SXY (except of the point S) and denote it by P', then by redoing our argumentation above backwards, we see that there is a point M on the line CD such that the point P' is the orthocenter of the triangle ABM (in fact, just call M the inverse of the point P' in the circle k; it is then easy to show that $SP^{\prime}\cdot SM = AS\cdot SB$, and applying Lemma 2 instead of Lemma 1 we get what we need). Hence, the locus of the point P while the point M traces the line CD is the circumcircle of triangle SXY. This proves assertion (a). For the proof of assertion (b), just look at the picture (you needn't draw the points A and B, just consider the points X, Y, C, D, M, S and P which all lie on the perbisplane of the segment AB). The point P lies on the arc XY of the circumcircle of triangle SXY if and only if the line SP divides the angle XSY internally, but this is clearly equivalent to saying that the point M lies inside the segment CD. So the proof is complete. Thanks for reading this (if you have ). Also thank you in advance if you find mistakes in the above solution (as I've said, I'm new to stereometry). Darij