Therefore,Īlternate solutions are always welcome. Because is cyclic and is isosceles with vertex angle, we have. Taking the quotient of the two equations yieldsīecause is an exterior angle to triangle, we have. Taking the quotient of the two equations and noting that, we findĪpplying the Law of Sines to triangles and, we find Īpplying the Law of Sines to triangles and gives Because is the perpendicular bisector of, it follows that triangle is an isosceles triangle with. The proof can be adapted to other configurations. This leads to configurations similar to the ones shown above. We may assume, without loss of generality, that. Indeed, the fact that the first two angles in are right is immediate because and are the perpendicular bisectors of and, respectively. It will then follow that lie on the circle with diameter. It follows that, , and are collinear, and then that the points, ,, and lie on a circle. Next note that, , and are collinear and are the images of, , and, respectively, under a homothety centered at and with ratio. Because the images under the inversion of lines and are circles that intersect in and, it follows that. Hence quadrilateral is cyclic and, by a similar argument, quadrilateral is also cyclic. Because is the midpoint of and is the midpoint of, we also have. Find point so that is a parallelogram and let denote the center of this parallelogram. Invert the figure about a circle centered at, and let denote the image of the point under this inversion.
Solution 8 (inversion) (Official Solution #2) Then is a median, use the lemma again to show that, and similarly, so you're done. Thus is taken to the circumcenter of and is the midpoint of, which is also the circumcenter of, so all lie on a circle. Now by the homothety centered at with ratio, is taken to and is taken to. If is the circumcenter of then so is cyclic. Therefore, it takes the midpoint of to the midpoint of, or to. Therefore, (by AA similarity), so a spiral similarity centered at takes to and to. From here, it is simple angle-chasing to show that satisfies the original construction for, showing we are done.īy the Law of Sines. It is easy to see (the intersection of ray and the circumcircle of ) is colinear with and, and because line is the diameter of that circle,, so is the point in the lemma hence, we may apply the lemma. This is fairly easy to prove (as H, O are isogonal conjugates, plus using SAS similarity), but the author lacks time to write it up fully, and will do so soon. If are points on circle with center, and the tangents to at intersect at, then is the symmedian from to. Let the circumcircles of triangles and intersect at and.
This solution utilizes the phantom point method. Hence, ,, and lie on a circle, as desired. ,, ,, and all lie on the circumcircle of. In other words, lies on the circumcircle of. Using essentially the same angle chasing, we can show that is directly similar to. It is easy to see that they are oriented such that they are directly similar.īy the similarity in the Lemma. Alsoīecause, and is the medial triangle of so. Notice that intercepts the minor arc in the circumcircle of, which is double. Note lies on the perpendicular bisector of, so. The intersection of and is, the circumcenter of. Now it remains to show the other angle equality would follow by symmetry. With the tangent formula and then taking the midpoint, we find that the center of is Then we want to find the reflection of 0 over the line through and Then we get We know that these two circles already intersect at so we can reflect over the line between their centers. We will find the ghost point the second intersection of and. This implies that bisects interior, making, as was required. Now, note that bisects the exterior and bisects exterior, making the -excentre of. Here, we used that, as is the midpoint of. It is sufficient to prove that, as this would immediately prove that are concyclic.īy applying the Menelaus' Theorem in the Triangle for the transversal, we have (in magnitude) Prove that points, ,, and all lie on one circle. Let the perpendicular bisectors of and intersect ray in points and respectively, and let lines and intersect in point, inside of triangle. ( Zuming Feng) Let be an acute, scalene triangle, and let, , and be the midpoints of, , and, respectively. 2.8 Solution 8 (inversion) (Official Solution #2).Amy Morton as Desk Sergeant Trudy Platt.LaRoyce Hawkins as Officer Kevin Atwater.Patrick John Flueger as Officer Adam Ruzek.Marina Squerciati as Officer Kim Burgess.Tracy Spiridakos as Detective Hailey Upton.
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