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Schiffler Point: Consider a triangle ABC. Let point I denote the incenter of triangle ABC. Now we can say that the Schiffler point is the concurrent point of intersection of the Euler lines of triangles ABI, BCI, and CAI. Call this point S. The Euler line of the original triangle ABC also passes through point S.Kurt Schiffler: meet the man behind the point Kurt Schiffler, an accomplished engineer gained fame among mathematicians when he stumbled upon the concurrent point of Euler lines for the triangles made with an original incenter as a vertex. Schiffler explored geometry as an amateur, but his discovery was no accident. Schiffler's life modeled success whatever his endeavors.Euler Line: A Euler line is the line through the centroid and circumcenter of a triangle. The Euler line also passes through the orthocenter.
Concurrent points: To say that two points P and Q are concurrent is to say that the two points are the same point.
Centroid: The concurrent point of the medians of a triangle.
Incenter: The concurrent point of the angle bisectors of a triangle.
Circumcenter: The concurrent point of the perpendicular bisectors of a triangle.
Orthocenter: The concurrent point of the alititudes of a triangle.
Exploration: We explored the movement of the Schiffler point as triangle ABC was an equilateral triangle, an isosceles triangle, and a right triangle. We noticed similarites and differences of the Euler lines with angle bisectors and other known features of the triangles as well.
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Conjecture: After exploring the Schiffler point we were able to make the following conjectures about the Schiffler point in each of the three distinct triangles.
THE GIVENS: Given a triangle ABC, with the incenter, which is the point of concurrency of the angle bisectors, labeled I. Let S be the Schiffler point. The Euler line in the line passing through the centroid, the circumcenter, and the orthocenter of a given triangle. Let E1, E2, and E3 be the Euler lines of the three triangles created from connecting each endpoint of triangle ABC to the incenter, triangles IAB, IBC, and ICA. Let E be the Euler line for the original triangle ABC.Equilateral triangle: When triangle ABC is equilateral, the Schiffler point is concurrent with the incenter of triangle ABC, as well as the orthocenter, the centroid, and the circumcenter. Also, three of the Euler lines that make up the Schiffler point, namely E1, E2, and E3 match up with the three angle bisectors of the original triangle ABC. They also act as perpendicular bisectors of the legs of triangle ABC. In an equilateral triangle, the Euler line E for triangle ABC is not a single visible line because the centroid, circumcenter, and orthocenter are all concurrent points.
Isosceles triangle: When triangle ABC is isosceles with AC and BC congruent, two of the Euler lines E3 and E are coincidental and are also coincidental with the bisector of angle ACB of triangle ABC. They are also coincidental with the perpendicular bisector of AB. This line is the Euler line of triangle ABC so the circumcenter, orthocenter and centroid all lie on line E. The Schiffler point also lies on line E.
Right triangle: When triangle ABC is a right triangle with angle CAB as the 90º angle. The Schiffler point S does not correspond to the angle bisector, or the altitude, but it is a point on the median between the 90º vertex and the midpoint of the hypotenuse. This is line AM where M is the midpoint of CB and AM is the Euler line of the original triangle ABC. While exploring the right triangle, we constructed some lines through the Schiffler point which led us to a new series of conjectures. We made a line m through S parallel to side AC and a line n through S parallel to side AB. We then observed the consequences of constructing these lines. These lines form a rectangle ALKS who shares the right angle vertex with triangle ABC. There is also an interesting relationship between the segment pieces formed when the parallel lines intersected AC and AB. Let L be the point of intersection with AC and K be the point of intersection with AB. Here is the ratio: AC/AL=AB/AK. This is a ratio that can be related to Euclid's observations about similar triangles and is an idea we will come back to in the proof section.
Proof: We wish to prove that in the isosceles triangle ABC, that the Euler line of triangle ABC is coincidental with the Euler line of the triangle AIB (called E3).
1) We know that since triangle ABC is isosceles that the sides AC and BC are congruent and that the base angles CAB and CBA are congruent angles also.
2) In an isosceles triangle, the angle bisector, which contains the incenter I of ACB (the vertex where the two congruent segments meet), is also the perpendicular bisector of segment AB, a median of triangle ABC, and the altitude of triangle ABC as well.
3) The line mentioned above is the Euler line E of triangle ABC because the Euler line contains the Centroid, the Circumcenter, and the Orthocenter. By the definition of a Euler line, because all of the points mentioned above lie on this line CD,this is the Euler line of ABC.
4) Because point I, the incenter, is on the perpendicular bisector of triangle ABC, then the incenter also lies on the Euler line E of triangle ABC.
5) Now consider triangle AIB. By the perpendicular bisector theorem, we know that AI is congruent to IB because all points on the perpendicular bisector of segment AB are equidistant from the endpoints A and B. Therefore, AIB is an isosceles triangle as well.
6) Since triangle ABC and AIB share the same base AB, we know from above that D is the midpoint and ID is the perpendicular bisector of triangle AIB, the altitude, and a median from vertex I to the opposite side AB.
7) Therefore, ID is a Euler line E3 of triangle AIB and so, Euler lines E and E3 are coincidental.
Extensions: We noticed while exploring the right triangle that lines parallel to the adjacent legs of the right triangle through the Schiffler point cut the legs proportionally. This notion has obvious connections to Euclid's ideas about similar triangles. It would be interesting to further explore this conjecture and show similar triangles wtih parallel hypotenuse. Also, it would be interesting to see if the Schiffler point is unique on Euler line E of triangle ABC in that it is the only point on this line with that property. Check the Euclid link below to see Euclid's propostion on similar triangles.
Links: find more information....
Kurt Schiffler | Schiffler point | Concurrent points | Euler Lines | more on Schiffler Point | Euclid's Elements | class page | Natalie's home pageor investigate for yourself... Yahoo
definitions GSP construction exploration conjecture proof biograghical sketch links extensions
this is the EMAT 4680 group project on the Schiffler point with contributions from :
Natalie Smith | Elizabeth Jones | Amy Terry | Ava Antone