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Incenter: The three angle bisectors of a triangle meet in one point called the incenter. It is the center of the incircle, the circle inscribed in the triangle. The area of the triangle is sr, where s=(a+b+c)/2, and r is the radius of the incircle.
Circumcenter: The three perpendicular bisectors of the sides of a triangle meet in one point called the circumcenter. It is the center of the circumcircle, the circle circumscribed about the triangle. The area of the triangle is abc/4R=2R^2sinAsinBsinC, where R is the radius of the circumcircle, and ^2 means squared. The distance between the incenter and the circumcenter is sqr(R(R-2r)). If the triangle is obtuse, then the circumcenter is outside the triangle. If it is a right triangle, then the circumcenter is the midpoint of the hypotenuse.
Centroid: The three medians (the lines drawn from the vertices to the bisectors of the opposite sides) meet in the centroid or center of mass (center of gravity). The centroid divides each median in a ratio of 2:1.
Orthocenter:The three altitudes of a triangle meet in one point called the orthocenter. If the triangle is obtuse, the orthocenter is outside the triangle. If it is a right triangle, the orthocenter is the vertex which is the right angle.
Euler line: The orthocenter O, centroid C, and circumcenter X all lie on a line, the Euler Line (if the triangle is equilateral, all four centers are the same point). The incenter does not lie on this line, unless the triangle is isosceles. The orthocenter is twice as far from the centroid as the circumcenter is.
Ceva's Theorem: Three cevians (a cevian is a line segment connecting a vertex with the opposite side), AX, BY, CZ, meet in a point iff (if and only if) (AZ/BZ)(BX/CX)(CY/AY) = 1. The proof is not particularly easy.
Try this on some of the above. It is particularly simple to use it on the centroid.
Question #1: Let's find the point P within a triangle ABC, so that if we draw lines from P to the three vertices, we get three equal (area) triangles. Is this P perhaps a different center of the triangle, from those mentioned above? The answer to this and other questions is at the bottom of this page.
Gergonne point: Another center of a triangle is the Gergone point. The lines connecting the tangent points of the incircle to the opposite vertices meet in this point.
Fermat point: Another center is the Fermat point (also called the isogonic center or the Rorricelli point). This is the point which minimizes the sum of the distances from the three vertices. This problem is called Fermat's problem or Steiner's problem. The three angles at this point, between the vertices, are all 120 degrees. This point has many practical uses, as it is often a good idea to minimize distances, so save money. Telephone lines between three cities, using the minimum amount of wire, will be three lines which meet at the Fermat point. More points result in connections between Fermat points. The Fermat point can be constructed by constructing equilateral triangles on each side of the triangle, and connecting their farthest vertices with the opposite vertex of the original triangle.
Congruent isoscelizer point: Another center is the "congruent isoscelizer point." An isoscelizer is a line which makes an angle into an isosceles triangle. Any point in a triangle produces three isoscelizers. In 1989, P. Yff proved that there is one unique such point, which makes all three isoscelizers equal in length.
Lemoine point: Another center is the Lemoine point. A line through a vertex of a triangle is isogonal to another such line if it is the reflection of that line about the angle bisector through that vertex. A symmedian line is isogonal to the median from that vertex. The three symmedian lines of a triangle meet in the Lemoine point (also called the symmedian point or the Grebe point). The isogonal lines of any three lines through the three vertices which meet in a point (like any of those above) also meet in a point.
Spieker center: The incenter of the median triangle is the Spieker center. It is the center of mass (center of gravity or centroid) of the perimeter of the original triangle.
There are other, more complicated points related to triangles. Also see The Nine Point Circle; the nine point center is on the Euler line, as is the tangential triangle circumcenter. The incircle and the nine-point circle are tangent at the Feuerbach point.
Question #2: In triangle ABC, the angle bisector of A meets the altitude from B, and the median from C in one point P. Does the triangle have to be equilateral? The answer to this and other questions is at the bottom of this page.
Answer #1: In our diagram, area A1=A2. They have a common base (CP) and must have equal heights. Therefore A3=A4, as they have a common base (PX) and have the same heights as A1 and A2 respectively. So A1+A3=A2+A4. Then CP must bisect AB at X, as triangles ACX and BCX have the same area and the same height (from C to AB) and their bases must be equal. Similarly, AP bisects BC and BP bisects AC. So P is the centroid, as described above. And all six small triangles in that diagram (see centroid above) are of equal area.
Answer #2: No, here is an example that is not an equilateral triangle.
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