Assignment # 8 Altitudes and Orthocenters by Victor L. Brunaud-Vega 1.     Construct any triangle ABC. 2.     Construct the Orthocenter H of the triangle ABC.  LetÕs remember: the orthocenter of a triangle is the common intersection of the three lines containing the altitudes. An altitude is a perpendicular segment from a vertex to the line of the opposite side. (Note: the foot of the perpendicular may be on the extension of the side of the triangle.) It should be clear that H does not have to be on the segments that are the altitudes. Rather, H lies on the lines extended along the altitudes. 3.     Construct the Orthocenter of triangle HBC. 4.     Construct the Orthocenter of triangle HAB. 5.     Construct the Orthocenter of triangle HAC. 6.     Construct the Circumcircles of triangles ABC, HBC, HAB and HAC. 7.     Conjectures?  Proofs? Conjecture 1: If triangle ABC is rectangle, then: á      Orthocenter H coincides with vertex B (where the catets of triangle ABC meet). á      Orthocenter H coincides with the Orthocenter of triangle HAC. á      The circumcircles of triangles ABC and HAC coincide. á      The circumcenters of triangles ABC and HAC coincide with the midpoint of the hipotenuse Conjecture 2: all four circumcircles are congruent.  Here is an animation of triangle ABC, showing how the radii of circles c1, c2, c3, and c4 is always the same, even if it is not constant. Conjecture 3: The orthocenters of the three triangles formed by creating the orthocenter of the original triangle, lie on the vertices of the original triangle.    As you can see in the picture, the orthocenter of triangle HAC is point B; the orthocenter of triangle HAB is point C; and the orthocenter of triangle HBC is point A.  This is valid for any triangle ABC, as you can explore here. Conjecture 4: Given three congruent circles with one common point, we can find a triangle ABC, its orthocenter H, and the triangles HAC, HAB, and HBC, and its orthocenters. Constructing a fourth congruent circle (c4), with a common point A with c3 and c1, and a common point B with c3 and c2 , and a common point C with c1 and c2 , we have defined the vertices of our main triangle.   Now, letÕs go for the orthocenter H, which is the intersection among c1, c2, and c3.   From this point H, orthocenter of the triangle ABC, we can construct the triangles HAB, HAC, and HBC. The centers of our original circles have became the circumcircles of the mentioned triangles: á      The circumcircle of triangle ABC is c4. á      The circumcircle of triangle HAC is c1. á      The circumcircle of triangle HAB is c3. á      The circumcircle of triangle HBC is c2.   The centers of the circles c2, c1, and c3 (P, Q, and R, respectively) are related to the perpendiculars to the sides of the original triangle ABC.   Here is an animation to see these relationships. As you can see in the picture below: á      P is part of the perpendicular to BC at its middle point; á      Q is part of the perpendicular to AC at its middle point; á      R is part of the perpendicular to AB at its middle point; á      and S is the intersection of the perpendiculars to the sides of triangle ABC at its middle points. As I said before, the orthocenters of the triangles HBC, HAC, and HAB are at points A, B, and C, respectively. Conjecture 5: When triangle ABC is acute or rectangle, the sum of the areas of triangles CBP, ARB, and ACQ (formed by the sides of ABC and the respective centers of the circumcircles of triangles HAC, HBC, and HAB), is equal to the area of triangle ABC.  Here is an animation showing this relationship. Return to my Class Page Return to EMAT 6680 Home Page.