Sarah Hofmann EMAT 6080 Summer 2006 Assignment 9 For assignment number 9, we have chosen to explore problems 8 and 9 having to do with the Simpson Line. First we constructed a triangle ABC and a point P. Then we constructed the petal triangle RST for triangle ABC and point P. Problem 8 asked what happens when the petal point P is one of the verticies of triangle ABC.  We can see that in all three cases, the petal triangle appears to be linear. Let's try to prove the conjecture: If P is a vertex of the triangle, then the petal triangle is linear. Let ABC be a triangle, P a point, and RST the petal triangle for point P and triangle ABC, where R is on line BC, S is on line AC, and T is on line AB, as shown below. Suppose P = A.  Then we can see that the intersection of the line through A perpendicular to line AB is A.  So S = A.  The intersection of the the line through A perpendicular to the line AC is A.  So T = A.  So, S = T.  Thus RST = RTT = RT, which is a segment, as needed. Problem 9 asks us to find all points in which the petal triangle is a segment (this is called Simpson Line).  Since all three points A, B, and C are points where the petal triangle is linear, and all three points A, B, and C are on the circumcircle.  The circumcircle would be a good place to look to find all points in which the petal triangle is linear.  Lets try. Looks like it worked!  Let's see if we can prove it. Conjecture:  The petal triangle is colinear if and only if, P is on the circumcircle of ABC. Proof: Let ABC be a triangle, C' its circumcenter, and P a point on its circumcircle. Construct the pedal triangle RST from the perpendicular intersections of lines through point P with the segments of ABC. From our construction of RST we know that segment PS is perpendicular to line AC, PT is perpeendicular to AB, and PR is perpendicular to BC. Since the above segments are perpendicular to the mentioned lines we can see angle(PTB) = 90 degrees angle(PRB) = 90 degrees angle(PSC) = 90 degrees angle(PRC) = 90 degrees So PTBR and PSCR are cyclic quadrilaterals (quadrilaterals whose opposite angles are equal - 90 degrees).  We know PTBR and PSCR are cyclic quadrilaterals if and only if angle(PRT) = angle(PBT) angle(PRS) = angle(PCS) We finally note that since points A, B, and T all lie on the same segment, angle(PBT) = angle(PBA) Similarly we can see that angle(PCA) = angle(PCS) Using the above equalities we can now see the following angle(PRT) = angle(PRS). So R,T, and S must be collinear, as needed.