|For assignment number 9, we
have chosen to explore problems 8 and 9 having to do with the Simpson
|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
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.
The petal triangle is colinear if and only if, P is on the circumcircle of ABC.
|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.