T is the
orthocenter of a triangle PQR and T is located within the triangle. S is the midpoint of the side QR and Angle QPR = 30
^{o}.
The line TS is extended to the point U such that S lies between T and U satisfying TS = SU.
Determine the length of PU given that QR = 1
For fixed points Q and R and a fixed angle
QPR of 30 degrees, all triangles PQR (with
vertex P on the same side of line QR) have
the same circumcircle. Let O be the center
of this circumcircle.
Let <XY> denote the vector from point X to
point Y.
<UP> = <OP>  <OU>
= <OP>  (<OT> + <TU>)
= <OP>  <OT>  2<TS>
= <OP>  <OT>  2(<OS>  <OT>)
= <OP> + <OT>  2<OS>
= <OP> + (<OP> + <OQ> + <OR>)
 2(<OQ> + <OR>)/2
= 2<OP>
The restriction that the orthocenter lie
within the triangle PQR is not needed.
If we take PQR to be a right triangle, it is
clear that the diameter of the circumcircle
is
PU = 2
Edited on November 6, 2007, 8:07 pm

Posted by Bractals
on 20071106 12:37:20 