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(Ortho)Centrally Seeking Length (Posted on 2007-11-06) Difficulty: 3 of 5
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 = 30o.

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

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 3

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 2007-11-06 12:37:20

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