Math 5-335     Fall 2004
Hints and supplementary exercises for the October 28 homework

Supplementary exercise #2   Given triangle   ABC,   let   P   be the midpoint of , let   Q   be the midpoint of , and let   R   be the midpoint of .
 
        

  1. Show that is parallel to .
    Suggestion: Using appropriate coordinate vectors for   P   and   Q,   find a direction indicator for and compare it with one of the usual direction indicators for .
  2. Similarly, show that is parallel to and that is parallel to .
  3. Using the vector formulas developed on parts a and b -- or by any other valid method -- verify the following equalities of lengths of segments:
    • || = 1/2 ||
    • || = 1/2 ||
    • || = 1/2 ||
  4. Show that the circumcenter of triangle   ABC   is equal to the orthocenter of triangle   PQR.
    Please note:   If we haven't covered both of the pertinent sections, then you'll have to look up the definition of orthocenter in §3.4 and the definition of circumcenter in §3.6,   but you should not need very much else from those sections. In particular, this problem is designed to be workable without use of the formulas from those sections.
     
    Further suggestion:   Suppose that we draw the perpendicular bisector of a side of triangle   ABC.   In what role does the line that we drew function relative to triangle   PQR?   ( This is intended to be a geometry problem. In particular, it is recommended not to use the formulas given in Theorem 7, Corollary 10, and Theorem 13 of Chapter 2.)

Hint for problems #2 and #3 in §5.9:   To answer the question about orientation of an isometry obtained by composition or conjugation, we need to find out whether the determinant of its matrix part is +1 or -1. The main linear algebra fact needed for this purpose is that if   U   and   V   are   n by n   matrices, then the determinant of the product is equal to the product of the determinants; thus:
           det(UV) = det(U)·det(V).
And also, in the case of an invertible matrix  U,   we have:
           det(U -1) = 1/det(U) ,
because the identity matrix has determinant = 1 and   U·U -1 = 1   (the identity matrix).   Well, maybe one could claim that the formula for   det(U -1)   isn't such a big deal in the case where   U   is orthogonal, since we have   det(U) = ±1   in this case anyway. Nonetheless, it still could be the right way to think about things when you're considering conjugate matrices.
 

Supplementary exercise #3   [Optional; please submit separately]   Given triangle   ABC   and triangle   PQR. Assume that there exists a nonzero real number   v   such that:
          || = v ||    || = v ||    and    || = v ||.

Show that if the orthocenter of triangle   ABC   is   (r,s,t)ABC,   then the orthocenter of triangle   PQR   is   (r,s,t)PQR.    (We use the notations   ABC   and   PQR   as part of the superscript to indicate the triangle relative to which we are taking the barycentric coordinates . . . )
 
Suggestion: Refer to the formula from Theorem 7 of Chapter 3 of the text.   What power of   v   can be factored from the numerators and from the denominators? ( It's useful to note here that the numerators and denominators in the formula given in Theorem 7 are homogeneous polynomials:   this means that all of the monomials occurring in one of these polynomials have the same total degree. For instance,   2x3 + 3y3 + 5z3 + 11xyz   is homogeneous of degree 3. )
 

Note:   The assignment also includes problem #7 from Chapter 3 of the text.   And problem #11 from Chapter 3 is optional for extra credit, just like Supplementary exercise #3.

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