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Perpendicular Area (Posted on 2005-12-08) Difficulty: 4 of 5
The altitude from one of the vertices of an acute-angled triangle ABC meets the opposite side at D. From D, perpendiculars DE and DF are drawn to the other two sides. Prove that the length EF is the same whichever vertex is chosen.

For extra credit: if a=|BC|, b=|CA|, c=|AB|, and d=|EF|, show that Area(ABC)=½√(abcd).

2002 British Mathematical Olympiad, Round 2, Problem 1.

  Submitted by Bractals    
Rating: 3.7500 (4 votes)
Solution: (Hide)
Pick C as the vertex with E and F on sides AC and BC
respectively. Since angles CED and CFD are supplementary,
CEDF is a cyclic quadrilateral. By Ptolemy's theorem we
have 

  |CD|*|EF| = |CE|*|DF| + |ED|*|FC|

            = [|CD|*cos(ECD)]*[|CD|*sin(DCF)] +
              [|CD|*sin(ECD)]*[|CD|*cos(DCF)]

            = |CD|2*[cos(ECD)*sin(DCF) + sin(ECD)*cos(DCF)]

            = |CD|2*sin(C)

                or

                                     sin(C)     Area(ABC)
  |EF| = |CD|*sin(C) = 2*Area(ABC)* -------- = -----------
                                       c            R

  , where R is the circumradius of triangle ABC.

A similar argument can be made for vertices A and B.

Bonus:

       Area(ABC)       Area(ABC)
  d = ----------- = ---------------
           R             a*b*c
                     -------------
                      4*Area(ABC)

               or

  Area(ABC) = sqrt(a*b*c*d)/2

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionNo extra creditowl2005-12-09 10:40:05
Some ThoughtsParticular caseFederico Kereki2005-12-08 19:58:14
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