Conference rusure::math

118.0. "5 Hard Geometry Problems" by HARE::STAN () Mon Aug 06 1984 20:57

I too, am an avid collector of geometry problems.  I have hundreds of
problems in my collection. (I guess I've just been collecting longer
than ecsvax!pizer.)

My idea of an elegant geometry problem is one that is easy to state,
but relatively hard to prove.  Here now are 5 "hard" problems from
my collection for your enjoyment:

1. In triangle ABC, AB=AC. M is the midpoint of BC.  Point D is on AC
   such that MD is perpendicular to AC.  E is the midpoint of MD.
   Prove that AE is perpendicular to BD.

2. In triangle ABC, P is any point on altitude AH. BP meets AC at D.
   CP meets AB at E.  Prove that angle EHA = angle DHA.

3. The circle inscribed in triangle ABC touches side BC at point D.
   DE is a diameter of this circle. AE meets BC at F. Prove BD=FC.

4. ABCD is a parallelogram. E is any point on side AD. F is any point on
   side BC. BE meets AF at P.  CE meets DF at Q.  PQ (extended) meets AB at R
   and meets CD at S. Prove AR=CS.

5. AB and BC are two chords of a circle (with B lying on minor arc AC) with
   BC > AB.  Let M be the midpoint of arc ABC.  Let D be the foot of the
   perpendicular dropped from M to chord BC.  Prove AB+BD=CD.
   [This is known as the Archimedes Broken Chord Theorem.]

enjoy,	Stanley Rabinowitz
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	ARPA:	stan%[email protected]
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	USPS:	6 Country Club Lane, Merrimack, NH 03054
		(603) 424-2616

    SOLUTION for 1):
                     
    Mark the midpoint of BD as N, connect NM to form triangle DMN. Now,
    I am going to prove that triangle ADM is similar to triangle DMN.
    
    First, since both M and N are midpoints of BC and BD, respectively, so
    MN // CD, since MD perpendicular to CD, so angle DMN is 90 degree.
    
    Second, I am going to prove that AD/DM = DE/MN. Since triangle ADM is
    similar to triangle MDC, so AD/DM = DM/DC, and DM/DC = (2DE)/DC =
    (2DE)/(2MN) because E is midpoint of MD, and MN // DC and BM = (1/2)BC.
    Therefore, AD/DM = DE/MN.
    
    Combining the above facts, we then have already proved that triangle
    ADM is similar to triangle DMN.
    
    Then, mark the intersecting point of AE and BD as F. Now, consider the
    triangle AFD. Since angle ADE = 90 degree, and angle FDE = angle FAD
    (by the previous result), so angle ADF + angle FAD = angle ADF + angle
    FDE = 90 degree, therefore, angle AFD = 90 degree.
                                                        Q.A.D.
    
    Yah-Kong Yang

There is a pretty proof of (5), the Broken Chord Theorem, in the Fall 
1991 issue of the MAPLE Technical Newsletter. The proof can be obtained in 
a few lines by using MAPLE's Geometry package and various 'simplify'-type 
commands.

The hardest part of the proof involves assuring that all the quantities 
involved are >0, i.e. do not involve negative square roots of positive 
quantities.