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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1216.0. "Math Puzzle from USENET" by AITG::DERAMO (Dan D'Eramo is on vacation. :-)) Sat Apr 07 1990 19:50

Path: shlump.nac.dec.com!decuac!haven!aplcen!samsung!usc!snorkelwacker!mit-eddie!ll-xn!rp
From: [email protected] (Richard Pavelle)
Newsgroups: sci.math,rec.puzzles
Subject: Trig problem
Keywords: Has a surprising answer (not given)
Message-ID: <[email protected]>
Date: 6 Apr 90 20:01:16 GMT
Organization: MIT Lincoln Laboratory, Lexington, MA
Lines: 17
Xref: shlump.nac.dec.com sci.math:10502 rec.puzzles:5622
 
 
       a --------------------------------------| g
                                               |
                                               |
       b                                       |
        |                                      |
        |                                      |
        |                                      |
        |                                        f
        |                                        e
        |                                      |
       c --------------------------------------  d
 
Given the rectangle acdg, two cuts are made of length ef and length ab.
Given ab, ac, de, ef and the angle aeb find cd. The complexity of the
answer surprised me. Perhaps it will surprise you too. -- 
Richard Pavelle         UUCP: ...ll-xn!rp
                        ARPANET: [email protected]
T.RTitleUserPersonal
Name
DateLines
1216.14GL::GILBERTOwnership ObligatesWed Apr 11 1990 12:4756
The problem is nearly equivalent to this (point c has been moved up |de|):

      a+
       |\
       | \
       |  \
      b+   \
       | \  \
       |   \ \
       |     \\
      c+-------+e
            
It's more useful to label the edges than the vertices, so let

	A = |ab|
	B = |bc|
	X = |ce|
	G = angle aeb

We are given A, B, and G, and want to determine X.

By the law of cosines, we have

	A� = |be|� + |ae|� - 2 |be| |ae| cos(G)

	A� = (B�+X�) + ((A+B)�+X�) - 2 cos(G) sqrt((B�+X�)((A+B)�+X�))

	(B�+X�) + ((A+B)�+X�) - A� = 2 cos(G) sqrt((B�+X�)((A+B)�+X�))

Squaring both sides, and further manipulations yields...

	   4      2
	T X  + T X  + T  = 0
	 4      2      0

	T  = 4 sin�(G)
	 4

	T  = 8 B (A+B) sin�(G) - 4 A� cos�(G)
	 2

	T  = 4 B� (A+B)� sin�(G)
	 0

Use the quadratic formula, take a sqquare root, and you have:


            A�cos�G - 2B(A+B)sin�G � AcosG sqrt(A�cos�G - 4B(A+B)sin�G) 
X = sqrt( --------------------------------------------------------------- )
                                    2sin�G

Nasty, but not hideous.

Note that we've introduced some extra roots.  In the final sqrt, a negative
is easily understood as the `mirror' point on ce to the left of abc.  But
what of the � from the quadratic formula?
1216.2the problem actually allows two values for X in generalCSSE::NEILSENI used to be PULSAR::WALLYWed Apr 11 1990 15:3724
I drew a figure much like that in .1, and then applied the rule for tangent of 
sum of two angles to the two angles with vertex at e.

I got the quadratic equation

	x^2 * Tan(G)/(B-A) + X * (1-B/(B-A)) + B * Tan(G) = 0

I did not bother to solve it, since I had gone far enough to see that the result
would indeed be nasty.  Anyway. my algebra usually has errors in it.


>Note that we've introduced some extra roots.  In the final sqrt, a negative
>is easily understood as the `mirror' point on ce to the left of abc.  But
>what of the � from the quadratic formula?

My opinion is that the problem is in general undetermined.  Visualize the 
triangle as you let x slide from 0 to infinity, keeping A and B constant.  The
angle G starts at 0, increases to some maximum, and then decreases to zero.  
For a given angle G, there will be two solutions provided the given G is below
the maximum.  At the maximum, the discriminant is zero, so the two answers are
the same.

So you could ask another question: what is the value of the angle G which 
allows you to compute a unique value for X?