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Non-Congruent Triangles

Date: 12/12/2001 at 21:41:47
From: Z
Subject: Proof

Construct and prove that there can be two non-congruent triangles in 
which five parts of one triangle are equal to five parts of another.

Date: 12/13/2001 at 12:58:18
From: Doctor Peterson
Subject: Re: Proof


The five parts will be all but one of the three sides and three 
angles. If two angles of each triangle equal two in the other, then 
the third will be the same in both triangles also (since their sum is 
180 degrees); so we can't have only two angles congruent, and the five 
parts must be two sides and all three angles.

If all three angles are the same (in some order), the triangles will 
be similar. So all you need is to find a way in which two (non-
corresponding) side lengths are the same in the two triangles. If 
corresponding sides were congruent (that is, a pair of sides for which 
the angle between them was also the same in both triangles), then the 
triangles would be congruent by SAS.

So suppose the triangles, placed with congruent angles in 
corresponding places, look like this:

          +           c /  \
      b /  \ a        /     \b
      /     \       /        \
    +--------+    +-----------+
         c              d

Here both triangles have a side b and a side c, but the third side is 
a in one and d in the other. We know the triangles are similar; can 
you find a set of lengths that work?

- Doctor Peterson, The Math Forum   

Date: 12/13/2001 at 13:36:25
From: Doctor Rob
Subject: Re: Proof

Thanks for writing to Ask Dr. Math.

If two triangles have five parts equal, then they have at least two 
angles equal. That implies that they have all three angles equal, 
since the sum of measures of the three angles of a triangle is 180 
degrees. (How does this imply that?)  Let the triangles be ABC and 
A'B'C'. Label the vertices so that AB = B'C' < BC = A'C'. (Why is this 
possible?) We know that <A = <A', <B = <B', and <C = <C'. That means 
that the triangles are similar. (Why?) That implies that corresponding 
sides are in proportion:

   AB/A'B' = BC/B'C' = AC/A'C',

and we let this common ratio be called r, where r > 1. (Why?) Thus 
AB = r*A'B', BC = r*B'C', and AC = r*A'C'. Then

   AC = r*A'C' = r*BC = r^2*B'C' = r^2*AB = r^3*A'B'.

Now the triangle inequality theorem and r > 1 together imply that

   AC < AB + BC,
   r^2*AB < AB + r*AB,
   r^2 < 1 + r,
   1 < r < (1+sqrt[5])/2 = 1.6180339887....

(Why?)  For any value of r satisfying this condition, and any x > 0, 
there is a pair of noncongruent triangles sharing their three angles 
and two side lengths, with side lengths given by the formulas

        A'B' =     x,
   AB = B'C' =   r*x,
   BC = A'C' = r^2*x,
   AC        = r^3*x.

(The four side lengths form a geometric progression.)

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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