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Trisected Hypotenuse of a Triangle


Date: 12/20/98 at 14:30:43
From: Will Nuse
Subject: Trisected hypotenuse of a triangle

Here's the problem:

In right triangle ABC, with C as the right angle, it is known that 
AD = DE = EB (the three segments of the hypotenuse) and that D and E 
are points on hypotenuse AB. It is also known that CD = 7 and that 
CE = 9.

What is the length of AB (the hypotenuse)?       

I tried to work this problem using the law of cosines because at first 
I thought that line segments CD and CE trisected the 90 degrees into 
30 degrees. However, my trig teacher showed me that there is more than 
one way to draw this problem, and therefore it would be almost 
impossible to get all three angles ACD, DCE, and ECB equal to 30 
degrees.

I'm stumped, along with my entire trig class.  Can you help me out?

Thank you very much.  I appreciate your help.

Will Nuse


Date: 12/22/98 at 17:05:13
From: Doctor Floor
Subject: Re: Trisected hypotenuse of a triangle

Hi Will,

Thanks for your question!

For naming, let a = BC, b = BC, and c = AB. Let's take a look at a 
sketch of the situation:

   B
   |\
   F E
   |  \
   G   D        I added the points F and G that trisect BC,
   |    \
   C-H-I-A      and H and I that trisect AC.


It is not difficult to see that, for instance, FE = CH. But that means 
that CE^2 = EF^2 + FC^2, and thus
         
          81 = ((1/3)a)^2 + ((2/3)b)^2

          81 = (1/9)a^2   + (4/9)b^2     [multiply by 9]

          a^2 + 4b^2 = 729  [1]

In the same way DC^2 = CI^2 + ID^2, and thus

          49 = ((2/3)a)^2 + ((1/3)b)^2

          49 = (4/9)a^2   + (1/9)b^2     [multiply by 9]

          4a^2 + b^2 = 441 [2]

Adding [1] and [2] gives that 

          5a^2 + 5b^2 = 1170, so then a^2 + b^2 = 234. 

So the hypotenuse must be sqrt(234).

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   

Date: 06/03/02 at 10:26:14
From: Doctor Ian
Subject: Re: Trisected hypotenuse of a triangle

Grant Woodward suggests another approach.  Here is a drawing of the triangle:

  

Start with a few simple observations:

  cosA = b/3x      
 
  cosB = a/3x

  a^2 + b^2 = 9x^2

Apply the Law of Cosines to triangles BCE and ACD, respectively:

  9^2 = x^2 + a^2 - 2ax cosB

  7^2 = x^2 + b^2 - 2bx cosA

Adding these two equations gives

  9^2 + 7^2 = 2x^2 + a^2 + b^2 - 2ax cosB - 2bx cosA

            = 2x^2 + a^2 + b^2 - 2x(a cosB + b cosA)

We noted earlier that a^2 + b^2 = 9x^2, so substituting gives

            = 2x^2 + 9x^2 - 2x(a cosB + b cosA)

            = 11x^2 - 2x(a cosB + b cosA)

Substituting the values we found earlier for the cosines gives

            = 11x^2 - 2x(a * a/(3x) + b * b/(3x))

            = 11x^2 - 2x((a^2 + b^2)/(3x))

Substituting again for a^2 + b^2 gives

            = 11x^2 - 2x(9x^2/(3x))

            = 11x^2 - 6x^2

            = 5x^2

So we end up with

  9^2 + 7^2 = 5x^2

                    9^2 + 7^2 
          x = sqrt( --------- )
                        5

            = sqrt(26)

So the length of the hypotenuse must be 3 sqrt(26). 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons
High School Trigonometry

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