Right triangle in a rectangle

 
 
An infinite number of rectangles may be generated, each with a right triangle inscribed as shown:
 
rect-rght-tringl-1
 
such that the lengths of the sides of the various right triangles are integers.
 
AE \; = \; 2 \, m \, n \, (m^2 \; - \; n^2)
ED \; = \; 2 \, m \, n \, (m^2 \; - \; n^2)
DF \; = \; (m^2 \; - \; n^2)^2
FC \; = \; 4 \, m^2 \, n^2 \; - \; (m^2 - n^2)^2
BC \; = \; 4 \, m \, n \, (m^2 \; - \; n^2)
AB \; = \; 4 \, m^2 \, n^2
BE \; = \; 2 \, m \, n \, (m^2 \; + \; n^2)
EF \; = \; m^4 \; - \; n^4
BF \; = \; (m^2 \; + \; n^2)^2

 

                            *****************************************                

 
Right triangle BEF:

BE \; = \; 2 \, m \, n \, (m^2 \; + \; n^2)
EF \; = \; m^4 \; - \; n^4
BF \; = \; (m^2 \; + \; n^2)^2

( \,2 \, m \, n \, (m^2 + n^2) \,)^2 \; + \; (m^4 - n^4)^2 \; = \; ( \,(m^2 + n^2)^2 \,)^2

Perimeter
P \; = \; 2 \, m \, (m+n) \, (m^2+n^2)

Area
A \; = \; m \, n \, (m^2-n^2) \, (m^2+n^2)^2

Inradius
r \; = \; n \, (m-n) \, (m^2+n^2)

 
Rectangle ADCB

AD \; = \; CB \; = \; 4 \, m \, n \, (m^2 - n^2)
DC \; = \; BA \; = \; 4 \, m^2 \, n^2

The diagonal is   4 \, m \, n \,  \sqrt {m^4-m^2 n^2+n^4}

If we want the diagonal to be an integer, then

m^4-m^2 n^2+n^4   needs to be a square

giving us   m = n = 1

 
For the continuation, see Paul’s solution below (in the comment section)

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Geometry and tagged . Bookmark the permalink.

3 Responses to Right triangle in a rectangle

  1. Paul says:

    It can be shown that in the arrangement above the main diagonal (BD or AC) can never be integer using the (m, n) notation. However there are arrangements that do have an integer main diagonal.

    In m, n notation the main diagonal is 4 Sqrt[m^2 n^2 (m^4 – m^2 n^2 + n^4)], solving for this to equal a square gives m = n = 1 to Infinity the value of the square follows 2 t^2 for all t. This means in all those where there are m – n would equate to zero. here are 2 solutions

    Format is first 3 numbers are the middle triangle (BEF)
    next 3 are (EDF), then (AEB) and then (BCF)
    964 is main diagonal for both.

    {507, 676, 845}, {{357, 360, 507},{476, 480, 676},{123, 836, 845}}, 964
    {595, 600, 845}, {{357, 476, 595},{360, 480, 600},{123, 836, 845}}, 964

    Paul.

  2. Paul says:

    However in that arrangement AE is not equal to ED.

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