1266. The mediant of two fractions

    A very common error with fractional sums is

    Although sometimes (or frequently) there are surprises: the mistake is not a mistake...

    We are talking about the mediant of two fractions... I'm using a heart to denote it:

   ( The mediant of  3/8  and  2/7  is  5/15 .)

    If the denominators are positive, the mediant of two fractions is between both of them on the real number line:

    The mediant of two fractions is a weighted arithmetic mean... The positive denominators are the weights...

a/b  ♡  c/d  =  (a+c) / (b+d)  =  (a/b · b + c/d · d) / (b+d)

    Mediants are related to the slope of the vector sum...

    Mediants appear in the Farey sequences... For each natural number  n  we have a Farey sequence consinsting of the irredutible fractions  p/q , such that  0≤p≤q≤n , arranged in order of increasing size. For  n = 6  the Farey sequence are

0/1    1/6    1/5    1/4    1/3    2/5    1/2    3/5    2/3    3/4    4/5    5/6    1/1

    It is easy to test that here every fraction (except the first one and the last one) is equivalent to the mediant of its neighbors. (This happens in all the Farey sequences.) In this, the mediant of the 8th fraction and the 10th fraction is equivalent to the 9th fraction:

3/5  ♡  3/4  =  6/9  =  2/3

    Mediants also appear with the Ford circles... Given an irreductible fraction  p/q , its Ford circle is on the fist quadrant, is tangent to the x-axis at the point  p/q , and its diameter measures  1/q² . Two different Ford circles are either disjoint or tangent to each other. If three Ford circles are tangent to one another, and if the irreductible fractions associated to the two largest ones are  a/b  and  c/d , then the irreductible fraction associated to the smallest one is  a/b  ♡  c/d . (This is a beautiful exercise...)

    This last drawing reminded Nina Guindilla of the third Japanese theorem. (A theorem of Mikami and Kobayashi's. The first and the second Japanese theorems are very beautiful too...)

    Dear Teacher:

    If three circles and a straight line are tangent to one another and if the radii of the circles are  r ,  s  and  t  (r<s≤t), then

1/√r  =  1/√s + 1/√t

    Pepe Chapuzas gave a proof:

    Dear Teacher:
    We have three right angled triangles and Pythagoras' theorem.
    The three sides of the blue triangle are

t + s

t – s

√ [ (t+s)² – (t–s)² ]  =  2√(ts)

    The three sides of the green triangle are

t + r

t – r

√ [ (t+r)² – (t–r)² ]  =  2√(tr)

    The three sides of the violet triangle are

s + r

s – r

√ [ (s+r)² – (s–t)² ]  =  2√(sr)

    So,

 2√(ts)  =   2√(tr) + 2√(sr)

    And dividing by  2√(tsr)

1/√r  =  1/√s + 1/√t    Q.E.D.

    (The proof is valid when  t = s .)
    Finally, Pepe also solved the exercise on the Ford circles...

    Dear Teacher:
    I may assume that b<d. (If you want, you may do it with b>d. It is not possible b=d, is it?)
    The diameters of the two largest circles are

2t  =  1/b²

2s  =  1/d²

    We just saw that

c/d – a/b  =  2√(ts)  =  √(2t·2s)  =  1/(bd)

    So, the determinant

bc – ad  =  1

    (If  bc – ad  =  0 , then the circles would be coincident.)

    (If  bc – ad  >  1 , then the circles would be disjoint.)

    We can do the determinant test with the mediant to prove the third circle to be tangent to the other two.

a(b+d) – b(a+c)  =  ab + ac – ba – bd  =  1

(a+c)d – (b+d)c  =  ad + cd – bc – dc  =  1

    So, the mediant  a/b  ♡  c/d  =  (a+c) / (b+d)  is an irreductible fraction. The diameter of its Ford circle must be

2r  =  1/(b+d)²

    This is true because according to the third Japanese theorem we'd have

1/√(2r)  =  1/√(2s) + 1/√(2t)  =  b + d