1411. Angles and fish

    Which angle is the greatest one? I admit that my question had a certain amount of malice and was overcome by the subtleness of Nina Guindilla's response...

    Dear Teacher:
    The big fish eats the little fish but angles are not fish...

    The exercise on angles was easy:

    Nina Guindilla commented that it was very easy from the formulas for the angles outside and inside a circle.

alpha + beta  =  gamma/2 + delta/2 + gamma/2 – delta/2  =  gamma

alpha – beta  =  gamma/2 + delta/2 – gamma/2 + delta/2  =  delta

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

1288. More about side midpoints

    Given a triangle, its inellipses are its inscribed ellipses, that is, the ellipses tangent to the three sides of the triangle. The Steiner inellipse of the triangle is the unique inellipse passing through the three side midpoints of the triangle, that is, the ellipse tangent to the three sides at their midpoints. (Furthermore, the Steiner inellipse has the largest area of any inellipse.)

    Imagine a triangle on the complex plane... Marden's theorem states that if the three vertices of that triangle are the zeroes of a cubic polynomial, then the zeroes of the derivative of the polynomial are the foci of the Steiner inellipse... 
    Assuming Marden's theorem, can you prove that the center  (k)  of the Steiner inellipse, the barycenter  (g)  of the given triangle and the zero  (h)  of the second derivative of the cubic polynomial are the same point? Pepe Chapuzas was able...

    Dear Teacher:
    How beautiful is this theorem!
    If  a, b, c  are the vertices of the triangle, then its barycenter is

g  =  (a+b+c) / 3

    And the cubic polynomial and its derivatives are

P (z)  =  (z–a)·(z–b)·(z–c) · d     with    d ≠ 0

P (z)  =  ( z³ – (a+b+c)·z² + (ab+ac+bc)·z – abc ) · d

P ' (z)  =  ( 3·z² – 2·(a+b+c)·z + ab + ac + bc ) · d
P '' (z)  =  ( 6·z – 2·(a+b+c) ) · d

    According to Marden's theorem, the foci of the Steiner inellipse are the zeroes of  P ' (z) . If

3·z² – 2·(a+b+c)·z + ab + ac + bc  =  0

and  Δ Δ Δ is the discriminant of this equation, its zeroes (the foci) are

f₁  =  ( 2·(a+b+c) + √Δ ) / 6

f₂  =  ( 2·(a+b+c) – √Δ ) / 6

    The center of the Steiner inellipse shall be the midpoint of the segment between the foci:

k  =  ( f₁ + f₂ ) / 2  =  (a+b+c) / 3

    Finally, if  h  is the zero of  P '' (z)  then...

P '' (h)  =  ( 6·h – 2·(a+b+c) ) · d  =  0
6·h – 2·(a+b+c)  =  0

h  =  (a+b+c) / 3 

g  =  k  =  h    Q.E.D.