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.

1261. Side midpoints...

    Varignon's theorem states that the side midpoints of any quadrilateral are vertices of a parallelogram, and the area of this parallelogram is half the area of that quadrilateral...

    Nina Guindilla explained it:

     Dear Teacher:

    If A, B, C, D are the vertices of the quadrilateral and E, F, G, H are its side midpoints, and if O is the origin, then the vector

EF  =  OF − OE  =  (OC+OB)/2 − (OA+OB)/2  = 

=  OC/2 − OA/2  =

=  (OC+OD)/2 − (OA+OD)/2  =  OG − OH  =  HG

    That is, EF and HG are equipollent and EFGH is a parallelogram.

    Pepe Chapuzas did it differently:

    Dear Teacher:

    AC and BF are the diagonals of the quadrilateral and we have the following pairs of similar triangles:

BEF and BAC

DAC and DHG

AEH and ABD

CBD and CFG

    So, we have the parallel segments

EF || AC || HG

EH || BD || FG

    So, EFGH is a parallelogram.
    The area of the parallelogram is obviously half the area of the quadrilateral if this is convex. If not, the following drawing helps to understand it:

    Furthermore, if the initial quadrilateral is a kite, a dart or a rhombus then the parallelogram is a rectangle. If the initial quadrilateral is an isosceles trapezoid or a rectangle then the parallelogram is a rhombus... If the quadrilateral is a square then the parallelogram is another square...

1274. Angle bisectors...

    Dear Teacher:

    ¿Do the angle bisectors of a quadrilateral form a cyclic quadrilateral?

    I answered Pepe Chapuzas that not always, because the angle bisectors of a square formed... a point! But if the angle bisectors did form a quadrilateral then this was really a cyclic quadrilateral!

    Nina Guindilla rushed to do a proof...

    Dear Teacher:

    Since

2α + 2β + 2γ + 2δ  =  360°

then

α + β + γ + δ  =  180°

    The upper blue angle is

180° − β − γ

and the lower blue angle is

180° − α − δ

and the sum of both blue angles sall be

360° − α − β − γ − δ  =  360° − 180°  =  180°

    That is, the quadrilateral formed by the angle bisectors is cyclic.

    Furthermore, the angle bisectors of squares, rhombi, kites and darts do not form any quadrilateral (but a point). Here are some examples of cyclic quadrilaterals formed by angle bisectors: 

    A square from a rectangle.

    A rectangle from a rhomboid.

    A kite from an isosceles trapezoid.

    An isosceles trapezoid from a...

1327. Even more special quadrilaterals

    If a polygon is inscriptible in a circle and circumscriptible about another circle then is a bicentric polygon. The radii of these circles are called inradius and circumradius, and their centers are called incenter and circumcenter... All the triangles are bicentric... If two circles allow a bicentric polygon between them then allow an infinity, all of them with the same number of sides (Poncelet's porism). 

    What condition must fulfill the inradius, the circumradius, the incenter and the circumcenter so that two circles allow a bicentric polygon between them? Nina Guindilla answered:

    Dear Teacher:
    That condition depends on the number of sides of the polygon. I've searched and I've found two theorems. Euler's theorem in Geometry provides the condition for triangles and Fuss's theorem provides the condition for bicentric quadrilaterals:

    If  r  is the inradius,  R  is the circumradius and  d  is the distance between the incenter and the circumcenter then...

Euler's theorem:     1/r = 1/(R+d) + 1/(R–d)

Fuss's theorem:     1/r² = 1/(R+d)² + 1/(R–d)²

1275. More special quadrilaterals

    If a quadrilateral is inscribed in a circle and circumscribed about another circle then is a bicentric quadrilateral.
    If  a ,  b ,  c  and  d  are the four sides of a bicentric quadrilateral then its area measures...

    Pepe Chapuzas reasoned as follows:

    Dear Teacher:
    Let  s  be the semiperimeter. Since the quadrilateral is circumscriptible...

a + c  =  b + d  =  s      (Pitot's theorem)

    Since the quadrilateral is inscriptible... 

Area  =  √ [(s−a)(s−b)(s−c)(s−d)]    (Brahmagupta's formula)

so

Area  =  √ (c·d·a·b)

1373. Special quadrilaterals

    A circumscriptible (or tangential) quadrilateral has the following property: its opposite sides add up to the same length (the semiperimeter).

    An inscriptible (or cyclic) quadrilateral has the following property: its opposite angles add up to the same width (a streight angle).

    Nina Guindilla proved both properties...

    Dear Teacher:

a + c = e + f + g + h

b + d = e + f + g + h

2α + 2γ = 360°
α + γ = 180°

and as well

 β + δ = 180°

1296. Oblong and oblate

    A spheroid is an ellipsoid having two equal diameters. If we rotate an ellipse about one of its axes, we get a spheroid: an oblong spheroid if it is about the major axis and an oblate spheroid if it is about the minor axis... The volume of the oblate spheroid is greater than that of the oblong one...

    Pepe Chapuzas asked...
    
    Dear Teacher:
    If the eccentricity of an ellipse is  e , what is the ratio between the volumes of both spheroids?

    And Nina Guindilla answered...

    Dear Teacher:
    Let  a  and  b  be the major semiaxis and the minor semiaxis of the ellipse.
    Let  v  and  V  be the volumes of the oblong spheroid and that of the oblate one.

v  =  4πab²/3

V  =  4πa²b/3

v / V  =  b / a  =  √ (1−e²)

1183. A triangular distribution

    A triangular distribution is a continuous probability distribution whose density function graph forms a triangle with the abscissa axis...

    Dear Teacher:
    The distribution mean is the abscissa of the triangle barycenter.
    The distribution mode is the abscissa of the triangle orthocenter.
    The distribution midrang is the abscissa of the triangle circumcenter.

    Pepe Chapuzas is right, of course...

1267. Apollonius' theorem

    Classical Geometry: unparalleled beauty...

    Apollonius' theorem states that the mean of the areas of the squares on any two sides of any triangle equals the sum of the area of the square on half the third side and the area of the square of the median bisecting the third side... See below:

 

    Pepe Chapuzas proved the theorem:

    Dear Teacher:
    The cosine law states that

a²  =  m² + c²/4 − m · c · cos θ

b²  =  m² + c²/4 − m · c · cos θ'  =  m² + c²/4 + m · c · cos θ

(θ and θ' are suplementary)

    So, the mean of  a²  and  b²  is

(a² + b²) / 2  =  m² + c²/4

    Mr. López, when the triangle is isosceles, Apollonius' theorem becomes Pithagoras' theorem, doesn't it?

1313. Pedal triangles

    Nina Guindilla has brought a beautiful proposition about pedal triangles. First, a definition:

    Dear Teacher:
    Given a triangle and a point, the pedal triangle is obtained by projecting the given point onto the sides of the given triangle: the vertices of the pedal triangle are the feet of the perpendiculars from the given point to the sides of the given triangle. (Sometimes the pedal triangle degenerates and collapses to a line...)

    Proposition. The pedal triangle of the pedal triangle of the pedal triangle is similar to the initial triangle (if no one degenerates).
    
    Nina Guindilla proved the proposition with an inner point to the given triangle.

    Dear Teacher:
    The three perpendicular segments from the point to the sides of the initial triangle divide this into three cyclic quadrilaterals (inscriptible in circles). Note now that angles with the same color are equal because intercept the same circle arc... 
    And observe below the dance of conguent angles: the small triangle is similar to the great one... QED.

1121. Mirage operations

    Dear Teacher:
    I've called mirage operation to any mathematical operation whose result doesn't change when the digits of its operands are reversed. So,  31 · 26  and  48 + 95  are mirage operations because

31 · 26  =  62 · 13    (=  806) 

48 + 95  =  59 + 84    (=  143)

    Who can find mirage multiplications and mirage additions?

    I wanted my pupils to solve this challenge of Pepe Chapuzas's... It was his classmate Nina Guindilla, again, who did it... She matched operands with lines...