1414. Real or imaginary?

    A problem:

    If the three green triangles  ABF ,  BCD  and  CAE  are similar, then... Have the triangles  ABC  and  DEF  the same barycenter  G ?
    Pepe Chapuzas imagined that these real triangles were imaginary, so he could work with complex numbers...

    Dear Teacher:
    Let's consider the complex numbers a, b, c, d, e, f, g associated with the points A, B, C, D, E, F, G... Since the green triangles are similar (same angles, proportional sides), then...

(d−b) / (c−b)  =  (e−c) / (a−c)  =  (f−a) / (b−a)  =  h

    Thusly

d  =  ch − bh + b

e  =  ah − ch + c

f  =  bh − ah + a

    Hence

(d+e+f) / 3  =  (a+b+c) / 3  =  g

    Another problem:

    In a circle of radius  r  we have three equidistant chords (red), and the distance between any two chords is  r  too, then... Are the midpoints of the chords the vertices of an equilateral triangle?

    Pepe Chapuzas solved it too...

    Dear Teacher:

    I have a circle... I can assume that its center is  0  and its radius is  2  to simplify, can I not? And I can assume that the endpoints of the chords are... (in polar coordinates):

2 _α   and   2 _β−60°

2 _β   and   2 _γ−60°

2 _γ   and   2 _α−60°

    The midpoints are...

a  =  1 _α + 1 _β−60°  =  1 _α + 1 _β · 1 _300°

b  =  1 _β + 1 _γ−60°  =  1 _β + 1 _γ · 1 _300°

c  =  1 _γ + 1 _α−60°  =  1 _γ + 1 _α · 1 _300°

    Let me calculate...

(a−b) / (c−b)  =
=  (1 _α + 1 _β · 1 _300° − 1 _β − 1 _γ · 1 _300°) / (1 _γ + 1 _α · 1 _300° − 1 _β − 1 _γ · 1 _300°)  =
=  (1 _α + 1 _β · 1 _240° + 1 _γ · 1 _120°) / (1 _α · 1 _300° + 1 _β · 1 _180° + 1 _γ · 1 _60°)  =
=  1 _60°

    If  A ,  B ,  C  are the corresponding real points to  a ,  b ,  c , then the sides  BA  and  BC  have the same length and the angle between them measures 60°, that is, the triangle ABC is equilateral.

1500. Summing infinite geometric series

1/2 + 1/4 + 1/8 + 1/16 + ... = 1

 1/3 + 1/9 + 1/27 + 1/81 + ... = 1/2

 1/4 + 1/16 + 1/64 + 1/256 + ... = 1/3

1287. Optimized pyramid

    Pepe Chapuzas has proposed an optimization challenge.

    Dear Teacher:
    Calculate the base side and the height of a square pyramid circumscribed on a cube whose edge measures 1 m in order that its lateral area be minimal. 

    This challenge was solved by Tita Batalla...

    Dear Teacher:

    If  x  and  y  are the base side and the height of our pyramid, then  1  and  y−1  are the base side and the height of the little pyramid that is above the cube. The two pyramids are similar, so... 

x = y/(y−1)  and  y = x/(x−1)

    The area we want to optimize can be calculated with Pythagoras' theorem:

Lateral area = 

= (base semiperimeter) · (slant height) =

= 2x · √ [ (x/2)² + y² ] =

= 2x ·  √ [ x²/4 + x²/(x−1)² ] =

= x · √ [ x⁴ − 2x³ + 5x² ] / (x−1)

    It's easier to optimize the function (Lateral area)²:

f(x) =  (x⁶ − 2x⁵ + 5x⁴) / (x−1)²

f ' (x) = [ (6x⁵ − 10x⁴ + 20x³) · (x−1) − 2 · (x⁶ − 2x⁵ + 5x⁴) ] / (x−1)³

f ' (x) = 4x³ · (x³ − 3x² + 5x − 5) / (x−1)³

    Since  x > 1 , I have to solve the equation  x³ − 3x² + 5x − 5 = 0 ... Here Ruffini's rule isn't useful... I need another tool: Cardano's method... If  x = z + 1 then...

(z+1)³ − 3(z+1)² + 5(z+1) − 5 = 0

z³ + 3z² + 3z + 1 − 3z² − 6z − 3 + 5z + 5 − 5 = 0

z³ + 2z − 2 = 0

    If  z = u + v 

(u+v)³ + 2(u+v) − 2 = 0

u³ + v³ + 3uv(u+v) = 2 − 2(u+v)

u³ + v³ = 2

uv = −2/3

u³ v³ = −8/27

    So,  u³  and  v³  are solutions to  

w² − 2w − 8/27 = 0

w = 1 ± √ (1+8/27) = 1 ± √ (35/27)

u = ³√ [ 1 + √ (35/27) ]

v = ³√ [ 1 − √ (35/27) ]

x = 1 + u + v = 1 + ³√ [ 1 + √ (35/27) ] + ³√ [ 1 − √ (35/27) ]

    I go on with a calculator:

x = 1.771 m

y = 2.297 m

    The reader can make the necessary checks...

1421. Distance between two parabolae

    Pepe Chapuzas had drawn two parabolae that didn't cut each other on the blackboard. One was convex and the other concave... It was a challenge...

    Dear Teacher:
    What's the distance between the parabolae if their equations are

f(x)  =  x²/3 + 47/3

g(x)  =  − x²/6 + 10x/3

    Susi Fogata did the calculations...

    Dear Teacher:
    These parabolae don't cut each other because the equation

 x²/3 + 47/3  =  − x²/6 + 10x/3

or better

3x² − 20x + 94  =  0

hasn't any solutions because its discriminant

20² − 4·3·94  =  − 728  <  0

    If  S (s, s²/3 + 47/3)  is a point of the parabola  f  and  T (t, − t²/6 + 10t/3)  is a point of the parabola  g , and are the nearest ones, then the slopes of  f  and  g  at  S  and  T  are equal...  

 f ' (s)  =  g ' (t)
2s/3  =  − t/3 + 10/3
t  =  10 − 2s

    On the other hand, the vector  TS  is perpendicular to  f  at  S ...

(s − t, f(s) − g(t)) · (1, f ' (s))  =  0
(s − t, s²/3 + 47/3 + t²/6 − 10t/3) · (1, 2s/3)  =  0
(s − (10−2s), s²/3 + 47/3 + (10−2s)²/6 − 10(10−2s)/3) · (1, 2s/3)  =  0
(3s − 10, s²/3 + 47/3 + 50/3 − 20s/3 + 2s²/3 − 100/3 + 20s/3) · (1, 2s/3)  =  0
(3s − 10, s² − 1) ·  (1, 2s/3)  =  0
3s − 10 + 2s³/3 − 2s/3  =  0
 2s³ + 7s − 30  =  0

    s = 2  is the only solution because  4² − 4·2·15  =  − 104  <  0 .

    So, we have the points  S (2, 17)  and  T (6, 14)  and the vector  TS = (−4, 3) , hence the distance is  |TS| = √(16+9) = √25 = 5 .

1277. The rainbow calendar

    This is a present from Pepe Chapuzas. It's a calendar for several years, since 2014 until 2030, although it can be extended. 

    Dear Teacher:
    If I want to know what day of the week will be October 13th 2027? First, I look for the color that relates October (X) with 2027 (27): it's green! Then, I look for the day of the week that is related with the 13th (13) by the color green: it's Wednesday (mi). (Spanish miércoles.)

    Leap years appear twice: in one black cell and in one white cell. The black one are reserved for January and February, so, February 6th 2024 will be Tuesday (ma). (Spanish martes.)

 Days of the week in Spanish:

lunes

martes

miércoles

jueves

viernes

sábado

domingo

1043. Seven colors

    One day I was talking about the four-color theorem demonstrated by K. Appel and W. Haken's computers in 1976. I commented that this theorem stated that 4 colors were enough to color a map demanding border regions have different colors. I pointed out that this theorem was true on planet Earth but not on planet Torus which was like a huge doughnut. On Torus there were maps that needed 7 colors.

    I left the subject here in the hope that some student would give some explanation... And it did happen... Pepe Chapuzas will never cease to amaze me! The next day he brought a set of dominoes. Well, there were only 21 tiles because he had removed the 7 doubles...

    Dear Teacher:

    In a rectangular world map of planet Earth, the left side and the right side represent the same meridian. In a rectangular world map of planet Torus, that happens too, but in addition, the top side and the down side represent the same parallel...
    If I arrange the tiles as follows and you imagine that this is a world map of Torus with 7 regions (from 0 to 6) that border to one another... We need 7 different colors...

1169. Parabolae for disoriented people

    We were correcting parabola exercises. In some of them, the parabola axis was vertical, and in others, it was horizontal. This is always an added difficulty for disoriented students... Pepe Chapuzas, who is usually quite oriented, proposed the following challenge...    

    Dear Teacher:

    If we have 2 parabolae, one with horizontal axis and the other with vertical axis, that have 4 intersection points..., then there is a circumference passing through those 4 points.

    My well oriented student Quica Garrucha solved the exercise...

    Dear Teacher:

    I'm taking the axes of the two parabolae as coordinate axes. If  p  is the parameter of the parabola  P  and  q  is the parameter of the parabola  Q , then their equations are:

P : y² − 2px − m² = 0

Q : x² − 2qy − n² = 0

(P passes through (0, ±m) and Q passes through (±n, 0).)

    If we sum both equations we obtain

E : x² + y²− 2px − 2qy − m² − n² = 0

    E is a circumference with center  G (p, q)  and radius  r = √ (p²+q²+m²+n²) . The intersection points A, B, C, D, of the two parabolae satisfy the equations P and Q and therefore also E, that is, the circumference  E  passes through A, B, C and D.

1102. Ellipses between hyperbolae

    I had drawn on the chalkboard the two conjugated equilateral hyperbolae  xy = ± 1  and it was easy to calculate the area of the circle that is tangent to the four hyperbola branches... Then Pepe Chapuzas interrupted me...

    Dear Teacher:

    My intuition tells me that all the ellipses that are tangent to the four branches have the same area. Am I wrong?

    Bravo for the intuition of Pepe! Was he right? Loli Palique answered...

    Dear Teacher:

    Intuition guides me often, but I'm aware that intuition often deceives... The first thing I'm going to do is calculate the area of the circle... Since its center is (0, 0) and the vertices of the hyperbolae are (±1, ±1), so its radius is

√(1²+1²)  =  √2

and its area is

πr²  =  2π . 

    An ellipse with center (0,0) and axes  x = 0  and  y = 0  has this equation:

x²/a² + y²/b² = 1 . 

    Whether the area of that ellipse  πab  were  2π  too, then  b = 2/a . In such case, the intersection  of this ellipse and the hyperbola branch  y = 1/x  on the first quadrant  (x>0, y>0)  shall furfill

x²/a² + a²/4/x²  =  1

x²/a² + a²/4/x²  −  1  =  0

(x/a − a/2/x)²  =  0

x/a  =  a/2/x

x²  =  a²/2

x  =  a/√2

y  =  √2/a

    So, there is only one instersection point, that is, the ellipse is tangent to that branche and, in the same way, to the other branches... 
    Of course, the ellipse and the hyperbola have the same slope in their tangency points...
    Calculating derivatives with the ellipse equation...

(x²/a² + a² y²/4) '  =  2x/a² + a² y y'/2  =  0

y'  =  −4x/a⁴/y  =  −2/a²

    ^{Calculating derivatives with the hyperbola equation...}

(x y) '  =   y + x y'  =  0

y'  =  −y/x  =  −2/a²

    Loli ended by saying...

    Dear Teacher:
    My intuition tells me that this proposition is true for conjugated hyperbolae even if they aren't equilateral.

    Is she right?