martes, 31 de enero de 2017

1031. The eye of Horus

    Nina Guindilla's notebook is a treasure. When Nina discovers some strange relationship she writes it, or rather, draws it in her notebook. Look at this strange drawing...

    I showed it to Pepe Chapuzas... and he told me that he didn't know if he looked at Horus' eye or if it was Horus who looked at him... Probably both.

    Dear Teacher:
    The various parts of an eye of Horus represented unit fractions in Egypt: more precisely, some powers of 1/2. Also music note values are unit fractions of the whole note: coincidentally, the same powers of 1/2...

lunes, 30 de enero de 2017

1002. Ruffini everywhere

    Pepe Chapuzas has calculated a limit... Is his procedure legal?

    Dear Teacher:

    I substituted  x  by    but I got an indeterminate form. Then I remembered the polynomial remainder theorem and used Ruffini's rule. Is it right? 
    Solution: 

viernes, 27 de enero de 2017

1041. Pythagorean triangles

    A Pythagorean triple  (a, b, c)  consists of three natural numbers  a < b < c  such that

a2 + b2 = c2

    A triangle is Pythagorean whether its three sides,  a, b, c,  form a Pythagorean triple, that is, a right angled triangle whose sides mesure natural numbers (the length unit doesn't matter). The most famous Pythagorean triangle is the Egyptian triangle, whose sides measure 3, 4 and 5. The Egyptian rope (with 12 knots) is a tool to make right angles...

    Pepe Chapuzas asked if there was a plane that formed 3 Pythagorean triangles with the planes x=0, y=0 and z=0. I proposed this question as a challenge. The next day Nina Guindilla said that she had found out one, but only gave as clues the hypotenuses of the three Pythagorean triangles formed:
    Pepe Chapuzas wrote this system of equations:

E:    x2 + y2 = 2672 = 71289
E:    x2 + z2 = 1252 = 15625
E:    y2 + z2 = 2442 = 59536
    And solved it:
E1 + E2 – E:    2x2 = 27378 ;   x = 117
E1 + E3 – E:    2y2 = 115200 ;   y = 240
E2 + E3 – E:    2x2 = 3872 ;   z = 44

    The plane was
x/117 + y/240 + z/44 = 1

1108. Right or wrong?




1208. Eccentricities

QUESTION: Which parabola has greater eccentricity? Which is more eccentric?


( Eccentric! )                             ( You more! )

ANSWER: Every parabola has eccentricity 1. They are all just as eccentric, ha-ha.

jueves, 26 de enero de 2017

1254. A set square

    The lenght of a leg of a 45 degree set square (isosceles right angled triangle) equals the sum of the inradius and the circumradius...

1249. Pythagoras' theorem

    Dear Teacher:
    It's easy to prove the Pythagorean theorem for a 45 degree set square with two tangram puzzles...

    I told Pepe Chapuzas that it was important to particularize but also generalize: there were other versions of this famous theorem:

    The sum of the areas of the two squares on the legs equals the area of the square on the hypotenuse.
    The sum of the areas of the two semicircles on the legs equals the area of the semicircle on the hypotenuse.
    The sum or the areas of the two regular pentagons on the legs equals the area of the regular pentagon on the hypotenuse.

    This happens with other figures...

miércoles, 25 de enero de 2017

1060. Vertices...

    Let  A (1, 3, 4) ,  B (3, 1, 3)  and  C (5, 2, 5)  be 3 of the 8 vertices of a cube. Find out the other 5 vertices knowing that they all are in the 1st octant (their coordinates are positive).
    Nina Guindilla wondered about the position of the 3 given vertices...
    Dear Teacher:
    The 3 given vertices can be vertices of an isosceles triangle, or a scalene triangle, or an equilateral triangle. According to the case, the problem is different...

|AB| = |(2, −2, −1)| = 3
|BC| = |(2, 1, 2)| = 3
|AC| = |(4, −1, 1)| = 18

    It's the 1st case! They are 3 of the 4 vertices of a face... The 4th vertex D can be calculated easily:
OD = OA + OC OB = (3, 4, 6)
D (3, 4, 6) 

    Now I'm calculating the vertices, E, F, G, H, of the opposite face:

AE = BF = CG = DH = BC x AB / 3 = (3, 6, −6) / 3 = (1, 2, −2)

    So
OE = (1, 3, 4) + (1, 2, −2) = (2, 5, 2)
OF = (3, 1, 3) + (1, 2, −2) = (4, 3, 1)
OG = (5, 2, 5) + (1, 2, −2) = (6, 4, 3)
OH = (3, 4, 6) + (1, 2, −2) = (4, 6, 4)
    So
E (2, 5, 2)
F (4, 3, 1)
G (6, 4, 3)
H (4, 6, 4)