jueves, 19 de enero de 2017

1286. No more data

    I was dictating a problem. I was giving the data of a system with 3 equations in 3 unknowns... A farmer had to buy pigs, ewes and hens. Datum for the 1st equation: he bought 100 animals. Datum for the 2nd equation: he spent 10 pesos per pig, 5 pesos per ewe, 1/2 peso per hen, 100 pesos in total... Then Pepe Chapuzas exclaimed: No more data, please!

    Dear Teacher:
    If the farmer bought  x  pigs,  y  ewes and  z  hens, the equations are...

x + y + z  =  100
10x + 5y + .5z  =  100
    And
x + y  =  100 − z
10x + 5y  =  100 − .5z
    And
x  =  .9z  80
y  =  180 − 1.9z

    Then  z  must be a multiple of 10 and

80/.9  <  z  <  180/1.9
88.88  <  z  < 94.74
    So
z  =  90 hens
x  =  81 − 10  =  1 pig
y  =  180 − 171  =  9 ewes

miércoles, 18 de enero de 2017

1247. When the sum equals the product

    In a certain exercise the sum of the tangents of the angles of a triangle was given, and the product of these three tangents was requested... Many students said they needed more data but Pepe Chapuzas calculated it.

    Dear Teacher:
    The sum equals the product...

tg α + tg β + tg γ  =
=  tg α + tg β + tg (180ºαβ)  =
=  tg α + tg β – tg (α+β)  =
=  tg α + tg β − (tg α + tg β) / (1 − tg α · tg β)  =
=  [ tg α + tg β – (tg α + tg β) · tg α · tg β − tg α − tg β ] / (1 − tg α · tg β)  =
=  − (tg α + tg β) · tg α · tg β / (1 − tg α · tg β)  =
=  − tg α · tg β · tg (α+β)  =
=  tg α · tg β · tg (180ºα β)  =
=  tg α · tg β · tg γ


... but not always, because the right angles...

martes, 17 de enero de 2017

1100. A pair of hyperbolae

    Two hyperbolae share their assymptotes in such a way that the two assymptotes separate the four hyperbola branches... Let  E  and  F  be the eccentricities of the hyperbolae. Prove that  

E2 + F2  =  E2 · F2 

(The two hyperbolae are not necessarily conjugated.)

    Yoyes Canasta did it easily...

    Dear Teacher:
    These two hyperbolae also share their axes. Both axes are perpendicular to each other and are bisectors of the angles between the assymptotes. Then...

E = sec 
α   and   F = sec β

    Since  
α  and  β  are complementary...

F  =  csc α
    And...
E2 + F2  =
=  sec2α + csc2α  =
=  1/cos2α + 1/sin2α  =
=  (sin2α + cos2α) / (cos2α · sin2α)  =
=  1/cos2α · 1/sin2α  =
=  E2 · F2

1193. To fly!

    Teacher, thanks for teaching me how to fly.

    I gave him back the compliment... I tell Pepe Chapuzas that it was very easy to teach to fly to children who were born with wings...
    Pepe said goodbye with a gift: a Geometry problem...

    Dear Teacher:
    The orange area is 10 m2, and the green area is...
    Nina Guindilla was born with wings too...
  
    Dear Teacher:
    There are 10 orange circles, so, one circle area is  1 m, and its radius measures  1/π .
    The radius of the blue decagon measures  csc 18° / π . I have no calculator but I have here a chart of cosecants...
csc 15°  =  6 + 2
csc 18°  =  5 + 1
csc 30°  =  2
csc 45°  =  2
csc 54°  =  − 1
csc 60°  =  √12 / 3
csc 75°  =  − 2
csc 90°  =  1

    So, the radius of the blue decagon measures

(5 + 1) / π .

    And the radius of the green circle measures

(√5 + 1) / π  −  1 / π  =  π .

    And the green area is... half the orange area

π · (π)2  =  5 m2 .

lunes, 16 de enero de 2017

1140. The windmill blades

    Pepe Chapuzas had drawn something like the blades of a windmill...

    The callenge was to calculate the sum of the blue angles... Chicho Madeja calculated it...

    Dear Teacher:
BLUE + RED = 5 · 180° = 900°
RED + GREEN = 360°
RED = GREEN
RED = 180°
BLUE = 900°  RED = 900°  180° = 720°

    

1084. Pingala's triangle

    "A lion killed Panini; an elephant killed Jaimini; a crocodile killed Pingala... What do senseless beasts, overcome with fury, care for intellectual virtues?" (Pachatantra.)

    Dear Teacher:
    Pascal's triangle is referred to as Tartaglia's triangle in Italy, Yang Hui's triangle in China and Khayyam's triangle in Iran, because these four mathematicians independently discovered it. However, long before them, Pingala, one of the three wise brethrin killed by beasts, already knew it... in India... May I call it Pingala's triangle?

    Pepe Chapuzas likes old stories... We had seen that in Pascal's (or Pingala's) triangle we could find the triangular numbers and the tetrahedral numbers... Pepe made a visual demonstration...

domingo, 15 de enero de 2017

1137. A dot on the blackboard

    Pepe Chapuzas was erasing the blackboard but there was a white dot that cannot be deleted...

    Dear Teacher:
    If  a ,  b ,  c  and  d  are the four distances from this point to the four corners of the blackboard... Is it true that
a+ c2  =  b + d2   ??

    Nina Guindilla (and Pythagoras) proved this equality:

    Dear Teacher:
    If  e ,  f ,  g  and  h  are the four distances from that point to the four sides of the blackboard... then...
a2  =  e + f 2
b2  =  f 2  + g2
c2  =  g + h2
d2  =  h + e2

     And therefore...
a+ c2  =  e + f 2 + g + h2
b+ d2  =  e + f 2 + g + h2

    Q.E.D.

1237. A radical equation with cube roots

    This equation was solved by a pupil of mine... Pepe Chapuzas again!


3√(1729 − x)  =  19 − 3√x
cubing...
1729 − x  =  193 − 3 · 192 · 3√x + 3 · 19 · 3√x2 − x
57 · 3√x2 − 1083 · 3√x + 5130  =  0
quadratic formula...
3√x  =  (1083 ± √3249) / 114 =  (1083 ± 57/ 114
x  =  (1083 ± 57)3 / 1143
x1  =  1000
x2  =  729
 checking...
3√(1729 − 1000) + 3√1000  =  3√729 + 3√1000  =  9 + 10  =  19
3√(1729 − 729) + 3√729  =  3√1000 + 3√729  =  10 + 9  =  19 

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...

(db) / (cb)  =  (ec) / (ac)  =  (fa) / (ba)  =  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):

α   and   2 β−60°
β   and   2 γ−60°
γ   and   2 α−60°
    The midpoints are...
a  =  1 α + 1 β−60°  =  α + 1 β · 1 300°
b  =  1 β + 1 γ−60°  =  β + 1 γ · 1 300°
c  =  1 γ + 1 α−60°  =  γ + 1 α · 1 300°
    Let me calculate...
(ab) / (cb)  =
=  (α + 1 β · 1 300° − β − 1 γ · 1 300°) / (γ + 1 α · 1 300° − β − 1 γ · 1 300°)  =
=  (α + 1 β · 1 240° + 1 γ · 1 120°(1 α · 1 300° + 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.