1144. L'Hôpital's limits

    L'Hôpital's rule has its limits and its limitations... Watch this example:

    Pepe Chapuzas calculated this limit without L'Hôpital's help...

1053. Laws

THE COSINE LAW

THE SINE LAW

1281. Equilibrist squares

    Five acrobats perform balancing feats. They are the equilibrist squares... Now, they realize that the square A and the triangle A have the same area... Here's Nina Guindilla's proof:

    Dear Teacher:
    I may take the side of the square A as length unit, may I not? So, the area of the square A is 1, right? I have to prove that the area of the triangle A is 1 too. Look at the angles α, β, σ, τ, and sides s, t, below...

Area  =

=  s · t · sinβ / 2  =

=  s · t · sin(90°−σ−τ) / 2  =

=  s · t · cos(σ+τ) / 2  =

=  s · t · (cosσ · cosτ − sinσ · sinτ ) / 2  =

=  (1 · s · cosσ) · (1 · t · cosτ) / 2  −  (s · sinσ) · (t · sinτ) / 2  =

    I apply the law of cosines and the law of sines...

=  (1² + s² − sin²α) · (1² + t² − cos²α) / 8  −  sinα · sin(90°+α) · cosα · sin(180°−α) / 2  =

=  (1+1²−2·sinα·cos(90°+α))·(1+1²−2·cosα·cos(180°−α))/8 − sinα·cosα·cosα·sinα/2  =

=  (2 + 2 · sinα · sinα) · (2 + 2 · cosα · cosα) / 8  −  sin²α · cos²α / 2  =

=  (1 + 1 · sin²α) · (1 + 1 · cos²α) / 2  −  sin²α · cos²α / 2  =

=  (1 + sin²α + cos²α +  sin²α · cos²α −  sin²α · cos²α) / 2  =

=  (1 + 1 + 0) / 2  =

=  1

1001. Divide and conquer

Right or wrong?

1128. Alpha. Beta. Gamma. Delta.

    The four internal angles of a quadrilateral, measured in degrees, are natural numbers and are in geometric progression: alpha, beta, gamma and delta. Calculate alpha, beta and delta when gamma is...

    a) ... 81°
    b) ... 90°
    c) ... 96°

    Pepe Chapuzas solved this problem (these three problems). He found the natural solutions and the other solutions...

    Dear Teacher:
    The sum of the four internal angles of a quadrilateral equals 360°

alpha + beta + gamma + delta = 360°

    If  R  is the common ratio of the geometric progression...

gamma/R² + gamma/R + gamma + gamma·R = 360

gamma + gamma·R + (gamma−360)·R² + gamma·R³ = 0

    Then...

    a) 

81R³ − 279R² + 81R + 81 = 0

9R³ − 31R² + 9R + 9 = 0

R = 3
alpha = 81°/9 = 9°

beta = 81°/3 = 27°

delta = 3 · 81° = 243°

    b)

90R³ − 270R² + 90R + 90 = 0

R³ − 3R² + R + 1 = 0

R = 1
alpha = beta = gamma = delta = 90°

    The progression is constant and the quadrilateral is a square. 

    c)

96R³ − 264R² + 96R + 96 = 0

4R³ − 11R² + 4R + 4 = 0

R = 2
alpha = 96°/4 = 24°

beta = 96°/2 = 48°

delta = 2 · 96° = 192°

    The other solutions are irrational (even negative).

    a)

    b) 

    c)

1007. The method of the shoes

    Shall I write the name of the method I've chosen?

    Pepe Chapuzas asked this during a systems test. I suposed he meant elimination, substitution or equalization... but not... In his test there's no system...

    Problem:

    In a farm there were rabbits and chickens. There were 62 heads and 168 legs. ¿How many rabbits and haw many chicken were there?

    Pepe Chapuzas wrote this:

    Method of the shoes: The farmer bought a pair of shoes to each animal...
    There were  62  heads, that is,  62  animals.
    There were  62 · 2 = 124  shoes.
    There were  168 − 124 = 44  legs without shoes, that is, 44 rabbit hands.
    There were  44 : 2 = 22  rabbits.
    There were  62 − 22 = 40  chickens.

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

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 (180º−α−β)  =

=  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 (180º−α −β)  =

=  tg α · tg β · tg γ

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

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 m² , 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°  =  √5 − 1

csc 60°  =  √12 / 3

csc 75°  =  √6 − √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  =  √5 / √π .

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

π · (√5 / √π)²  =  5 m² .

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...

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 each corner of the blackboard... Is it true that

a² + c²  =  b²  + d²   ??

    Nina Guindilla (and Pythagoras) proved this equality:

    Dear Teacher:
    If  e ,  f ,  g  and  h  are the four distances from that point to each side of the blackboard... then...

a²  =  e²  + f ²
b²  =  f ²  + g²
c²  =  g²  + h²
d²  =  h²  + e²

     And therefore...

a² + c²  =  e²  + f ² + g²  + h²
b² + d²  =  e²  + f ² + g²  + h²

    Q.E.D.