viernes, 24 de febrero de 2017

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...


jueves, 16 de febrero de 2017

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 furfill 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  =   [(sa)(sb)(sc)(sd)]    (Brahmagupta's formula)
so
Area  =  √ (c·d·a·b)

miércoles, 15 de febrero de 2017

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°

miércoles, 8 de febrero de 2017

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)

martes, 7 de febrero de 2017

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.

lunes, 6 de febrero de 2017

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...

sábado, 4 de febrero de 2017

1295. How many fractions must be added?

    Dear Teacher:
    If  there were  n  fractions and we rationalized all of them then

− √1 + √2 − √2 + √3 − √3 + √4 − √4 + √5 − √5 + √6 ...  =  − 1 + √(n+1)  =  1000
√(n+1)  =  1000 + 1  =  1001
n + 1  =  10012  =  1002001
n  =  1002001 − 1  =  1002000  =  One million two thousand fractions.

viernes, 3 de febrero de 2017

1135. Tangram cat

    Pepe Chapuzas had drawn a silhouette that looked like the cat of the tangram. If green dots represent points with integer coordinates... What's the area of the silhouette?
    Nina Guindilla calculated it:



    Dear Teacher:
    I'm using Pick's theorem...
area  =  i + b/2 − 1

where  i  is the number or interior points and  b  is the number of boundery points... 
    Let me count:
area  =  38 + 60/2 − 1  =  67  square units.

1265. Carnot's theorem

    Nina Guindilla has proved Carnot's theorem for acute angled triangles. Will you demonstrate it for other triangles?

    Dear Teacher:
    In an acute angled triangle... If  a ,  b  and  c  are the distances from the circumcenter to the sides, and  r  and  R  are the inradius and the circumradius, then


a + b + c = r + R
    Proof:
    Let  A ,  B  and  C  be the sides of the triangle...
    If I divide the triangle in this way, its area is


area  =  Ar/2 + Br/2 + Cr/2  =  (A+B+C) r/2
    So
r  =  2 · area / (A+B+C)



    But if I divide the triangle in this other way, thus, the area is


area  =  Aa/2 + Bb/2 + Cc/2  =  (Aa+Bb+Cc) / 2
    So
r  =  (Aa+Bb+Cc) / (A+B+C)
    Finally, dividing the triangle again... but now into three cyclic quadrilaterals (they are cyclic because have two opposite right angles), I can apply Ptolemy's theorem:

RA/2  =  bC/2 + cB/2
RB/2  =  aC/2 + cA/2
RC/2  =  aB/2 + bA/2
    Multiplying by 2 and summing:

RA + RB + RC  =  bC + cB + aC + cA + aB + bA
R (A+B+C)  =  bC + cB + aC + cA + aB + bA
R  =  (bC+cB+aC+cA+aB+bA) / (A+B+C)
    Hence
r + R  =
 (Aa+Bb+Cc) / (A+B+C)  +  (bC+cB+aC+cA+aB+bA) / (A+B+C)  =
=  (Aa+Bb+Cc+bC+cB+aC+cA+aB+bA) / (A+B+C) =
=  (a+b+c) (A+B+C) / (A+B+C)  =
=  a + b + c

miércoles, 1 de febrero de 2017

1006. Out of tune?

    Dear Teacher:
    I've studied musical harmonics... a little. If the 1st harmonic is C4 (do4) then the 2nd harmonic is C5 (do5) which is found after 12 semitones along a pitch scale with audible frequencies in geometric progression until the double-frequency sound wave. I can calculate its common ratio:

    The 3rd harmonic is G5 (sol5) which is found after 19 semitones towards the triple-frequency wave, so the ratio would be:

    Are you sure that Music is based on Mathematics?

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)