Sharygin equation

In this note we will show how to find strictly positive integer solutions to the following equation

(1)ab+c+bc+a=ca+b

Some of the algebraic manipulations are rather tedious, so we will use a symbolic algebra package to justify some of the steps.

>>> from sympy import *
>>> init_printing(use_unicode=True)

Since we are only looking for positive

a,
b
and
c
we can multiply both sides of (1) by
(a+b)(b+c)(c+a)
and obtain polynomial equation.

(2)a(a+b)(c+a)+b(a+b)(b+c)c(b+c)(c+a)=0

With the help of sympy we expand (2) as follows:

>>> a, b, c = symbols('a b c')
>>> P = a*(a+b)*(c+a)+b*(a+b)*(b+c)-c*(b+c)*(c+a)
>>> P.expand()
 3    2      2        2              2    3    2        2    3
a  + a ⋅b + a ⋅c + a⋅b  + a⋅b⋅c - a⋅c  + b  + b ⋅c - b⋅c  - c

(3)a3+a2b+a2c+ab2+abcac2+b3+b2cbc2c3=0

We will later explain that equation (1) comes from a geometrical problem were

a,
b
and
c
are the lengths of the sides of a triangle.

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The following variable substitution is well known in planimetry

(4)a=y+zb=x+zc=x+y

and turns (3) into

>>> x, y, z = symbols('x y z')
>>> Q = P.subs(dict(a=y+z, b=x+z, c=x+y)).expand()
>>> Q
     2        2          2                  2      2          2      3
- 3⋅x ⋅y + 5⋅x ⋅z - 3⋅x⋅y  + 6⋅x⋅y⋅z + 9⋅x⋅z  + 5⋅y ⋅z + 9⋅y⋅z  + 4⋅z

3x2y+5x2z3xy2+6xyz+9xz2+5y2z+9yz2+4z3=0

The last equation is homogeneous, so instead of looking for three integer values, we can set

z=1 and look for two rational values of
x
and
y
:

>>> R = Q.subs(z, 1)
>>> R
​    2        2        2                    2
- 3⋅x ⋅y + 5⋅x  - 3⋅x⋅y  + 6⋅x⋅y + 9⋅x + 5⋅y  + 9⋅y + 4

3x2y+5x23xy2+6xy+9x+5y2+9y+4=0

This equation, while qubic overall, is only quadratic in

x and
y
. Thus for fixed
y
it has the form

(5)α(y)x2+β(y)x+γ(y)=0

where

α(y)=53yβ(y)=3y2+6y+9γ(y)=5y2+9y+4

which while easily derived by hand can be found in sympy as follows:

>>> R.coeff(x, 2)
5 - 3⋅y
>>> R.coeff(x, 1)
     2
- 3⋅y  + 6⋅y + 9
>>> R.coeff(x, 0)
   2
5⋅y  + 9⋅y + 4

If

x1 is one solution of (5) then the other can be found by Vieta theorem

x1+x2=β(y)α(y)=3y26y953y

In other words, if

(x1,y) satisfy (4) then so will
(x2,y)
with

x2=3y26y953yx1

Furthermore, since (4) is symmetric with respect to

x and
y
,
(x2,y)
being a solution implies that so will be
(y,x2)
.

This explans the role of the following function in our solution:

def a(p):
    x, y = p
    return y, (3 * y * (y - 2) - 9) / (5 - 3 * y) - x

It maps one rational solution

(x1,y) of (4) into another -
(y,x2)
. And thus, since
P=(1,3)
is a solution so will be
a(a(a(a(P))))
which happens to be positive.

x=13387391y=18951391067039

In order to find an integer solution we need to rescale

(x,y,z)=(13387391,18951391067039,1)

by a common denominator of

x and
y
which happens to be
1067039=3912729
.

(x,y,z)=(133872729,1895139,1067039)=(36533123,1895139,1067039)
Finally, substituting these values back in (4) we obtain the sought values for
a
,
b
, and
c
.

a=y+z=1895139+1067039=2962178b=x+z=36533123+1067039=37600162c=x+y=1895139+36533123=38428262

Since all values that we just found are even, we can make a smaller solution by dividing each value by

2:

(a,b,c)=(1481089,18800081,19214131)

Sharygin triangle

Sharygin triangle is a scalene triangle whose bisectoral triangle is isosceles. In the following diagram,

ABC is a Sharygin triangle and
A1B1C1
is its bisectoral triangle.

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We will use this diagram to prove that the lengths

a,
b
and
c
of the sides of a Sharygin triangle satisfy equation (1).

Consider triangles

CC1A1 and
CC1B1
. They have two pairs of mutually congruent sides (
CC1
is common and
C1A1C1B1
by assumption) and congruent angles (
A1CC1B1CC1
because
CC1
is a bisector). This can happen only in two cases: when the two triangles are congruent or when they have supplementary angles
(6)CA1C1+CB1C1=180.

The first case is realised when
ABC
is isosceles which is contrary to our assumption, so we will focus on the second case. Since
AB1C1
is supplementary to
CB1C1
is supplementary to
CA1C1
, we can conclude that

(7)AB1C1CA1C1.

Now, let's rotate

AB1C1 clockwise around point
C1
until
B1C1
consides with
A1C1
and point
A
becomes
D
. Note that congruence of the angles (7) requres that
D
lies on the linear extension of the segment
BC
beyond the point
C
. The resulting triangle
DC1A1
together with the adjacent
BC1A1
form
DBC1
that is similar to
ABC
and the lengths of its sides can be expresed as follows:

|BC1|=aca+b, [derive this]|BD|=|BA1|+|A1D|=|BA1|+|B1A|=acb+c+bca+c.

The similarity

DBC1ABC

implies

|BC1||BC|=|BD||AB|

or after substitution and cancellation

ab+c+bc+a=ca+b.

Sharygin's original

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This problem was first published in issue 17 of Kvant's Library in 1982. Kvant's Library (Библиотечка "Квант") was a series of books for schoolchildren popular in the Soviet Union but largely unknown in the West. These days books from this series are difficult to find and Sharygin's "Problems in Geometry" is not an exception. Only medium quality scanned copies [3] are available оn the Internet and only in Russian. For this reason, we will reprint Sharygin’s original formulation and his solution here together with English translation.

Problem 58

A given triangle is such that the triangle formed by the bases of its bisectors is isosceles. Is it true that the given triangle is also isosceles? [3, pp. 154, 157–158]

Let

ABC - given triangle.
AA1
,
BB1
,
CC1
- bisectors. If
|A1B1|=|A1C1|
, then either
A1B1C^=A1C1B^
(in this case
ABC
will be isosceles), or
A1B1C^+A1C1B^=180
.
In the second case, we will rotate
A1B1C
around point
A1
by angle
B1A1C1^
. As a result, triangles
A1C1B
and
A1B1C
will become adjacent and form a triangle similar to
ABC
. If the sides of
ABC
are
a
,
b
и
c
, then the sides of the resulting triangle will be
acb+c
,
abb+c
and
aca+b+aba+c
. Taking similarity into account, we find the following relation between
a
,
b
and
c
:
ca+b+ba+c=abb+c 

(1)b3+c3a3+b2c+b2a+c2b+c2aa2ba2c+abc=0.

Let's denote
cosBAC^=x
; by cosine theorem,
b2+c2a2=2bcx
. Multiplying the last equation by
a
,
b
and
c
in sequence and subtracting results from (1), we obtain
2x(a+b+c)+a=0a=2(b+c)x2x+1.
Since
0<a<b+c
,
(2)14<x<0.

Expressing
a
in the cosine theorem expression in terms of
b
,
c
and
x
and denoting
bc=λ
, we get the following equation for
λ
:
(4x+1)λ22λ(4x3+8x2+x)+4x+1=0.

In order for this equation to have solution
λ>0
under conditions (2), the following inequalities must hold:

4x3+8x2+x>0,
(4)14D=(4x3+8x2+x)2==(2x+1)2(x+1)(2x1)(2x2+5x+1)>0.

The system of inequalities (2), (3), (4) holds when
14<x<1754
.

Therefore, the original triangle is not necessarily isosceles. However, we have just proven that this can only happen in such case when one of the angles of the original triangle is obtuse and its cosine falls in the interval
(14,1754)
, which for the angle corresponds to the interval
(10240;10428)
. At one end of the interval
(14)
the constructed triangle will degenerate and the other end
(1754)
corresponds to the equality
A1B1C^=A1C1B^=90
, in other words, the two cases that we identified at the start of the solution coincide for this value of the angle.

Задача 58

Про данный треугольник известно что треугольник образованный основаниями его биссектрис, является равнобедренным. Будет ли верно что и данный треугольник является равнобедренным? [3, стр. 154, 157158]

Пусть

ABC - данный треугольник.
AA1
,
BB1
,
CC1
- биссектрисы. Если
|A1B1|=|A1C1|
, то или
A1B1C^=A1C1B^
(в этом случае
ABC
будет равнобедренным), или
A1B1C^+A1C1B^=180
. Во втором случае повернём
A1B1C
вокруг точки
A1
на угол
B1A1C1^
. В результате треугольники
A1C1B
и
A1B1C
окажутся приложенными друг к другу и образуют треугольник подобный
ABC
. Если стороны
ABC
есть
a
,
b
и
c
, то стороны получившегося треугольника быдут равны
acb+c
,
abb+c
и
aca+b+aba+c
. Учитывая подобие, получим между
a
,
b
и
c
соотношение
ca+b+ba+c=abb+c 

(1)b3+c3a3+b2c+b2a+c2b+c2aa2ba2c+abc=0.

Обозначим
cosBAC^=x
; по теореме косинусов
b2+c2a2=2bcx
. Умножая последнее равенство последовательно на
a
,
b
и
c
и вычитая из (1), получим
2x(a+b+c)+a=0a=2(b+c)x2x+1.

Поскольку
0<a<b+c
,
(2)14<x<0.

Заменив в теореме косинусов
a
через
b
,
c
и
x
и обозначив
bc=λ
, получим для
λ
уравнение
(4x+1)λ22λ(4x3+8x2+x)+4x+1=0.

Для того чтобы это уравнение при условиях (2) имело решение
λ>0
,
λ0
, должны выполняться неравенства

4x3+8x2+x>0,
(4)14D=(4x3+8x2+x)2==(2x+1)2(x+1)(2x1)(2x2+5x+1)>0.

Система неравенств (2), (3), (4) удовлетворяется при
14<x<1754
.

Таким образом, исходный треугольник не обязательно равнобедренный. Однако мы доказали, что это может иметь место только в том случае, когда один из углов исходного треугольника тупой и его косинус находится в интервале
(14,1754)
, что соответствует для угла интервалу приблизительно
(10240;10428)
. Для одного конца интервала
(14)
построенный нами треугольник будет вырождаться, другой же конец
(1754)
соответствует равенству
A1B1C^=A1C1B^=90
, т. е. два случая, которые мы выделели в начале решения, для этого значения угла совпадают.

References

  1. I. V. Netay and A.V. Savvateev, "Sharygin triangles and elliptic curves", (October 2016).
  2. Oliver Nash, "Sharygin's group of triangles" (December 2016).
  3. I. F. Sharygin, "Problems in Geometry (Planimetry)" [in Russian], Nauka, 1982.