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Sharygin equation
In this note we will show how to find strictly positive integer solutions to the following equation
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 , and we can multiply both sides of (1) by and obtain polynomial equation.
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
We will later explain that equation (1) comes from a geometrical problem were , and are the lengths of the sides of a triangle.
If is one solution of (5) then the other can be found by Vieta theorem
In other words, if satisfy (4) then so will with
Furthermore, since (4) is symmetric with respect to and , being a solution implies that so will be .
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 of (4) into another –- . And thus, since is a solution so will be which happens to be positive.
In order to find an integer solution we need to rescale
by a common denominator of and which happens to be .
Finally, substituting these values back in (4) we obtain the sought values for , , and .
Since all values that we just found are even, we can make a smaller solution by dividing each value by :
Sharygin triangle
Sharygin triangle is a scalene triangle whose bisectoral triangle is isosceles. In the following diagram, is a Sharygin triangle and is its bisectoral triangle.
We will use this diagram to prove that the lengths , and of the sides of a Sharygin triangle satisfy equation (1).
Consider triangles and . They have two pairs of mutually congruent sides ( is common and by assumption) and congruent angles ( because is a bisector). This can happen only in two cases: when the two triangles are congruent or when they have supplementary angles The first case is realised when is isosceles which is contrary to our assumption, so we will focus on the second case. Since is supplementary to is supplementary to , we can conclude that
Now, let's rotate clockwise around point until consides with and point becomes . Note that congruence of the angles (7) requres that lies on the linear extension of the segment beyond the point . The resulting triangle together with the adjacent form that is similar to and the lengths of its sides can be expresed as follows:
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 –- given triangle. , , –- bisectors. If , then either (in this case will be isosceles), or . In the second case, we will rotate around point by angle . As a result, triangles and will become adjacent and form a triangle similar to . If the sides of are , и , then the sides of the resulting triangle will be , and . Taking similarity into account, we find the following relation between , and : Let's denote ; by cosine theorem, . Multiplying the last equation by , and in sequence and subtracting results from (1), we obtain Since , Expressing in the cosine theorem expression in terms of , and and denoting , we get the following equation for : In order for this equation to have solution under conditions (2), the following inequalities must hold:
The system of inequalities (2), (3), (4) holds when .
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 , which for the angle corresponds to the interval . At one end of the interval the constructed triangle will degenerate and the other end corresponds to the equality , 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, 157–158]
Пусть –- данный треугольник. , , –- биссектрисы. Если , то или (в этом случае будет равнобедренным), или . Во втором случае повернём вокруг точки на угол . В результате треугольники и окажутся приложенными друг к другу и образуют треугольник подобный . Если стороны есть , и , то стороны получившегося треугольника быдут равны , и . Учитывая подобие, получим между , и соотношение Обозначим ; по теореме косинусов . Умножая последнее равенство последовательно на , и и вычитая из (1), получим Поскольку ,
Заменив в теореме косинусов через , и и обозначив , получим для уравнение Для того чтобы это уравнение при условиях (2) имело решение , , должны выполняться неравенства
Система неравенств (2), (3), (4) удовлетворяется при .
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 , что соответствует для угла интервалу приблизительно . Для одного конца интервала построенный нами треугольник будет вырождаться, другой же конец соответствует равенству , т. е. два случая, которые мы выделели в начале решения, для этого значения угла совпадают.