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PHYSICO-MATHEMATICAL SCIENCES
ABOUT THE ISSUES OF GEOMETRICAL INEQUALITIES AND THE METHODS OF THEIR SOLUTION Soatov U.A.1, Dzhonuzokov U.A.2 (Republic of Uzbekistan) Email: Soatov456@scientifictext.ru
1Soatov Ulugbek Abdukadirovich - PhD in Physics and Mathematics, Senior Lecturer; 2Dzhonuzokov Ulugbek Abduganievich - Teacher, DEPARTMENT OF HIGHER MATHEMATICS, JIZZAKH POLYTECHNIC INSTITUTE, JIZZAKH, REPUBLIC OF UZBEKISTAN
Abstraat: in practice, in the process of solving mathematical problems, we are faced with many professions related to geometric inequalities. Solving them requires students to have a thorough enough mathematical knowledge. It is one of the important tasks to teach mathematics teachers methods of proving geometrical inequalities in the process of forming the ability to apply their mathematical knowledge to solving various issues. This article explores some issues regarding geometric inequalities and the pure analytical and geometric ways to solve them.
Keywords: geometric inequalities, triangular, inequality, analytical method, geometric method, median, perimeter, angl, restangl, circle, radius.
О ВОПРОСАХ ГЕОМЕТРИЧЕСКОГО НЕРАВЕНСТВА
И СПОСОБАХ ИХ РЕШЕНИЯ 12 Соатов У.А. , Джонузоков У.А. (Республика Узбекистан)
1Соатов Улугбек Абдукадирович - кандидат физико-математических наук, старший преподаватель; 2Джонузоков Улугбек Абдуганиевич - преподаватель, кафедра высшей математики, Джизакский политехнический институт, г. Джизак, Республика Узбекистан
Аннотация: в процессе решения математических задач на практике мы сталкиваемся со многими задачами, относящимися к геометрическим неравенствам. Их решение требует от учащихся достаточно основательных математических знаний. Одной из важных задач является обучение способам доказательства геометрических неравенств в процессе формирования навыков одаренных обучающихся применять математические знания к решению различных задач. В этой статье изучены задачи, относящиеся геометрическим неравенствам, а также чисто аналитические и геометрические методы их решения.
Ключевые слова: геометрические неравенства, неравенство треугольника, аналитический метод, геометрический метод, медиана, периметр, угол, треугольник, четырехугольник, окружность, радиус.
We are faced with geometrical inequalities in several issues related to proof. Teaching them in elective classes or circle classes will help to formulate the skills of strengthening and applying mathematical knowledge to the students. Several geometrical issues related to the joint application of the basic theorem and formulas of igeometrianin [1] and many algebraic inequalities [2-3] were studied in the study. Below are some of the issues related to geometrical inequalities and we aimed to study the methods of their solution. The
simplest view of geometrical inequalities is the inequality of triangles [6], in solving many issues, this fundamental inequality is used.
Issue 1. The sum of the medians of the triangle is equal to s , its perimeter is 2p .
3 p < s < 2p prove inequality.
Proof. Let's assume BD — mc ABC mediana, which is lowered to the side of the
triangle AC, triangular icons BC — a, CA — b, AB — c and F location BC let it be the middle of the side (figure 1).
Fig. 1. m-ABC is the mediana lowered to the side of the triangle AC,triangle toms a,b, c and f are the
middle of the sides of the point BC
Thus BFD for a triangle, according to the inequality of the triangle mb < BF + FD —1(a + c) tensile fitting and ma va mc for medians, too, we write about similar inequalities:
ma < < (b + c), mc < < (a + b) . In the result of the addition of these inequalities
mb + ma + mc < 1 (a + c + b + c + a + b) — 1 (2a + 2b + 2c) — a + b + c
much we form. According to the condition of the matter ma + mb + mc — c and
a + b + c — 2p from that s < 2p it turns out.
Let's assume that the point M is the point at the intersection of the medians of the ABC triangle. In that case, as a result of applying the inequality of triangles to ABM, BCM and CAM triangles, we will have the following:
2 2 2
3(ma + mbb> c , 3(mb + mc)> a , 3(mc+maa>b .
If we add these inequalities to the limit, 3(ma + mb + mb + mc + mc + ma)> a + b + c or 4(mb + ma + mc) > 2p
4 o 3
or we will have. From the condition of matter — S > 2 p or s > P ■ And so on,
3
— p < s < 2 p inequality proved.
Issue 2. If BC=a, AC=b, AB=c-the sides of the acute-angled triangle, R--the outer drawn circle to it, then a) a '+ b + c > 8R . b) a+b + c > 4R prove that you are. Proof. a) let's assume mc = CD , ABC let it be the median of the triangle (figure 2).
Fig. 2. BC=a, AC=b, AB=c-the sides of an acute-angled triangle,R - the outer drawn circle to it,
m-the median of an ABC triangle
According to the formula of the length of the mediana m =— (2a + 2b - 2c ) . As
c 4
1
3
a result, we get a + b + c = 2mc + — c . AABC since it is a sharp angle, the
center of the circle lies in a triangle with O points. AAC1B we look at a triangle with an acute angle.
ZCOD > ZC\OD because it is C1D < CD we find that it is. As a result,
a2 + b2 + c2 = 2m2 + 3c2 > 2C.D2 + 3c2 = AC,2 + C,B2 + BA2 = 8R2 c 2 1 2 11
b) Since
a, b
and
are
smaller 2 R
than
2R(a + b + c) > a2 + b2 + c2 > 8R2. All in all, a + b + c > 4R.
Issue 3. Prove inequality if a, b ,c and S are triangular integers and surfaces,
a
+ b2 + c2 > 4S>/3 + (a - b)2 + (b - c)2 + (c - a)2
respectively.
Proof. On the right side of the given inequality, we leave only the first suffixes, we shift the rest to the left and group them in the form of a pair of square brackets. Then we divide
into multipliers in pairs and X = a + b - c, y = a - b + c, z =-a + b + c
will
add
new
variables.
we Thus
xy + xz + yz = (a2 - (b - c)2) + (b2 - (c - a)2) + (c2 - (a - b)2)
Now
11 1 j 1
p = — (a + b + c) = — (x + y + z) va p — a = -z, p — b = —y,
p — c = — x if we take into account the equations, we form
according to the formula of Geron. As a result, our perceived inequality will have xy + xz + yz 3(x + y + z)xyz appearance. We divide both sides of this
1 1 _ 1
inequality into xyz and get u = , v = , w = new variables. We will have
xyz
u + V + w >yj3(uv + vw + wu) without it. After the last inequality is squared and
2 2 2
simplified, we come to a certain U + V + W > UV + UW + WV inequality.
Comments. it is always useful to switch from a,b,c triangular integers to x,y,z variables by formulas x = a + b — C , y = a - b + C , z = -a + b + C . The main thing is that a,b, c are provided that together with the variables being positive, the triangle satisfies the inequality, in this case the new variables will also be positive.
Issue 4. Prove that R and r are
R > 2r , if there are radiuses of the outer and inner drawn circles to the triangle respectively.
Proof. Suppose, A1 ,B1 and C1 are the middle of the ABC triangle integers (figure 3).
Fig. 3. Radius of the outer and inner drawn circles R and r respectively into the triangle, A1 ,B1 and Cj are the middle of the triangle integers ABC
R
A1B1C1 is equal to the radius — of the outer drawn circle to the triangle, and this circle
2
R
in general "goes out" from the border of the triangle ABC. Therefore — > r place. We pass
2
the parallel strokes to the sides of the ABC triangle into the circle we are looking at. As a
R
result, we get an inner drawn circle with — radi, A2B2C2 triangle, similar to the ABC
2
triangle, which contains the ABC triangle itself.
Issue 5. On the AC side of the ABC triangle, the points K and M are obtained so that AK=MC. If AB > BC . then prove that there will be ZABK < ZMBC (figure 4).
A K D M C
Fig. 4. On the AC side of the ABC triangle, the points K and M are obtained so that AK=MC. BD -
mediana
Proof. We can assume that AK = MC < 1 AC . We pass the BD media. It does not
2
pass from AB > BC to ABDA, and the projection of point B to AC lies in the DC light. So, the projection of KB is larger than the projection of BMand BK > BM . But, triangles of equal size AABK and ACBM, that is
AB ■ BK sin AABK = CB ■ BM sin ACBM . As a result,
sin AMBC > sin AKBA. However AABK sharp angle. All in all, AABK <AMBC.
Issue 6. a, b, c, d - the lengths of the sides of the rectangle, and S -let it be his face, then prove S < i (ac + bd) inequality.
Proof. If a , c and b , d were the bases of the quadrilateral army, then the inequality is seen to be reasonable (the face of the triangle does not exceed half the multiples of its two integers). That's why we make them an army. To do this, we change the ABCD rectangle to the ABC1D rectangle (drawing 5).
C
Fig. 5. a, b, c, d - the lengths of the sides of the rectangle, and S -its face
There ABC1D = ABCD . Now the ABC1D rectangle is equal in size with the ABCD
rectangle. In it we pass AC1 dioganal and get two triangles, the sides of which are a, c and b, d are equal, respectively.
Conclusion. Problems with geometric inequalities are also common in mathematics. By solving the above problems, the applications of triangular inequalities, as well as pure analytical and geometric methods of proving inequalities were studied. Issues like these can be used as a ready-made material in a circle for gifted students in mathematics.
References / Список литературы
1. Soatov U.A., Djonuzoqov. U.A. "Problems of geometry with the help of joint application of basic theorems and formulas" // Scientific-methodical journal of "Physics, Mathematics and Informatics". № 4, 2018. 40-49 pages.
2. Soatov U.A., Dusmatov Е. "The use of Cauchy inequality in proving inequalities" // "Scientific-methodological journal of " Physics, Mathematics and Informatics", 2012. № 4. Pages 59-63.
3. Soatov U.A., Rakhimov R., Jurakulov М. "On the main methods of proving inequalities" // Republican scientific-practical conference. JizPI.April, 2016.
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