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9PROBLEMS OF SOLVING GEOMETRIC EQUATIONS USING MATHEMATICAL EQUATIONS
Комилжон Аскарович Ражапов
Фаргона политехника институти п.ф.н доцент
ABSTRACT
This article describes how to solve a rectangular triangle with only one hypotenuse, or to solve all of its second catheters and hypotenuse, and to solve all three-dimensional equations when given the diagonal length of a rectangular parallel.
Keywords: triangle, catheter, hypothetical, diagonal, parallelopiped, theorem, proof, natural, presence, uniqueness
The article explores the algorithm for solving geometric problems in a simple algebraic way.
Issue 1: The hypotenuse of a right-angled triangle is equal to с. Determine the length of its a, b catheters in integers.
Here are three cases.
10. If x2 + = f , (f = c2, f > 2, f 6 N) equation (1) has a uniform natural solution, the a, b catheters of Xj — a, x — b,(a, b 6 N) right-angled triangle are determined by X = a, x = b, (a, b 6 N) one value.
Example: The hypotenuse of a right-angled triangle is с = 17 cm. Find all the values of its a, b cathode lengths.
Solution: Let's find the natural solutions of x2 + x2 = 289 the equation.
289 = 172 + O, n = 17, k = O, N(x; x ) 6 0 289 = 162 + 33 = 162 + 52 + 8, n = 16, k2 = 5, щ > k2, r22 = 8,N(x;x2) 6 0. 289 = 152 + 64 = 152 + 82,щ = 15,k3 = 8,щ > k3,r23 = 0,N(x;x) = (15;8)
289 = 132 + 120 = 132 + 102 + 20, щ = 13, k4 = 10, щ > k4, r24 = 20 ^ O, N(хг; x ) 6 0. equation (15; 8) has a natural solution of one binary pair.
Thus, the triangles of a triangle that satisfy the condition of the matter are the integers of a single length a = 15 sm, b = 8 sm.
20. If equation (1) has several natural solutions, catheters a, b of a right-angled triangle are not uniformly defined.
Example: The hypotenuse of a right-angled triangle is с = Тб5 sm in length. Find the entire length of its a, b catheter.
Solution: Let's find the natural solutions of x2 + = 65 the equation
65 = 82 +1 = 82 +12, n = 8, K = 1, n > K, r2 i = 0,N(x; x2) = (8;1)
65 = 72 +16 = 72 + 42, n = 7, K = 4, n > K, r2.2 = 0,N(x; x2 ) = (7; 4)
65 = 62 + 29 = 62 + 52 + 4, n = 6, K = 5, n > K, r2.3 = 4 ^ 0, N(x; x2 ) e 0 equa^on
(8;1>, (7;4> has a natural solution of two binary pairs
Hence, the triangles of triangles satisfying the condition of the problem are integers ax = 8 sm, b = 1 sm and a2 = 7 sm, b2 = 4 sm .
30 If equation (1) does not have a natural solution, the whole values of catheters a, b of a right-angled triangle will not exist.
Example: The hypotenuse of a right-angled triangle is C=7cm in length. Find the lengths of its A, B catheters for all values.
Solution: Let's find the natural solutions of x2 + = 49 the equation
49 = 72 + 0, n = 7, K = 0, n > K, r2.i = 0, N(x; x2) e 0
49 = 62 +13 = 62 + 32 + 4, n = 6, K = 3, n > K, r2.2 = 4 ^ 0, N(x; x2 ) e 0 49 = 52 + 24 = 52 + 42 + 8, n = 5, K = 4, n > K, r2.3 = 6 ^ 0, N(x; X ) e 0 49 = 42 + 33 = 42 + 42 + 17, n4 = 4, K4 = 4, n4 = K, r2.4 = 17 ^ 0,N(x; X ) e0 the equation does not have a natural solution.
50 the triangle that satisfies the condition of the problem does not have the values of the integer lengths.
Issue 2. Given the length of one catheter of a rectangular triangle b, the length of the hypotenuse c and its second catheter a.
Theorem 1. If one catheter of a right-angled triangle is b = p > 2 prime number, its second catheter a and hypotenuse c are defined by the integers. Proof. We use the following theorem to prove the above theorem. Theorem 2. If p > 2 is a prime number, then n is the only natural number that
satisfies equality (2n + 1)2 = (2n)2 + p2 (2)
Proof. Availability. (2) If we open and compact both sides of the equation, it p 2 -1
became 4n = p2 -1, n = (3).
The expression image to the right of equation is divisible by 4 because
p 2 -1-,
n =
4
is the even number p2-1 = (p-1)-(p +1). p-1,p +1 in the equation image.
p2 -1 p2 -1
It can be showed that n is unique by assuming ^ = n and = n. If we
say 2n+1 = c, 2n = a theorem 1 proves.
Example. One catheter of a rectangular triangle is b = p = 31 cm in length. Determine the second catheter A and the length of the hypotenuse C in integers.
Since b = p = 31 according to the theorem 2 is n =
p2 -1 _312 -1
= 240
4 4
according to the theorem 1 is c = 2n+1 = 2 ■ 240+1 = 48lsm, a = 2n = 2 ■ 240 = 480 sm
Example. ABC The height of the rectangular triangle at the right angle of the triangle to the hypotenuse is BD = h = 5 cm. The catheter and its hypotenuse projection are known to be integers of CD lengths. Find the lengths of the sides of the triangle.( Photo 1)
B
C
(1-photo) .
Since h = p = 5 according to the theorem 2 is n =
p2 -1 _ 52 -1
= 6 according to
4 4
the theorem 1 is BC = 2n+1 = 2 ■ 6+1 = l3sm, CD = 2n = 2 ■ 6 = 12 sm We determined that.
According to the metric relationship in a right-angled triangle, AD 12 = 52
equals. Using AD = —, we find that AB = — AC =169. M 12 12 12
The diagonal length of a rectangular parallellopiped is D, determine all three
dimensions in integers:
Here are three instances below.
10. If equation xj2 + x2 + x32 = g, (g = d2,g > 3,g e N) (3) has only one natural solution, the measurements A,B,C of a right-angled parallelalipipede x1 = a,x2 = b,x3 = c(a,b,c e N) are determined by a single value.
Example: The diagonal length of a rectangular parallellopiped is d = 7cm, find all the values of its three dimensions.
Solution: Let us find the natural solutions of equation xf + xl + x32 = 49. equation (2;3;6) has the only natural solution in triplicate pairs.
Therefore, the criteria for satisfying the condition are a = 2 sm, b = 3 sm, c = 6 sm dimensional parallellopiped dimensions.
20. If equation (3) has several natural solutions, the a, b, c measurements of a right-angled parallelalipiped are not uniformly defined.
Example: The diagonal length of a rectangular parallellopiped is A = 12 cm, find all the values of its three dimensions
Solution: Let us find the natural solutions of Equation x2 + x2 + x2 =441. Equation (20;5;4);(19;8;4);(18;9;6);(16;13;4);(16;11;8);(14;14;7)has natural solutions of six to three pairs.
So there are six parallellopeds, all three dimensions, satisfying the condition of the problem.
30. If equation (3) does not have natural solutions, the a, b, c measurements of the right-angled parallellopiped will not be integers.
Example: The diagonal length of a rectangular parallellopiped is d = >/10 cm find all the values of its three dimensions.
Solution: Let natural solutions of the equation x2 + x2 + x2 = 10. 10 = 32 +1 = 32 +12 + 0 m = 3,n = 1,k = 0,(0 й N) ,N(x; x; x3 ) e 0 . 10 = 22 + 6 = 22 + 22 + 2 = 22 + 22 +12 +1, m = 2,n2 = 2,k2 = 1,m2 = Щ > К, гз.г * 0,
N(x1; x2; x3 ) e0 .the equation does not have a natural solution.
Therefore, there are no parallellopipedes of all the three dimensions that satisfy the condition of the matter.
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