Every student knows that there is always a square of the hypotenuse. is equal to the sum legs, each of which is squared. This statement is called the Pythagorean theorem. It is one of the most famous theorems in trigonometry and mathematics in general. Let's consider it in more detail.

The concept of a right triangle

Before proceeding to the consideration of the Pythagorean theorem, in which the square of the hypotenuse is equal to the sum of the legs that are squared, one should consider the concept and properties of a right-angled triangle for which the theorem is valid.

A triangle is a flat shape with three corners and three sides. A right-angled triangle, as its name implies, has one right angle, that is, this angle is 90 o.

From general properties for all triangles it is known that the sum of all three angles of this figure is 180 o, which means that for a right-angled triangle the sum of two angles that are not straight is 180 o - 90 o = 90 o. The latter fact means that any angle in right triangle which is not straight will always be less than 90 o.

The side that lies against right angle, it is customary to call the hypotenuse. The other two sides are the legs of the triangle, they can be equal to each other, or they can differ. It is known from trigonometry that the larger the angle against which the side in the triangle lies, the greater the length of this side. This means that in a right-angled triangle, the hypotenuse (lies opposite the 90 o angle) will always be greater than any of the legs (lie opposite the angles< 90 o).

Mathematical notation of the Pythagorean theorem

This theorem states that the square of the hypotenuse is equal to the sum of the legs, each of which is previously squared. To write this formulation mathematically, consider a right-angled triangle in which sides a, b, and c are two legs and a hypotenuse, respectively. In this case, the theorem, which is formulated as the square of the hypotenuse is equal to the sum of the squares of the legs, can be represented by the following formula: c 2 = a 2 + b 2. From this, other important formulas for practice can be obtained: a = √ (c 2 - b 2), b = √ (c 2 - a 2) and c = √ (a 2 + b 2).

Note that in the case of a right-angled equilateral triangle, that is, a = b, the formulation: the square of the hypotenuse is equal to the sum of the legs, each of which is squared, is mathematically written as follows: c 2 = a 2 + b 2 = 2a 2, which implies the equality: c = a√2.

Historical reference

The Pythagorean theorem, which says that the square of the hypotenuse is equal to the sum of the legs, each of which is squared, was known long before the famous Greek philosopher... Many papyri Ancient egypt and the clay tablets of the Babylonians confirm that these peoples used the noted property of the sides of a right-angled triangle. For example, one of the first Egyptian pyramids, the pyramid of Khafre, whose construction dates back to the XXVI century BC (2000 years before the life of Pythagoras), was built based on the knowledge of the aspect ratio in a right-angled triangle 3x4x5.

Why, then, is the theorem now named after the Greek? The answer is simple: Pythagoras was the first to prove this theorem mathematically. The surviving Babylonian and Egyptian written sources speak only of its use, but no mathematical proof is given.

It is believed that Pythagoras proved the theorem under consideration by using the properties of similar triangles, which he obtained by drawing the height in a right-angled triangle from an angle of 90 o to the hypotenuse.

An example of using the Pythagorean theorem

Consider simple task: it is necessary to determine the length of the inclined staircase L, if it is known that it has a height of H = 3 meters, and the distance from the wall against which the staircase rests to its foot is P = 2.5 meters.

In this case, H and P are legs, and L is hypotenuse. Since the length of the hypotenuse is equal to the sum of the squares of the legs, we get: L 2 = H 2 + P 2, whence L = √ (H 2 + P 2) = √ (3 2 + 2.5 2) = 3.905 meters or 3 m and 90, 5 cm.

Pythagorean theorem: The sum of the areas of the squares resting on the legs ( a and b) is equal to the area of ​​the square built on the hypotenuse ( c).

Geometric formulation:

Initially, the theorem was formulated as follows:

Algebraic formulation:

That is, denoting the length of the hypotenuse of a triangle by c, and the lengths of the legs through a and b :

a 2 + b 2 = c 2

Both statements of the theorem are equivalent, but the second statement is more elementary, it does not require the concept of area. That is, the second statement can be checked without knowing anything about the area and measuring only the lengths of the sides of a right-angled triangle.

The reverse Pythagorean theorem:

Proof

On this moment in the scientific literature, 367 proofs of this theorem have been recorded. Probably the Pythagorean theorem is the only theorem with such an impressive number of proofs. This variety can be explained only by the fundamental meaning of the theorem for geometry.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them are: proofs by the area method, axiomatic and exotic proofs (for example, using differential equations).

Through similar triangles

The following proof of the algebraic formulation is the simplest of the proofs built directly from the axioms. In particular, it does not use the concept of the area of ​​a figure.

Let be ABC there is a right-angled triangle with a right angle C... Let's draw the height from C and denote its base by H... Triangle ACH like a triangle ABC in two corners. Similarly, triangle CBH is similar ABC... Introducing the notation

we get

What is the equivalent

Adding, we get

Areas proof

The proofs below, despite their seeming simplicity, are not at all so simple. All of them use the properties of area, the proof of which is more difficult than the proof of the Pythagorean theorem itself.

Equal complementarity proof

1. Place four equal right-angled triangles as shown in Figure 1.
2. Quadrilateral with sides c is a square, since the sum of two acute angles is 90 °, and the unfolded angle is 180 °.
3. The area of ​​the entire figure is, on the one hand, the area of ​​a square with sides (a + b), and on the other hand, the sum of the areas of four triangles and two inner squares.

Q.E.D.

Evidence through scattering

Elegant proof by permutation

An example of one of such proofs is shown in the drawing on the right, where a square built on the hypotenuse is transformed by permutation into two squares built on the legs.

Euclid's proof

Drawing for Euclid's proof

Illustration for Euclid's proof

The idea behind Euclid's proof is as follows: let's try to prove that half of the area of ​​the square built on the hypotenuse is equal to the sum of the halves of the areas of the squares built on the legs, and then the areas of the large and two small squares are equal.

Consider the drawing on the left. On it, we built squares on the sides of a right-angled triangle and drawn a ray s from the vertex of the right angle C perpendicular to the hypotenuse AB, it cuts the square ABIK, built on the hypotenuse, into two rectangles - BHJI and HAKJ, respectively. It turns out that the areas of these rectangles are exactly equal to the areas of the squares built on the corresponding legs.

Let's try to prove that the area of ​​the square DECA is equal to the area of ​​the rectangle AHJK To do this, let's use an auxiliary observation: The area of ​​a triangle with the same height and base as this rectangle is equal to half the area of ​​the given rectangle. This is a consequence of defining the area of ​​a triangle as half the product of the base and the height. From this observation it follows that the area of ​​the triangle ACK is equal to the area of ​​the triangle AHK (not shown in the figure), which, in turn, is equal to half the area of ​​the rectangle AHJK.

Let us now prove that the area of ​​the triangle ACK is also equal to half the area of ​​the square DECA. The only thing that needs to be done for this is to prove the equality of the triangles ACK and BDA (since the area of ​​the triangle BDA is equal to half the area of ​​the square according to the above property). Equality is obvious, the triangles are equal on two sides and the angle between them. Namely - AB = AK, AD = AC - the equality of the angles CAK and BAD is easy to prove by the method of motion: we rotate the triangle CAK 90 ° counterclockwise, then it is obvious that the corresponding sides of the two triangles under consideration will coincide (since the angle at the apex of the square is 90 °).

The reasoning about the equality of the areas of the square BCFG and the rectangle BHJI is completely analogous.

Thus, we have proved that the area of ​​the square built on the hypotenuse is the sum of the areas of the squares built on the legs. The idea behind this proof is further illustrated with the animation above.

Proof of Leonardo da Vinci

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and motion.

Consider the drawing, as seen from the symmetry, the segment CI cuts the square ABHJ into two identical parts (since the triangles ABC and JHI are equal by construction). Rotating 90 degrees counterclockwise, we see that the shaded shapes are equal CAJI and GDAB ... Now it is clear that the area of ​​the shaded figure is equal to the sum of the halves of the areas of the squares built on the legs and the area of ​​the original triangle. On the other hand, it is equal to half the area of ​​the square built on the hypotenuse plus the area of ​​the original triangle. The final step in the proof is left to the reader.

Proof by the method of infinitesimal

The following proof using differential equations is often attributed to the famous English mathematician Hardy, who lived in the first half of the 20th century.

Looking at the drawing shown in the figure and observing the change of the side a, we can write the following relation for infinitely small increments of the sides with and a(using the similarity of triangles):

Proof by the method of infinitesimal

Using the method of separating variables, we find

A more general expression for changing the hypotenuse in the case of increments of both legs

Integrating this equation and using initial conditions, we get

c 2 = a 2 + b 2 + constant.

Thus, we arrive at the desired answer

c 2 = a 2 + b 2 .

As it is easy to see, the quadratic dependence in the final formula appears due to the linear proportionality between the sides of the triangle and the increments, while the sum is related to independent contributions from the increments of different legs.

A simpler proof can be obtained if we assume that one of the legs does not experience an increment (in this case, the leg b). Then for the constant of integration we obtain

Variations and generalizations

• If instead of squares we construct other similar figures on the legs, then the following generalization of the Pythagorean theorem is true: In a right-angled triangle, the sum of the areas of similar figures built on the legs is equal to the area of ​​the figure built on the hypotenuse. In particular:
  • The sum of the areas of regular triangles built on the legs is equal to the area of ​​a regular triangle built on the hypotenuse.
  • The sum of the areas of the semicircles built on the legs (as in the diameter) is equal to the area of ​​the semicircle built on the hypotenuse. This example is used to prove the properties of figures bounded by arcs of two circles and bearing the name of hippocratic lunes.

History

Chu-pei 500-200 BC. Left inscription: the sum of the squares of the lengths of the height and base is the square of the length of the hypotenuse.

The ancient Chinese book Chu-pei speaks of Pythagorean triangle with sides 3, 4 and 5: In the same book, a drawing is proposed that coincides with one of the drawings of the Hindu geometry of Baskhara.

Cantor (the largest German historian of mathematics) believes that the equality 3 ² + 4 ² = 5 ² was already known to the Egyptians around 2300 BC. e., during the time of King Amenemhat I (according to papyrus 6619 of the Berlin Museum). According to Cantor, the harpedonapts, or "rope pulls", built right angles using right-angled triangles with sides 3, 4, and 5.

It is very easy to reproduce their way of building. Take a rope 12 m long and tie it to it along a colored strip at a distance of 3 m. from one end and 4 meters from the other. The right angle will be enclosed between the sides 3 and 4 meters long. The Harpedonapts might argue that their way of building becomes superfluous, if you use, for example, the wooden square used by all carpenters. Indeed, there are known Egyptian drawings in which such a tool is found, for example, drawings depicting a carpentry workshop.

Somewhat more is known about the Babylonian Pythagorean theorem. In one text dating back to the time of Hammurabi, that is, to 2000 BC. BC, an approximate calculation of the hypotenuse of a right-angled triangle is given. From this we can conclude that in Mesopotamia they knew how to perform calculations with right-angled triangles, at least in some cases. Based, on the one hand, on the current level of knowledge about Egyptian and Babylonian mathematics, and on the other, on a critical study of Greek sources, Van der Waerden (Dutch mathematician) made the following conclusion:

Literature

In Russian

• Skopets Z.A. Geometric miniatures. M., 1990
• Yelensky Sch. In the footsteps of Pythagoras. M., 1961
• Van der Waerden B.L. Awakening science. Mathematics of Ancient Egypt, Babylon and Greece. M., 1959
• Glazer G.I. History of mathematics at school. M., 1982
• V. Litzman, "The Pythagorean Theorem" M., 1960.
  • A site about the Pythagorean theorem with a large number of proofs, the material is taken from the book of V. Litzman, a large number of drawings are presented in the form of separate graphic files.
• The Pythagorean theorem and Pythagorean triplets a chapter from the book by DV Anosov "A Look at Mathematics and Something From It"
• On the Pythagorean theorem and methods of its proof G. Glazer, Academician of the Russian Academy of Education, Moscow

In English

• The Pythagorean Theorem at WolframMathWorld
• Cut-The-Knot, a section on the Pythagorean theorem, about 70 proofs and a wealth of additional information

Wikimedia Foundation. 2010.

(according to papyrus 6619 of the Berlin Museum). According to Cantor, the harpedonapts, or "rope tensioners", built right angles using right-angled triangles with sides 3, 4, and 5.

It is very easy to reproduce their way of building. Take a rope 12 m long and tie it to it along a colored strip at a distance of 3 m from one end and 4 meters from the other. The right angle will be enclosed between the sides 3 and 4 meters long. The Harpedonapts might argue that their method of construction becomes superfluous if we use, for example, the wooden square used by all carpenters. Indeed, there are known Egyptian drawings in which such a tool is found, for example, drawings depicting a carpentry workshop.

Somewhat more is known about the Babylonian Pythagorean theorem. In one text dating back to the time of Hammurabi, that is, to 2000 BC. NS. , an approximate calculation of the hypotenuse of a right triangle is given. From this we can conclude that in Mesopotamia they knew how to perform calculations with right-angled triangles, at least in some cases. Based, on the one hand, on the current level of knowledge about Egyptian and Babylonian mathematics, and on the other, on a critical study of Greek sources, Van der Waerden (Dutch mathematician) concluded that it is highly probable that the theorem on the square of the hypotenuse was known in India already around the 18th century BC. NS.

Around 400 BC. e., according to Proclus, Plato gave a method for finding Pythagorean triplets, combining algebra and geometry. Around 300 BC. NS. the oldest axiomatic proof of the Pythagorean theorem appeared in Euclid's "Elements".

The wording

Geometric formulation:

Initially, the theorem was formulated as follows:

Algebraic formulation:

That is, denoting the length of the hypotenuse of the triangle through, and the lengths of the legs through and:

Both statements of the theorem are equivalent, but the second statement is more elementary, it does not require the concept of area. That is, the second statement can be checked without knowing anything about the area and measuring only the lengths of the sides of a right-angled triangle.

The reverse Pythagorean theorem:

Proof

At the moment, 367 proofs of this theorem have been recorded in the scientific literature. Probably the Pythagorean theorem is the only theorem with such an impressive number of proofs. This variety can be explained only by the fundamental meaning of the theorem for geometry.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them: proofs by the area method, axiomatic and exotic proofs (for example, using differential equations).

Through similar triangles

The following proof of the algebraic formulation is the simplest of the proofs built directly from the axioms. In particular, it does not use the concept of the area of ​​a figure.

Let be ABC there is a right-angled triangle with a right angle C... Let's draw the height from C and denote its base by H... Triangle ACH like a triangle ABC in two corners. Similarly, triangle CBH is similar ABC... Introducing the notation

we get

What is the equivalent

Adding, we get

, which was required to prove

Areas proof

The proofs below, despite their seeming simplicity, are not at all so simple. All of them use the properties of area, the proof of which is more difficult than the proof of the Pythagorean theorem itself.

Equal complementarity proof

1. Place four equal right-angled triangles as shown in Figure 1.
2. Quadrilateral with sides c is a square, since the sum of two acute angles is 90 °, and the unfolded angle is 180 °.
3. The area of ​​the entire figure is, on the one hand, the area of ​​a square with sides (a + b), and on the other hand, the sum of the areas of four triangles and the area of ​​the inner square.

Q.E.D.

Euclid's proof

The idea behind Euclid's proof is as follows: let's try to prove that half of the area of ​​the square built on the hypotenuse is equal to the sum of the halves of the areas of the squares built on the legs, and then the areas of the large and two small squares are equal.

Consider the drawing on the left. On it, we built squares on the sides of a right-angled triangle and drawn a ray s from the vertex of the right angle C perpendicular to the hypotenuse AB, it cuts the square ABIK, built on the hypotenuse, into two rectangles - BHJI and HAKJ, respectively. It turns out that the areas of these rectangles are exactly equal to the areas of the squares built on the corresponding legs.

Let's try to prove that the area of ​​the square DECA is equal to the area of ​​the rectangle AHJK To do this, let's use an auxiliary observation: The area of ​​a triangle with the same height and base as this rectangle is equal to half the area of ​​the given rectangle. This is a consequence of defining the area of ​​a triangle as half the product of the base and the height. From this observation it follows that the area of ​​the triangle ACK is equal to the area of ​​the triangle AHK (not shown in the figure), which, in turn, is equal to half the area of ​​the rectangle AHJK.

Let us now prove that the area of ​​the triangle ACK is also equal to half the area of ​​the square DECA. The only thing that needs to be done for this is to prove the equality of the triangles ACK and BDA (since the area of ​​the triangle BDA is equal to half the area of ​​the square according to the above property). Equality is obvious: the triangles are equal on two sides and the angle between them. Namely - AB = AK, AD = AC - the equality of the angles CAK and BAD is easy to prove by the method of motion: we rotate the triangle CAK 90 ° counterclockwise, then it is obvious that the corresponding sides of the two triangles under consideration will coincide (since the angle at the apex of the square is 90 °).

The reasoning about the equality of the areas of the square BCFG and the rectangle BHJI is completely analogous.

Thus, we have proved that the area of ​​the square built on the hypotenuse is the sum of the areas of the squares built on the legs. The idea behind this proof is further illustrated with the animation above.

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and motion.

Consider the drawing, as can be seen from the symmetry, the segment cuts the square into two identical parts (since the triangles and are equal in construction).

By rotating 90 degrees counterclockwise around a point, we see that the shaded figures and are equal.

Now it is clear that the area of ​​the shaded figure is equal to the sum of the halves of the areas of the small squares (built on the legs) and the area of ​​the original triangle. On the other hand, it is equal to half the area of ​​the large square (built on the hypotenuse) plus the area of ​​the original triangle. Thus, half of the sum of the areas of small squares is equal to half of the area of ​​the large square, and therefore the sum of the areas of the squares built on the legs is equal to the area of ​​the square built on the hypotenuse.

Proof by the method of infinitesimal

The following proof using differential equations is often attributed to the famous English mathematician Hardy, who lived in the first half of the 20th century.

Looking at the drawing shown in the figure and observing the change of the side a, we can write the following relation for infinitely small increments of the sides with and a(using the similarity of triangles):

Using the method of separating variables, we find

A more general expression for changing the hypotenuse in the case of increments of both legs

Integrating this equation and using the initial conditions, we obtain

Thus, we arrive at the desired answer

As it is easy to see, the quadratic dependence in the final formula appears due to the linear proportionality between the sides of the triangle and the increments, while the sum is related to independent contributions from the increments of different legs.

A simpler proof can be obtained if we assume that one of the legs does not experience an increment (in this case, the leg). Then for the constant of integration we obtain

Variations and generalizations

Similar geometric shapes on three sides

Generalization for similar triangles, area of ​​green shapes A + B = area of ​​blue C

Pythagorean theorem using similar right triangles

The generalization of the Pythagorean theorem was made by Euclid in his work Beginnings, expanding the areas of the squares on the sides to areas of similar geometric shapes :

If you build similar geometric shapes (see Euclidean geometry) on the sides of a right-angled triangle, then the sum of the two smaller figures will be equal to the area of ​​the larger figure.

The main idea of ​​this generalization is that the area of ​​such a geometric figure is proportional to the square of any of its linear dimensions, and in particular to the square of the length of any side. Therefore, for similar figures with areas A, B and C built on sides with length a, b and c, we have:

But, according to the Pythagorean theorem, a 2 + b 2 = c 2, then A + B = C.

Conversely, if we can prove that A + B = C for three similar geometric figures without using the Pythagorean theorem, then we can prove the theorem itself, moving in the opposite direction. For example, the starting center triangle can be reused as a triangle C on the hypotenuse, and two similar right-angled triangles ( A and B), built on the other two sides, which are formed as a result of dividing the central triangle by its height. The sum of the two smaller areas of the triangles is then obviously equal to the area of ​​the third, thus A + B = C and, performing the previous proofs in reverse order, we obtain the Pythagorean theorem a 2 + b 2 = c 2.

Cosine theorem

The Pythagorean theorem is special case a more general cosine theorem, which relates the lengths of the sides in an arbitrary triangle:

where θ is the angle between the sides a and b.

If θ is 90 degrees then cos θ = 0 and the formula is simplified to the usual Pythagorean theorem.

Arbitrary triangle

To any selected corner of an arbitrary triangle with sides a, b, c we inscribe an isosceles triangle in such a way that equal angles at its base, θ were equal to the chosen angle. Suppose that the chosen angle θ is opposite the side marked c... As a result, we got a triangle ABD with an angle θ, which is located opposite the side a and parties r... The second triangle is formed by the angle θ, which is opposite the side b and parties with the length s, as it shown on the picture. Thabit Ibn Qurrah argued that the sides in these three triangles are connected as follows:

As the angle θ approaches π / 2, the base of the isosceles triangle decreases and the two sides r and s overlap less and less. When θ = π / 2, ADB becomes a right triangle, r + s = c and we get the initial Pythagorean theorem.

Let's consider one of the reasons. Triangle ABC has the same angles as triangle ABD, but in reverse order. (Two triangles have a common angle at the vertex B, both have an angle θ and also have the same third angle, according to the sum of the angles of the triangle.) Accordingly, ABC is similar to the ABD reflection of triangle DBA, as shown in the lower figure. Let us write down the ratio between opposite sides and adjacent to the angle θ,

Also a reflection of another triangle,

Let's multiply the fractions and add these two ratios:

Q.E.D.

Generalization for arbitrary triangles via parallelograms

Generalization for arbitrary triangles,
area of ​​green plot = area blue

Proof of the thesis that in the picture above

Let us generalize further to non-rectangular triangles by using parallelograms on three sides instead of squares. (squares are a special case.) The upper figure demonstrates that for acute-angled triangle the area of ​​the parallelogram on the long side is equal to the sum of the parallelograms on the other two sides, provided that the parallelogram on the long side is constructed as shown in the figure (the dimensions indicated by the arrows are the same and define the sides of the lower parallelogram). This replacement of squares with parallelograms bears a clear resemblance to the initial theorem of Pythagoras, it is believed that it was formulated by Pappus of Alexandria in 4 AD. NS.

The bottom figure shows the progress of the proof. Let's look at the left side of the triangle. The left green parallelogram has the same area as the left side of the blue parallelogram because they have the same base b and height h... In addition, the left green parallelogram has the same area as the left green parallelogram in the upper figure because they share a common base (upper left side of the triangle) and overall height perpendicular to this side of the triangle. Arguing similarly for the right side of the triangle, we prove that the lower parallelogram has the same area as the two green parallelograms.

Complex numbers

The Pythagorean theorem is used to find the distance between two points in a Cartesian coordinate system, and this theorem is true for all true coordinates: distance s between two points ( a, b) and ( c, d) equals

There is no problem with the formula if you treat complex numbers as vectors with real components x + i y = (x, y). ... For example distance s between 0 + 1 i and 1 + 0 i we calculate as the modulus of the vector (0, 1) − (1, 0) = (−1, 1), or

Nevertheless, for operations with vectors with complex coordinates, it is necessary to make a certain improvement to the Pythagorean formula. Distance between points with complex numbers ( a, b) and ( c, d); a, b, c, and d all complex, we will formulate using absolute values. Distance s based on vector difference (a − c, b − d) in the following form: let the difference a − c = p+ i q, where p- the real part of the difference, q is the imaginary part, and i = √ (−1). Similarly, let b − d = r+ i s... Then:

where is the complex conjugate number for. For example, the distance between points (a, b) = (0, 1) and (c, d) = (i, 0) , we will calculate the difference (a − c, b − d) = (−i, 1) and as a result we would get 0 if complex conjugates were not used. Therefore, using the improved formula, we get

The module is defined as follows:

Stereometry

A significant generalization of the Pythagorean theorem for three-dimensional space is de Gua's theorem, named after J.-P. de Gua: if the tetrahedron has a right angle (as in a cube), then the square of the area of ​​the face lying opposite the right angle is equal to the sum of the squares of the areas of the other three faces. This conclusion can be summarized as “ n-dimensional Pythagorean theorem ":

Pythagoras' theorem three-dimensional space connects the diagonal AD with three sides.

Another generalization: The Pythagorean theorem can be applied to stereometry in the following form. Consider a rectangular parallelepiped, as shown in the figure. Let us find the length of the diagonal BD by the Pythagorean theorem:

where the three sides form a right-angled triangle. We use the horizontal diagonal BD and the vertical edge AB to find the length of the diagonal AD, for this we again use the Pythagorean theorem:

or, if everything is written in one equation:

This result is a 3D expression for determining the magnitude of a vector v(diagonal AD) expressed in terms of its perpendicular components ( v k) (three mutually perpendicular sides):

This equation can be viewed as a generalization of the Pythagorean theorem for multidimensional space. However, the result is in fact nothing more than a repeated application of the Pythagorean theorem to a sequence of right-angled triangles in successively perpendicular planes.

Vector space

In the case of an orthogonal system of vectors, the equality holds, which is also called the Pythagorean theorem:

If is the projection of the vector onto the coordinate axes, then this formula coincides with the Euclidean distance - and means that the length of the vector is equal to the square root of the sum of the squares of its components.

An analogue of this equality in the case of an infinite system of vectors is called Parseval's equality.

Non-euclidean geometry

The Pythagorean theorem is derived from the axioms of Euclidean geometry and, in fact, is not valid for non-Euclidean geometry, in the form in which it is written above. (That is, the Pythagorean theorem turns out to be a kind of equivalent to Euclid's postulate of parallelism) In other words, in non-Euclidean geometry, the ratio between the sides of a triangle will necessarily be in a form different from the Pythagorean theorem. For example, in spherical geometry, all three sides of a right triangle (say a, b and c), which limit the octant (eighth part) of the unit sphere, have length π / 2, which contradicts the Pythagorean theorem, because a 2 + b 2 ≠ c 2 .

Consider here two cases of non-Euclidean geometry - spherical and hyperbolic geometry; in both cases, as in the Euclidean space for right-angled triangles, a result that replaces the Pythagorean theorem follows from the cosine theorem.

However, the Pythagorean theorem remains valid for hyperbolic and elliptic geometry, if the requirement for the rectangularity of the triangle is replaced by the condition that the sum of the two angles of the triangle must equal the third, say A+B = C... Then the ratio between the sides looks like this: the sum of the areas of circles with diameters a and b equal to the area of ​​a circle with a diameter c.

Spherical geometry

For any right-angled triangle on a sphere of radius R(for example, if the angle γ in a triangle is a straight line) with sides a, b, c the relationship between the parties will look like this:

This equality can be derived as a special case of the spherical cosine theorem, which is true for all spherical triangles:

where cosh is the hyperbolic cosine. This formula is a special case of the hyperbolic cosine theorem, which is valid for all triangles:

where γ is the angle whose vertex is opposite to the side c.

where g ij is called the metric tensor. It can be a function of position. Such curvilinear spaces include Riemannian geometry as general example... This formulation is also suitable for Euclidean space when using curvilinear coordinates. For example, for polar coordinates:

Vector product

The Pythagorean theorem connects two expressions for the magnitude of a vector product. One approach to defining a cross product requires that it satisfy the equation:

this formula uses dot product. The right side of the equation is called the Gram determinant for a and b, which is equal to the area of ​​the parallelogram formed by these two vectors. Based on this requirement, as well as the requirement for the perpendicularity of the vector product to its components a and b it follows that, with the exception of trivial cases from 0- and 1-dimensional space, the vector product is defined only in three and seven dimensions. We use the definition of the angle in n-dimensional space:

this property of the vector product gives its value in the following form:

Through fundamental trigonometric identity Pythagoras, we get another form of recording its value:

An alternative approach to defining a cross product uses an expression for its magnitude. Then, arguing in reverse order, we get a connection with the dot product:

see also

Notes (edit)

1. History topic: Pythagoras’s theorem in Babylonian mathematics
2. (, P. 351) p. 351
3. (, Vol I, p. 144)
4. Discussion historical facts given in (, p. 351) p. 351
5. Kurt Von Fritz (Apr. 1945). "The Discovery of Incommensurability by Hippasus of Metapontum." The Annals of Mathematics, Second Series(Annals of Mathematics) 46 (2): 242–264.
6. Lewis Carroll, "A Story with Knots", M., Mir, 1985, p. 7
7. Asger aaboe Episodes from the early history of mathematics. - Mathematical Association of America, 1997. - P. 51. - ISBN 0883856131
8. Pythagorean Proposition, by Elisha Scott Loomis
9. Euclid's Elements: Book VI, Proposition VI 31: "In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle."
10. Lawrence S. Leff cited work... - Barron's Educational Series. - P. 326. - ISBN 0764128922
11. Howard whitley eves§4.8: ... generalization of Pythagorean theorem // Great moments in mathematics (before 1650). - Mathematical Association of America, 1983. - P. 41. - ISBN 0883853108
12. Tâbit ibn Qorra (full name Thābit ibn Qurra ibn Marwan Al-Ṣābiʾ al-Ḥarrānī) (826-901 AD) was a physician living in Baghdad who wrote extensively on Euclid's Elements and other mathematical subjects.
13. Aydin Sayili (Mar. 1960). "Thâbit ibn Qurra" s Generalization of the Pythagorean Theorem. " Isis 51 (1): 35–37. DOI: 10.1086 / 348837.
14. Judith D. Sally, Paul Sally Exercise 2.10 (ii) // Cited work. - P. 62. - ISBN 0821844032
15. For the details of such a construction, see George jennings Figure 1.32: The generalized Pythagorean theorem // Modern geometry with applications: with 150 figures. - 3rd. - Springer, 1997. - P. 23. - ISBN 038794222X
16. Arlen Brown, Carl M. Pearcy Item C: Norm for an arbitrary n-tuple ... // An introduction to analysis. - Springer, 1995. - P. 124. - ISBN 0387943692 See also pages 47-50.
17. Alfred Gray, Elsa Abbena, Simon Salamon Modern differential geometry of curves and surfaces with Mathematica. - 3rd. - CRC Press, 2006. - P. 194. - ISBN 1584884487
18. Rajendra Bhatia Matrix analysis. - Springer, 1997. - P. 21. - ISBN 0387948465
19. Stephen W. Hawking cited work... - 2005. - P. 4. - ISBN 0762419229
20. Eric W. Weisstein CRC concise encyclopedia of mathematics. - 2nd. - 2003. - P. 2147. - ISBN 1584883472
21. Alexander R. Pruss

Pythagorean theorem Is one of the fundamental theorems of Euclidean geometry, establishing the relation

between the sides of a right-angled triangle.

It is believed to have been proven by the Greek mathematician Pythagoras, after whom it was named.

Geometric formulation of the Pythagorean theorem.

Initially, the theorem was formulated as follows:

In a right-angled triangle, the area of ​​the square built on the hypotenuse is equal to the sum of the areas of the squares,

built on legs.

Algebraic formulation of the Pythagorean theorem.

In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.

That is, denoting the length of the hypotenuse of a triangle by c, and the lengths of the legs through a and b:

Both formulations Pythagorean theorems are equivalent, but the second formulation is more elementary, it is not

requires the concept of area. That is, the second statement can be verified without knowing anything about the area and

by measuring only the lengths of the sides of a right-angled triangle.

The converse theorem of Pythagoras.

If the square of one side of the triangle is equal to the sum of the squares of the other two sides, then

rectangular triangle.

Or, in other words:

For every three positive numbers a, b and c such that

there is a right-angled triangle with legs a and b and hypotenuse c.

Pythagoras' theorem for an isosceles triangle.

Pythagoras' theorem for an equilateral triangle.

Proofs of the Pythagorean theorem.

At the moment, 367 proofs of this theorem have been recorded in the scientific literature. Probably the theorem

Pythagoras is the only theorem with such an impressive number of proofs. Such diversity

can only be explained by the fundamental meaning of the theorem for geometry.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them:

proof area method, axiomatic and exotic evidence(for example,

by using differential equations).

1. Proof of the Pythagorean theorem through similar triangles.

The following proof of the algebraic formulation is the simplest of the proofs under construction

directly from the axioms. In particular, it does not use the concept of the area of ​​a figure.

Let be ABC there is a right-angled triangle with a right angle C... Let's draw the height from C and denote

its foundation through H.

Triangle ACH like a triangle AB C in two corners. Similarly, triangle CBH is similar ABC.

Introducing the notation:

we get:

,

which corresponds to -

By adding a 2 and b 2, we get:

or, as required to prove.

2. Proof of the Pythagorean theorem by the area method.

The proofs below, despite their seeming simplicity, are not at all so simple. All of them

use the properties of the area, the proof of which is more difficult than the proof of the Pythagorean theorem itself.

• Proof through equal complementarity.

Place four equal rectangular

triangle as shown in the figure

on right.

Quadrilateral with sides c- square,

since the sum of two acute angles is 90 °, and

expanded angle - 180 °.

The area of ​​the entire figure is, on the one hand,

area of ​​a square with side ( a + b), and on the other hand, the sum of the areas of the four triangles and

Q.E.D.

3. Proof of the Pythagorean theorem by the method of infinitesimal.

​

Considering the drawing shown in the figure, and

watching the side changea, we can

write the following relation for infinitely

small side incrementswith and a(using the similarity

triangles):

Using the variable separation method, we find:

A more general expression for changing the hypotenuse in the case of increments of both legs:

Integrating this equation and using the initial conditions, we get:

Thus, we arrive at the desired answer:

As it is easy to see, the quadratic dependence in the final formula appears due to the linear

proportionality between the sides of the triangle and the increments, while the sum is related to independent

contributions from the increment of different legs.

A simpler proof can be obtained if we assume that one of the legs does not experience an increment

(in this case, the leg b). Then for the constant of integration we get:

Average level

Right triangle. The Complete Illustrated Guide (2019)

RIGHT TRIANGLE. FIRST LEVEL.

In tasks, a right angle is not at all necessary - the lower left, so you need to learn how to recognize a right-angled triangle in this form,

and in such,

and in such

What good is there in a right triangle? Well ... first, there are special beautiful names for his parties.

Attention to the drawing!

Remember and don't confuse: legs - two, and the hypotenuse - only one(the one and only and the longest)!

Well, the names have been discussed, now the most important thing: the Pythagorean theorem.

Pythagorean theorem.

This theorem is the key to solving many problems involving a right-angled triangle. It was proved by Pythagoras in completely immemorial times, and since then it has brought many benefits to those who know it. And the best thing about her is that she is simple.

So, Pythagorean theorem:

Do you remember the joke: "Pythagorean pants are equal on all sides!"?

Let's draw these same Pythagorean pants and look at them.

Doesn't it look like some kind of shorts? Well, on which sides and where are they equal? Why and where did the joke come from? And this joke is connected precisely with the Pythagorean theorem, more precisely, with the way Pythagoras himself formulated his theorem. And he formulated it as follows:

"Sum squares built on legs is equal to square area built on the hypotenuse ”.

Doesn't it sound a little different? And so, when Pythagoras drew the statement of his theorem, just such a picture turned out.

In this picture, the sum of the areas of the small squares is equal to the area of ​​the large square. And so that the children better remember that the sum of the squares of the legs is equal to the square of the hypotenuse, someone witty and invented this joke about Pythagorean pants.

Why are we now formulating the Pythagorean theorem

Did Pythagoras suffer and talk about squares?

You see, in ancient times there was no ... algebra! There were no designations and so on. There were no inscriptions. Can you imagine how awful it was for the poor ancient disciples to memorize everything with words ??! And we can be glad that we have a simple formulation of the Pythagorean theorem. Let's repeat it again to remember it better:

It should be easy now:

The square of the hypotenuse is equal to the sum of the squares of the legs.

Well, the most important theorem about a right-angled triangle has been discussed. If you are interested in how it is proved, read the next levels of the theory, and now let's go further ... in dark forest... trigonometry! To the terrible words sine, cosine, tangent and cotangent.

Sine, cosine, tangent, cotangent in a right triangle.

In fact, it's not that scary at all. Of course, the "real" definitions of sine, cosine, tangent and cotangent should be found in the article. But I really don't want to, right? We can rejoice: to solve problems about a right-angled triangle, you can simply fill in the following simple things:

Why is it all about the corner? Where is the corner? In order to understand this, you need to know how statements 1 - 4 are written in words. Look, understand and remember!

1.
In fact, it sounds like this:

And what about the corner? Is there a leg that is opposite the corner, that is, the opposite (for the corner) leg? Of course have! This is a leg!

But what about the angle? Look closely. Which leg is adjacent to the corner? Of course, the leg. Hence, for the angle, the leg is adjacent, and

Now, attention! Look what we got:

You see how great:

Now let's move on to tangent and cotangent.

How can I write it down in words now? What is the leg in relation to the corner? Opposite, of course - it "lies" opposite the corner. And the leg? Adjacent to the corner. So what did we do?

See the numerator and denominator are reversed?

And now again the corners and made the exchange:

Summary

Let's briefly write down everything we have learned.

Pythagorean theorem:

The main theorem about a right-angled triangle is the Pythagorean theorem.

Pythagorean theorem

By the way, do you remember well what legs and hypotenuse are? If not, then look at the picture - refresh your knowledge

It is possible that you have already used the Pythagorean theorem many times, but have you ever wondered why such a theorem is true? How can I prove it? Let's do like the ancient Greeks. Let's draw a square with a side.

You see how cleverly we divided its sides into lengths and!

Now let's connect the marked points

Here we, however, have noted something else, but you yourself look at the drawing and think about why this is so.

What is the area of ​​the larger square? Right, . A smaller area? Of course, . The total area of ​​the four corners remains. Imagine that we took them two at a time and leaned them against each other with hypotenuses. What happened? Two rectangles. This means that the area of ​​the "scraps" is equal to.

Let's put it all together now.

Let's transform:

So we visited Pythagoras - we proved his theorem in an ancient way.

Right triangle and trigonometry

For a right-angled triangle, the following relationships hold:

The sine of an acute angle is equal to the ratio of the opposite leg to the hypotenuse

The cosine of an acute angle is equal to the ratio of the adjacent leg to the hypotenuse.

The tangent of an acute angle is equal to the ratio of the opposite leg to the adjacent leg.

The cotangent of an acute angle is equal to the ratio of the adjacent leg to the opposite leg.

And once again, all this is in the form of a plate:

It is very convenient!

Equality tests for right-angled triangles

I. On two legs

II. On the leg and hypotenuse

III. By hypotenuse and acute angle

IV. On a leg and a sharp corner

a)

b)

Attention! It is very important here that the legs are "appropriate". For example, if it is like this:

THEN TRIANGLES ARE NOT EQUAL, despite the fact that they have one of the same acute angle.

Need to in both triangles, the leg was adjacent, or in both triangles, opposite.

Have you noticed how the signs of equality of right triangles differ from the usual signs of equality of triangles? Take a look at the topic “and pay attention to the fact that for equality of“ ordinary ”triangles you need the equality of their three elements: two sides and an angle between them, two angles and a side between them or three sides. But for the equality of right-angled triangles, only two corresponding elements are enough. Great, isn't it?

The situation is approximately the same with the signs of the similarity of right-angled triangles.

Signs of the similarity of right-angled triangles

I. On a sharp corner

II. On two legs

III. On the leg and hypotenuse

Median in a right triangle

Why is this so?

Consider a whole rectangle instead of a right-angled triangle.

Let's draw a diagonal and consider a point - the point of intersection of the diagonals. What is known about the diagonals of a rectangle?

And what follows from this?

So it turned out that

1. - median:

Remember this fact! Helps a lot!

What's even more surprising is that the converse is also true.

What good can you get from the fact that the median drawn to the hypotenuse is equal to half of the hypotenuse? Let's look at the picture

Look closely. We have:, that is, the distances from the point to all three vertices of the triangle turned out to be equal. But in a triangle there is only one point, the distances from which about all three vertices of the triangle are equal, and this is the CENTER of the DESCRIBED CIRCLE. So what happened?

Let's start with this "besides ..."

Let's look at and.

But in such triangles all angles are equal!

The same can be said about and

Now let's draw it together:

What benefit can be derived from this "triple" similarity.

Well, for example - two formulas for the height of a right triangle.

Let's write down the relationship of the respective parties:

To find the height, we solve the proportion and get the first formula "Height in a right triangle":

So, let's apply the similarity:.

What happens now?

Again we solve the proportion and get the second formula:

Both of these formulas must be very well remembered and whichever is more convenient to apply. Let's write them down again

Pythagorean theorem:

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs:.

Signs of equality of right-angled triangles:

• on two legs:
• on the leg and hypotenuse: or
• along the leg and adjacent acute angle: or
• along the leg and the opposite acute angle: or
• by hypotenuse and acute angle: or.

Signs of the similarity of right-angled triangles:

• one sharp corner: or
• from the proportionality of the two legs:
• from the proportionality of the leg and the hypotenuse: or.

Sine, cosine, tangent, cotangent in a right triangle

• The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse:
• The cosine of an acute angle of a right triangle is the ratio of the adjacent leg to the hypotenuse:
• The tangent of an acute angle of a right-angled triangle is the ratio of the opposite leg to the adjacent one:
• The cotangent of an acute angle of a right-angled triangle is the ratio of the adjacent leg to the opposite one:.

Height of a right triangle: or.

In a right-angled triangle, the median drawn from the vertex of the right angle is half the hypotenuse:.

Area of ​​a right triangle:

• through the legs: