Английская Википедия:Geometric mean theorem

Содержание

Текст

Текст

Текст

Текст

Текст

Текст

Текст

Текст

Текст

Файл:Hoehensatz2.svg

area of grey square = area of grey rectangle: <math> h^2=pq \Leftrightarrow h=\sqrt{pq}</math>

In Euclidean geometry, the right triangle altitude theorem or geometric mean theorem is a relation between the altitude on the hypotenuse in a right triangle and the two line segments it creates on the hypotenuse. It states that the geometric mean of the two segments equals the altitude.

Theorem and applications

Файл:Root construction geometric mean5.svg

Construction of Шаблон:Tmath by setting Шаблон:Mvar to 1

If Шаблон:Mvar denotes the altitude in a right triangle and Шаблон:Mvar and Шаблон:Mvar the segments on the hypotenuse then the theorem can be stated as:

<math>h=\sqrt{pq} </math>

or in term of areas:

<math>h^2=pq.</math>

Файл:Am gm half circle3.svg

AM-GM inequality

The latter version yields a method to square a rectangle with ruler and compass, that is to construct a square of equal area to a given rectangle. For such a rectangle with sides Шаблон:Mvar and Шаблон:Mvar we denote its top left vertex with Шаблон:Mvar. Now we extend the segment Шаблон:Mvar to its left by Шаблон:Mvar (using arc Шаблон:Mvar centered on Шаблон:Mvar) and draw a half circle with endpoints Шаблон:Mvar and Шаблон:Mvar with the new segment Шаблон:Math as its diameter. Then we erect a perpendicular line to the diameter in Шаблон:Mvar that intersects the half circle in Шаблон:Mvar. Due to Thales' theorem Шаблон:Mvar and the diameter form a right triangle with the line segment Шаблон:Mvar as its altitude, hence Шаблон:Mvar is the side of a square with the area of the rectangle. The method also allows for the construction of square roots (see constructible number), since starting with a rectangle that has a width of 1 the constructed square will have a side length that equals the square root of the rectangle's length.

Another application of provides a geometrical proof of the AM–GM inequality in the case of two numbers. For the numbers Шаблон:Mvar and Шаблон:Mvar one constructs a half circle with diameter Шаблон:Math. Now the altitude represents the geometric mean and the radius the arithmetic mean of the two numbers. Since the altitude is always smaller or equal to the radius, this yields the inequality.

Файл:Sehnensatz hoehensatz.svg

\Leftrightarrow h^2=pq</math>

The theorem can also be thought of as a special case of the intersecting chords theorem for a circle, since the converse of Thales' theorem ensures that the hypotenuse of the right angled triangle is the diameter of its circumcircle.

The converse statement is true as well. Any triangle, in which the altitude equals the geometric mean of the two line segments created by it, is a right triangle.

History

The theorem is usually attributed to Euclid (ca. 360–280 BC), who stated it as a corollary to proposition 8 in book VI of his Elements. In proposition 14 of book II Euclid gives a method for squaring a rectangle, which essentially matches the method given here. Euclid however provides a different slightly more complicated proof for the correctness of the construction rather than relying on the geometric mean theorem.

Proof

Based on similarity

Файл:Hoehensatz.svg

Шаблон:Center

Proof of theorem:

The triangles Шаблон:Math are similar, since:

• consider triangles Шаблон:Math; here we have <math display=block>\angle ACB=\angle ADC=90^\circ, \quad \angle BAC=\angle CAD;</math> therefore by the AA postulate <math display=block>\triangle ABC \sim \triangle ACD .</math>
• further, consider triangles Шаблон:Math; here we have <math display=block>\angle ACB=\angle BDC= 90^\circ, \quad \angle ABC=\angle CBD;</math> therefore by the AA postulate <math display=block>\triangle ABC \sim \triangle BCD.</math>

Therefore, both triangles Шаблон:Math are similar to Шаблон:Math and themselves, i.e. <math display=block>\triangle ACD \sim \triangle ABC \sim \triangle BCD.</math>

Because of the similarity we get the following equality of ratios and its algebraic rearrangement yields the theorem:

<math> \frac{h}{p}=\frac{q}{h}\,\Leftrightarrow\,h^2=pq\,\Leftrightarrow\,h=\sqrt{pq}\qquad (h,p,q> 0)</math>

Proof of converse:

For the converse we have a triangle Шаблон:Math in which <math>h^2=pq</math> holds and need to show that the angle at Шаблон:Mvar is a right angle. Now because of <math>h^2=pq</math> we also have <math>\tfrac{h}{p}=\tfrac{q}{h}.</math> Together with <math>\angle ADC=\angle CDB </math> the triangles Шаблон:Math have an angle of equal size and have corresponding pairs of legs with the same ratio. This means the triangles are similar, which yields:

<math>\begin{align}

\angle ACB &= \angle ACD +\angle DCB \\

          &= \angle ACD+(90^\circ-\angle DBC) \\
          &= \angle ACD+(90^\circ-\angle ACD) \\
          &= 90^\circ

\end{align}</math>

Based on the Pythagorean theorem

Файл:Hoehensatz beweis pythagoras.svg

Proof with the Pythagorean theorem

In the setting of the geometric mean theorem there are three right triangles Шаблон:Math, Шаблон:Math and Шаблон:Math in which the Pythagorean theorem yields:

<math>\begin{align}

h^2 &= a^2-q^2 \\ h^2 &= b^2-p^2 \\ c^2 &= a^2+b^2 \end{align}</math> Adding the first 2 two equations and then using the third then leads to:

<math>\begin{align}

2h^2 &= a^2+b^2-p^2-q^2 \\

    &= c^2-p^2-q^2 \\
    &= (p+q)^2-p^2-q^2 \\
    &= 2pq \\

\therefore \ h^2 &= pq. \end{align}</math> which finally yields the formula of the geometric mean theorem.

Based on dissection and rearrangement

Файл:Geometrischer Höhensatzbeweis.svg

Dissecting the right triangle along its altitude Шаблон:Mvar yields two similar triangles, which can be augmented and arranged in two alternative ways into a larger right triangle with perpendicular sides of lengths Шаблон:Math and Шаблон:Math. One such arrangement requires a square of area Шаблон:Math to complete it, the other a rectangle of area Шаблон:Mvar. Since both arrangements yield the same triangle, the areas of the square and the rectangle must be identical.

Based on shear mappings

The square of the altitude can be transformed into an rectangle of equal area with sides Шаблон:Mvar and Шаблон:Mvar with the help of three shear mappings (shear mappings preserve the area):

Файл:Scherungen alle2.svg

Shear mappings with their associated fixed lines (dotted), starting with the original square as preimage each parallelogram displays the image of a shear mapping of the figure left of it

External links

• Geometric Mean at Cut-the-Knot