Regular Polygons

Regular polygon and circumscribed circle

Definition:

A polygon is termed inscribed in a circle if all its vertices lie on a common circle.

Theorem:

About the regular polygon, it is possible to circumscribe a circle and, moreover, only one.

Given:

A₁A₂A₃...A_n is the regular polygon

Prove:

there exists a unique circle of center at O and radius R such that the vertices of the regular polygon lie on this circle

! circle (O; R): A₁; A₂; A₃;...A_n circle (O; R)

Proof:

1) We draw bisectors of the angle A₁ and the angle A₂.

Since the polygon is regular, we see that A₁ = A₂

1 =2 =3 =4.

From the condition 1 =3 it follows that the triangle ΔA₁OA₂ - is isosceles, so A₁O = OA₂

Consider the triangle ΔA₂OA₃:

1) A₂O - common side

2) A₁A₂ = A₂A₃

3) 3 =4

Then, by the first theorem of the equality of the triangles

Δ A₁OA₂ = Δ A₂OA₃. Therefore, A₂O = A₃O.

By connecting each remaining vertex to the point O, it can be shown that all the triangles are equal.

Then A₁O =A₂O = A₃O = ... = A_nO

Since the point O is the center of the circle and the radius is R = A₁O =A₂O = A₃O = ... = A_nO , we see that,

! circle (O; R): A₁; A₂; A₃;...A_n circle (O; R)

Uniqueness:

Take any three vertices of the regular polygon; they form a triangle such that only one circle can be circumscribed about this triangle, it follows that only one circle can be circumscribed about the given polygon.

Given:

the regular polygon,

the arc AB= 60° ( AB = 60°)

AB is the side of the regular polygon

Find:

number of sides of the regular polygon, i.e. n = ?

Solution:

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Given:

the regular polygon,

1) AB = 36°

2) AB = 18°

AB is the side of the regular polygon

Find:

the number of sides of the polygon, i.e. n = ?

Solution:

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The inscribed and circumscribed circle in the regular polygon

Conclusion:

In regular polygons, the centers of the inscribed and circumscribed circles coincide.

Given:

A₁A₂A₃A₄A₅ is the regular pentagon

Prove:

A₁A₃ = A₁A₄

Proof:

By the definition of the regular polygon in a given pentagon, all sides and angles are equal.

Consider the triangles ΔA₁A₂A₃ and ΔA₁A₄A₅.

A₁A₂ = A₁A₅

A₂A₃ = A₅A₄

A₂ =A₅

Then, according to the first theorem of the equality of the triangles (ΔA₁A₂A₃ = ΔA₁A₄A₅) it follows that A₁A₃ = A₁A₄ as the corresponding sides.

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Given:

ABCD is the square

AM : MK : KD = 1 : : 1

Prove:

MNOZLFEK is the regular polygon

Proof:

ΔAMN = ΔOBZ = ΔLCF = ΔEKD (by the first theorem of the equality of the triangles)

By the Pythagorean theorem:

MN = OZ = LF = EK =

By the condition NO = ZL = EF = MK =

Therefore, all sides are equal.

Since 1=2 = 45°, then NMK=MKE = KEF=EFL = FLZ=ZON = ... = 180° - 45° = 135°

It follows that MNOZLFEK is the regular octagon.

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The regular polygon and inscribed circle

Theorem:

In any regular polygon, it is possible to inscribe a circle, and moreover only one.

Given:

A₁A₂A₃...A_n is the regular polygon

Prove:

There is a unique inscribed circle of center at O and radius R

! circle (O; R): H₁; H₂; H₃;...H_n circle (O; R)

Proof:

1) We draw the heights of the triangles, i.e. OH₁; OH₂; ...; OH_n

It is known that the triangles are equal, i.e. ΔOA₁A₂ = ΔOA₂A₃ = ... = ΔOA_nA₁.

Therefore, OH₁ = OH₂ = ... = OH_n . Then H₁; H₂; H₃;...H_n circle (O; R).

Uniqueness:

2) Let us suppose that as well as the circle (O; R) there is another circle inscribed in a given polygon.

Then its center O₁ is equidistant from the sides of the polygon and coincides with the point O of the intersection of the bisectrixes lying on each angle of the polygon.

Therefore, the radius of this circle is equal to OH₁ and from this it follows that the circles coincide.

***

the regular polygon,

the arc AB= 72° ( AB = 72°)

AB is the side of the regular polygon

Find:

the number of sides of the polygon, i.e. n = ?

Solution:

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Formulas for calculating the area of the regular polygon, its side and the radius of the inscribed circle.

Given:

A₁A₂A₃...A_n is the regular polygon

R is the radius of the circumscribed circle

r is the radius of the inscribed circle

a_n is the side of the polygon

Prove:

1) the area of the regular polygon is equal to half the product of the perimeter of the polygon by the radius of the inscribed circle

S_n = P_n • r

2) the side of the regular polygon is equal to twice the product of the radius of the circumscribed circle by the sine of the angle equal to the number from the division of 180° by n - the number of sides of the polygon

a_n = 2R • Sin ()

3) the radius of the inscribed circle is equal to the product of the radius of the circumscribed circle by the cosine of the angle equal to the number from the division of 180° by n - the number of sides of the regular polygon

r = R • Cos ()

Proof:

1) By connecting the point O to the vertices of the regular polygon, we obtain triangles Δ A₁A₂O = Δ A₂A₃O = ... = Δ A₁A_nO, where the number of all triangles in the polygon = n.

The area of a triangle

S (Δ A₁A₂O) = • A₁A₂ • OH₁ = • a_n • r.

Then the area of the polygon is equal to the sum of the areas of all triangles

S_n = S (Δ A₁A₂O) • n = • a_n • r • n = (• a_n • n) • r = P_n • r

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2)

Since the angle in the polygon is given by formula

α_n = • 180°, we see that the angle A₁ in the triangle A₁H₁O is half the angle of the polygon.

A₁ = •(• 180°) = • 90° = = 90° -

Cos A₁ = =

Then A₁H₁ = Cos A₁ • R = Cos (90° - ) • R = R • Sin()

Knowing that a_n = A₁H₁ • 2, we obtain

a_n = 2R • Sin ()

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Examples:

If n=3, then a₃ = 2R • = R•

If n=4 (the square), then a₄ = 2R • Sin45° = 2R • = R•

If n=6 (the regular hexagon), then a₆ = 2R • 0,5 = R

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3)

Sin A₁ =

Then r = R • Sin A₁ = R • (90° - ) = R • Cos ()

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Given:

The regular quadrilateral is the square

what are the radii of the inscribed and circumscribed circumference, the sides, perimeter and area of the regular polygon – the square

R, r, a₄, P, S = ?

Solution:

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Given:

ABCD is the square,

the side of the square a₄ = 6

Calculate: what is the doubled radius of the inscribed circle 2r = ?

Solution:

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Given:

the square, the regular hexagon are inscribed in the circle (O;r) of the center at the point O and the radius r ;

the perimeter of the hexagon P₆ = 48 cm

Find: the perimeter of the square P₄ = ?

Solution:

Given:

ABCDEF is the regular hexagon

BF = 1.5 cm

Find: the area of the hexagon

S_ABCDEF = ?

Solution:

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Given:

A₁A₂A₃A₄A₅A₆ is a regular hexagon inscribed in a circle(O; R)

B₁B₂B₃B₄B₅B₆ is a regular hexagon, circumscribed about the circle(O; R)

OH = r

OH₁ = R

= ?

Solution:

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