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.

***

 

Problem 113.

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:

Since the degree measure AB = 60° < 180°, then the arc is equal to the angle, i.e.

AB = AOB.

ΔAOB - is isosceles, where OAB = OBA =

= (180° - 60°) : 2 = 60°.

Then ΔAOB is equilateral.

The radii of a circle circumscribed about the regular polygon are the bisectors of its angles, so each angle of the polygon is 60° • 2 = 120°.

Knowing that

α_n = • 180°

120° •n = (n - 2) • 180°

120° •n - 180°•n = - 360°

-60° •n = - 360°

n = 6

Answer: the number of sides of the regular polygon is n = 6.

***

 

Problem 114.

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:

1)

Since the degree measure AB = 36° < 180°, we see that the arc is equal to the angle, i.e.

AB = AOB.

ΔAOB is isosceles, where OAB = OBA =

= (180° - 36°) : 2 = 72°.

The radii of the circle circumscribed about the polygon are the bisectors of its angles, so each angle of the polygon is 72° • 2 = 144°.

Knowing that

α_n = • 180°

144° •n = (n - 2) • 180°

144° •n - 180°•n = - 360°

-36° •n = - 360°

n = 10

 

2)

Since the degree measure AB = 18° < 180°, then the arc is equal to the angle, i.e.

AB = AOB, where the angle AOB is central.

ΔAOB is isosceles (OA = OB = r), where OAB =

=OBA = (180° - 18°) : 2 = 81°.

The radii of the circle circumscribed about the polygon are the bisectors of its angles, so each angle of the polygon is 81° • 2 = 162°.

Knowing that

α_n = • 180°

162° •n = (n - 2) • 180°

162° •n - 180°•n = - 360°

-18° •n = - 360°

n = 20

Answer: 1) n = 10; 2) n = 20.

***

 

The inscribed and circumscribed circle in the regular polygon

Conclusion:

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

 

Problem 115.

Given:

A₁A₂A₃A₄A₅ is the regular pentagon

α_n = • 180° = 108°

 

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.

***

 

Problem 116.

Given:

ABCD is the square

The ratio of the segments

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.

***

 

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.

***

 

Problem 117.

Given:

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:

Since the degree measure AB = 72° < 180°, then the arc is equal to the angle, i.e.

AB = AOB, where the angle AOB is central.

ΔAOB is isosceles (OA = OB = r), where

OAB = OBA = (180° - 72°) : 2 = 54°.

Then ΔAOB is equilateral.

The radii of the circle circumscribed about the polygon are the bisectors of its angles, so each angle of the polygon is 54° • 2 = 108°.

Knowing that

α_n = • 180°

108° •n = (n - 2) • 180°

108° •n - 180°•n = - 360°

-72° •n = - 360°

n = 5

Answer: n = 5.

***

 

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

***

 

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 ()

***

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

***

3)

Sin A₁ =

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

***

 

Problem 118.

Given:

The regular quadrilateral is the square

 

	R	r	a4	P	S
1	-	-	6	-	-
2	-	2	-	-	-
3	-	-	-	-	16

 

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:

1) If the side of the square a₄ = 6 and since given is the square, we see that

the perimeter of the square P = 4 • a₄ = 4 • 6 = 24

The area of the square S = a₄² = 6² = 36

Since the area of the square can be found by the formula S₄ = P₄ • r, we see that the radius of the inscribed circle

r = = = 3

Since the number of sides n=4, we see that a₄ = R•

The radius of the circumscribed circle R== ==

 

2) If r=2, then the radius of the circumscribed circle

r = R • Cos () = R • Cos 45° = R •

R = 2 • = 2

Since the number of sides n=4, then the side of the square

a₄ = R•= 2• = 4

Then the perimeter of the square P = 4 • a₄ = 4 • 4 = 16

The area of the square S = a₄² = 4² = 16

 

3) If the area of the square S = 16, then the side of the square

a₄² = 16 a₄ = ±. But a₄ = - does not satisfy the condition of the problem.

Therefore, a₄ = = 4

Then the perimeter of the square P = 4 • a₄ = 4 • 4 = 16

Since the number of sides n=4, then the radius of the circumscribed circle

a₄ = R• 4 = R• R = = 2

Then the radius of the inscribed circle

r = R • Cos () = R • Cos 45° = 2 • = 2

***

 

Problem 119.

Given:

Δ ABC is equilateral,

the perimeter of the triangle

P_Δ = 18 cm

 

Find: the side of a square inscribed in the same circle

a₄ = ?

Solution:

Since P_n = n • a_n , we see that the side of the triangle

18 = 3 • a₃ a₃ = 6

Then the radius of the circumscribed circle

a₃ = R• R = = 2

Then the side of the square a₄ = • R = • 2= 2

Answer: 2.

***

 

Problem 120.

Given:

ABCD is the square,

the side of the square a₄ = 6

 

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

Solution:

Since the area of the square can be found by the formula S₄ = P₄ • r, then the radius of the inscribed circle

r = = = 3 (cm)

Therefore, 2r = 2 • 3 = 6 (cm).

Answer: 6 cm.

***

 

Problem 121.

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:

Since the formula of the perimeter of the hexagon P₆ = 6 • a₆ , then the side of the hexagon a₆ = 48 : 6 = 8 (cm)

Since given is a regular hexagon for n=6, then

a₆ = 2R • 0.5 = R

Since given is one circle, the radii of the circumscribed circle are equal, i.e. R₄ = R₆ = 8 cm

Then the side of the square a₄ = R•= 8cm

Therefore, the perimeter of the square P₄ = 4 • 8= 32(cm)

Answer: P₄ = 32cm.

***

 

Problem 122.

Given:

ABCDEF is the regular hexagon

BF = 1.5 cm

 

Find: the area of the hexagon

S_ABCDEF = ?

Solution:

We draw the diagonal AD and consider the isosceles triangle ΔABF.

By the property of an isosceles triangle AO - is the median.

Therefore, FO = OB = BF : 2 = 1.5 : 2 = 0.75 (cm)

 

Since the hexagon is regular, we see that using the formula to calculate the angles of the regular polygon

α_n = • 180°, we obtain

α₆ = • 180° = 120°

Then the angle BAO = 60°

Consider the triangle ΔAOB, where the angle O = 90°.

Sin 60° = = AB = = (cm)

Then the perimeter of the hexagon

P₆ = 6 • = 3(cm)

Then the radius of the inscribed circle

r₆ = OF = cm

Therefore, the area of the hexagon

S₆ = • P₆ • r₆ = = (см²)

Answer: S_ABCDEF = cm².

***

 

Problem 123.

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

 

Find: the proportion of areas of hexagons

= ?

Solution:

1)

Consider the hexagon A₁A₂A₃A₄A₅A₆

The sides of the hexagon are equal to the radii of the circumscribed circle A₁A₂ = A₂A₃ = A₆A₁ = R

Knowing that S (A₁A₂A₃A₄A₅A₆) = P • r = 6Rr = 3R•r

Consider the triangle ΔOHA₆ is right, since the radius of the inscribed circle is perpendicular to the side of the hexagon r A₁A₆

Since the point O - is the point of intersection of bisectors, we see that R - is the bisector of the angle A₁A₆A₅

Knowing that OA₆A₁ = 60°, we obtain

Sin 60° = r = Sin 60° • R =

Therefore, the area of the hexagon

S (A₁A₂A₃A₄A₅A₆) = 3R •=R²

2)

Consider the hexagon B₁B₂B₃B₄B₅B₆

B₁B₂ = ... = B₆B₁ = OB₅ , where OB₅ - is the radius of the circumscribed circle about a given regular polygon, namely the hexagon B₁B₂B₃B₄B₅B₆

S (B₁B₂B₃B₄B₅B₆) = • (6 • OB₅) • R

Consider ΔH₁OB₅ - is a right triangle, since R - of the inscribed circle in the polygon B₁B₂B₃B₄B₅B₆:

R B₅B₆

Then OB₅H₁= 60°. Therefore

Sin 60° = OB₅ = 2R

Then the area of the hexagon is B₁B₂B₃B₄B₅B₆

S (B₁B₂B₃B₄B₅B₆) = • 6 • 2R • R = 2R²

 

Then the proportion of the areas of the hexagons

= 2R² • =

Answer:

***