geometry theorems

AMIYA

GEOMETRY CONCEPTS &

THEOREMS

3 E L E A R N I N G , 3 R D F L O O RH . B . R O A D

GEOMETRY CONCEPTS &

REMS COMPILATION MATHS BY AMIYA

Amiya

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Barbier's theorem

In geometry, Barbier's theorem states that every curve of constant width

regardless of its precise shape.

Curve of constant width :- In geometry, a curve

(defined as the perpendicular distance between two distinct parallel lines each having at least one point in

common with the shape's boundary but none with the shape's interior) is the same regardless

orientation of the curve.

More generally, any compact convex planar body D has one pair of parallel supporting lines in any given

direction. A supporting line is a line that has at least one point in common with the boundary of D but no

points in common with the interior of D. The width of the body is defined as before. If the width of D is the

same in all directions, the body is said to have constant width and its boundary is a curve of constant width;

the planar body itself is called an orbiform.

The width of a circle is constant: its diameter. On the other hand, the width of a square varies between the

length of a side and that of a diagonal, in the ratio

constant in all directions, is it necessarily a circle? The surprising answer is that there are many non

circular shapes of constant width. A nontrivial example is the Reuleaux triangle. To construct this, take an

equilateral triangle with vertices ABC and draw the arc BC on the circle cen

circle centered at B, and the arc AB on the circle centered at C. The resulting figure is of constant width.

The Reuleaux triangle lacks tangent continuity at three points, but constant

constructed without such discontinuities (as shown in the second illustration on the right). Curves of

constant width can be generated by joining circular arcs centered on the vertices of a regular or irregular

convex polygon with an odd number of sides (triangle, penta

The most familiar examples of curves of constant width are the circle

and the Reuleaux triangle. For a circle, the width is the same as the

diameter; a circle of width w has perimeter w. A Reuleaux triangle

of width w consists of three arcs of circles of radius w. Each of these

arcs has central angle /3, so the perimeter of the Reuleaux triangle

of width w is equal to half the perimeter of a circle of radius w and

therefore is equal to w. A similar analysis of other simple exampl

such as Reuleaux polygons gives the same answer.

The analogue of Barbier's theorem for

false. In particular, the unit sphere has surface area

while the surface of revolution of a Reuleaux triangle

constant width has surface area

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These Reuleaux polygons have

constant width, and all have the

same width; therefore by

Barbier's theorem they also have

equal perimeters.

In geometry, Barbier's theorem states that every curve of constant width* has perimeter times its width,

In geometry, a curve of constant width is a convex planar shape whose width


common with the shape's boundary but none with the shape's interior) is the same regardless



common with the interior of D. The width of the body is defined as before. If the width of D is the


the planar body itself is called an orbiform.


length of a side and that of a diagonal, in the ratio

. Thus the question arises: if a given shape's width is

necessarily a circle? The surprising answer is that there are many non


equilateral triangle with vertices ABC and draw the arc BC on the circle centered at A, the arc CA on the


The Reuleaux triangle lacks tangent continuity at three points, but constant-width curves can also be

thout such discontinuities (as shown in the second illustration on the right). Curves of


convex polygon with an odd number of sides (triangle, pentagon, heptagon, etc.).

The most familiar examples of curves of constant width are the circle

and the Reuleaux triangle. For a circle, the width is the same as the

diameter; a circle of width w has perimeter w. A Reuleaux triangle

three arcs of circles of radius w. Each of these

arcs has central angle /3, so the perimeter of the Reuleaux triangle

of width w is equal to half the perimeter of a circle of radius w and

therefore is equal to w. A similar analysis of other simple examples

such as Reuleaux polygons gives the same answer.

The analogue of Barbier's theorem for surfaces of constant width is

has surface area ,

Reuleaux triangle with the same

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These Reuleaux polygons have

constant width, and all have the

same width; therefore by

Barbier's theorem they also have

equal perimeters.

has perimeter times its width,

of constant width is a convex planar shape whose width


common with the shape's boundary but none with the shape's interior) is the same regardless of the



common with the interior of D. The width of the body is defined as before. If the width of D is the



. Thus the question arises: if a given shape's width is

necessarily a circle? The surprising answer is that there are many non-


tered at A, the arc CA on the


width curves can also be

thout such discontinuities (as shown in the second illustration on the right). Curves of


gon, heptagon, etc.).


Brahmagupta theorem

In geometry, Brahmagupta's theorem

has perpendicular diagonals), then the perpendicular to a side from the point of intersection of the

diagonals always bisects the opposite side.

More specifically, let A, B, C and D

lines AC and BD are perpendicular. Denote the intersection of

AC and BD by M. Drop the perpendicular from

the intersection E. Let F be the intersection of the line

edge AD. Then, the theorem states that

Proof :-

We need to prove that AF = FD. We will prove that both

To prove that AF = FM, first note that the angles

because they are inscribed angles that intercept the same arc of the circle.

Furthermore, the angles CBM and CME

angle BCM (i.e., they add up to 90), and are therefore equal. Finally,

the angles CME and FMA are the same. Hence,

triangle, and thus the sides AF and FM

The proof that FD = FM goes similarly: the

angles FDM, BCM, BME and DMF

triangle, so FD = FM. It follows that

Brahmagupta's formula

Brahmagupta's formula gives the area

where s, the semiperimeter, is defined to be


Brahmagupta theorem

Brahmagupta's theorem states that if a cyclic quadrilateral is orthodiagonal

diagonals), then the perpendicular to a side from the point of intersection of the

the opposite side.

be four points on a circle such that the

are perpendicular. Denote the intersection of

. Drop the perpendicular from M to the line BC, calling

be the intersection of the line EM and the

e theorem states that F is the midpoint AD.

. We will prove that both AF and FD are in fact equal to

, first note that the angles FAM and CBM are equal,

that intercept the same arc of the circle.

CME are both complementary to

(i.e., they add up to 90), and are therefore equal. Finally,

are the same. Hence, AFM is an isosceles

FM are equal.

goes similarly: the

DMF are all equal, so DFM is an isosceles

. It follows that AF = FD, as the theorem claims.

Brahmagupta's formula

Brahmagupta's formula gives the area A of a cyclic quadrilateral whose sides have lengths

semiperimeter, is defined to be


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orthodiagonal (that is,

diagonals), then the perpendicular to a side from the point of intersection of the

are in fact equal to FM.

whose sides have lengths a, b, c, d as


This formula generalizes Heron's formula

regarded as a quadrilateral with one side of length zero. From this perspective, as

approaches zero, a cyclic quadrilateral converges into a cyclic triangle (all triangles are

cyclic), and Brahmagupta's formula simplifies to Heron's formula.

If the semiperimeter is not used, Brahmagupta's formula is

Another equivalent version is

Extension to non-cyclic quadrilaterals

In the case of non-cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the

measures of two opposite angles of the

where is half the sum of two opposite angles. (The choice of which pair of opposite angles is irrelevant: if

the other two angles are taken, half their

cos2(180 ) = cos

2.) This more general formula is known as

It is a property of cyclic quadrilaterals

quadrilateral sum to 180. Consequently, in the case of an inscribed quadrilater

It follows from the latter equation that the area of a cyclic quadrilateral is the maximum possible area for

any quadrilateral with the given side lengths.

A related formula, which was proved by

is

where p and q are the lengths of the diagonals of the quadrilateral. In a

according to Ptolemy's theorem, and the formula of Coolidge reduces


Heron's formula for the area of a triangle. A triangle may be

a quadrilateral with one side of length zero. From this perspective, as


cyclic), and Brahmagupta's formula simplifies to Heron's formula.

not used, Brahmagupta's formula is

cyclic quadrilaterals

cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the

measures of two opposite angles of the quadrilateral:

the sum of two opposite angles. (The choice of which pair of opposite angles is irrelevant: if

their sum is the supplement of . Since cos(180

.) This more general formula is known as Bretschneider's formula

cyclic quadrilaterals (and ultimately of inscribed angles) that opposite angles of a

quadrilateral sum to 180. Consequently, in the case of an inscribed quadrilateral, = 90, whence the term

giving the basic form of Brahmagupta's formula.


any quadrilateral with the given side lengths.

hich was proved by Coolidge, also gives the area of a general convex quadrilateral. It

are the lengths of the diagonals of the quadrilateral. In a cyclic uadrilateral

, and the formula of Coolidge reduces to Brahmagupta's formula.


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. A triangle may be

a quadrilateral with one side of length zero. From this perspective, as d


cyclic quadrilaterals, Brahmagupta's formula can be extended by considering the

the sum of two opposite angles. (The choice of which pair of opposite angles is irrelevant: if

sum is the supplement of . Since cos(180 ) = cos, we have

Bretschneider's formula.

) that opposite angles of a

al, = 90, whence the term

giving the basic form of Brahmagupta's formula.


, also gives the area of a general convex quadrilateral. It

cyclic uadrilateral,

to Brahmagupta's formula.


Butterfly theorem

The butterfly theorem is a classical result in

which can be stated as follows:

Let M be the midpoint of a chord

which two other chords AB and

drawn; AD and BC intersect chord

at X and Y correspondingly. Then

Proof:-

Let the perpendiculars and

and respectively. Similarly, let

from the point perpendicular to the straight lines

respectively.

Now, since

From the preceding equations, it can be easily seen that

since =

Now,

So, it can be concluded that


is a classical result in Euclidean geometry,

chord PQ of a circle, through

and CD are

chord PQ

correspondingly. Then M is the midpoint of XY.

be dropped from the point on the straight lines

respectively. Similarly, let and be dropped

perpendicular to the straight lines and

From the preceding equations, it can be easily seen that

or is the midpoint of


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on the straight lines


Casey's theorem

In mathematics, Casey's theorem, also known as the

generalized Ptolemy's theorem, is a theorem in

geometry named after the Irish mathematician

Let be a circle of radius . Let

that order) four non-intersecting circles that

tangent to it. Denote by the length of the exterior common

tangent of the circles .

Then:

Note that in the degenerate case, where all four circles reduce

to points, this is exactly Ptolemy's theorem

De Gua's theorem

De Gua's theorem is a three-dimensional analog of the

Gua de Malves.

If a tetrahedron has a right-angle corner (like the corner of a

square of the area of the face opposite the right

the squares of the areas of the other three faces.

The Pythagorean theorem and de Gua's theorem are special cases (

a general theorem about n-simplices


, also known as the

Ptolemy's theorem, is a theorem in Euclidean

mathematician John Casey.

be (in

intersecting circles that lie inside and

the length of the exterior common

Note that in the degenerate case, where all four circles reduce

Ptolemy's theorem.

dimensional analog of the Pythagorean theorem and named for

angle corner (like the corner of a cube), then the

square of the area of the face opposite the right-angle corner is the sum of

the squares of the areas of the other three faces.

and de Gua's theorem are special cases (n = 2, 3) of

simplices with a right angle corner.


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and named for Jean Paul de


Descartes' theorem

In geometry, Descartes' theorem states that for every four

the circles satisfy a certain quadratic equation. By solving this equation, one can construct a fourth circle

tangent to three given, mutually tangent circles. The theor

in 1643.

Definition of curvature

Kissing circles. Given three mutually tangent circles (

fourth tangent circle have? There are in general two possible answers (

Descartes' theorem is most easily stated in terms of the circles'

The curvature (or bend) of a circle is defined as

radius. The larger a circle, the smaller is the

and vice versa.

The plus sign in k = 1/r applies to a circle that is

other circles, like the three black circles in the image. For an

internally tangent circle like the big red circle, that

circles, the minus sign applies.

If a straight line is considered a degenerate

theorem also applies to a line and two circles that are all three mutually tangent, giving the radius of a third

circle tangent to the other two circles and the line.

If four circles are tangent to each other at six distinct points, and the circles have

curvatures ki (for i = 1, ..., 4), Descartes' theorem says:

When trying to find the radius of a fourth circle tangent to three given

rewritten as:

The sign reflects the fact that there are in general

line, one solution is positive and the other is either

negative; if negative, it represents a circle that circumscribes the

first three (as shown in the diagram above).

favor one solution over the other in any given problem.

SPECIAL CASE

If one of the three circles is replaced

one ki, say k3, is zero and drops out of

(2) then becomes much simpler:


Descartes' theorem

states that for every four kissing, or mutually tangent,

quadratic equation. By solving this equation, one can construct a fourth circle

tangent to three given, mutually tangent circles. The theorem is named after Ren Descartes, who stated it

Kissing circles. Given three mutually tangent circles (black), what radius can a

fourth tangent circle have? There are in general two possible answers (red).

rem is most easily stated in terms of the circles' curvatures.

) of a circle is defined as k = 1/r, where r is its

radius. The larger a circle, the smaller is the magnitude of its curvature,

applies to a circle that is externally tangent to the

other circles, like the three black circles in the image. For an

tangent circle like the big red circle, that circumscribes the other

degenerate circle with zero curvature (and thus infinite radius), Descartes'


circle tangent to the other two circles and the line.


4), Descartes' theorem says:

..........................

When trying to find the radius of a fourth circle tangent to three given kissing circles, the equation is best

..........................

The sign reflects the fact that there are in general two solutions. Ignoring the degenerate case of a straight

line, one solution is positive and the other is either positive or

negative; if negative, it represents a circle that circumscribes the

first three (as shown in the diagram above). Other criteria may

favor one solution over the other in any given problem.

If one of the three circles is replaced by a straight line, then

, is zero and drops out of equation (1). Equation


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tangent, circles, the radii of

quadratic equation. By solving this equation, one can construct a fourth circle

Ren Descartes, who stated it

circle with zero curvature (and thus infinite radius), Descartes'



..........................(1)

kissing circles, the equation is best

..........................(2)

solutions. Ignoring the degenerate case of a straight


Japanese theorem for cyclic polygons

In geometry, the Japanese theorem

the sum of inradii of triangles is constant.

sum of the radii of the green circles = sum of the radii of the red circles

Conversely, if the sum of inradii independent from the triangulation, then the polygon is cyclic.

Carnot's theorem

In Euclidean geometry, Carnot's theorem

after Lazare Crarnot (17531823), is as follows. Let

an arbitrary triangle. Then the sum of the

distances from the circumcenter D to the sides of

triangle ABC is

where r is the inradius and R is the circumradius. Here the

sign of the distances is taken negative if and only if the

segment DX (X = F, G, H) lies completely outside the

triangle. In the picture DF is negative and

both DG and DH are positive.


Japanese theorem for cyclic polygons

Japanese theorem states that no matter how one triangulates a cyclic

constant.

sum of the radii of the green circles = sum of the radii of the red circles


Carnot's theorem, named

1823), is as follows. Let ABC be

triangle. Then the sum of the signed

to the sides of

circumradius. Here the

is taken negative if and only if the line

) lies completely outside the

is negative and


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cyclic polygon,



Pitot theorem

A tangential quadrilateral ABCD is a closed figure of four straight

sides that are tangent to a given circle

inscribed in the quadrilateral ABCD.

This conclusion follows from the equality of the tangent segments

from the four vertices of the quadrilateral. Let the tangent points be

denoted as P (on segment AB), Q (on segment BC),

CD) and S (on segment DA). The symmetric tangent segments about

each point of ABCD are equal, e.g., BP=BQ=

DR=DS=d, and AS=AP=a. But each side of the quadrilateral is composed of two such tangent segments

The converse is also true: a circle can be inscribed into every quadrilateral in which the lengths of opposite

sides sum to the same value.

This theorem and its converse have various uses. For example, they show immediately that no rectangle

can have an inscribed circle unless it is a

a general parallelogram does not.

Japanese theorem for cyclic quadrilaterals

In geometry, the Japanese theorem

the incircles of certain triangles inside a

quadrilateral are vertices of a rectangle.

Triangulating an arbitrary concyclic quadrilateral by its

diagonals yields four overlapping triangle

diagonal creates two triangles). The centers of the

incircles of those triangles form a rectangle.

Specifically, let be an arbitrary concyclic

quadrilateral and let be

incenters of the triangles

Then the quadrilateral formed by


ABCD is a closed figure of four straight

sides that are tangent to a given circle C. Equivalently, the circle C is

in the quadrilateral ABCD.

This conclusion follows from the equality of the tangent segments

ces of the quadrilateral. Let the tangent points be

(on segment BC), R (on segment

(on segment DA). The symmetric tangent segments about

each point of ABCD are equal, e.g., BP=BQ=b, CQ=CR=c,

But each side of the quadrilateral is composed of two such tangent segments

a circle can be inscribed into every quadrilateral in which the lengths of opposite

various uses. For example, they show immediately that no rectangle

can have an inscribed circle unless it is a square, and that every rhombus has an inscribed circle, whereas


Japanese theorem states that the centers of

inside a cyclic

are vertices of a rectangle.

Triangulating an arbitrary concyclic quadrilateral by its

diagonals yields four overlapping triangles (each

diagonal creates two triangles). The centers of the

incircles of those triangles form a rectangle.

be an arbitrary concyclic

the

.

is a rectangle.


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But each side of the quadrilateral is composed of two such tangent segments

a circle can be inscribed into every quadrilateral in which the lengths of opposite

various uses. For example, they show immediately that no rectangle

square, and that every rhombus has an inscribed circle, whereas



Pick's theorem

Given a simple polygon constructed on a grid of equal

with integer coordinates) such that all the polygon's vertices are grid points,

simple formula for calculating the area

interiorlocated in the polygon and the number

boundary placed on the polygon's perimeter

In the example shown, we have i = 7 interior points and

points, so the area is A = 7 + 8/2 1 =

Note that the theorem as stated above is only valid for

i.e., ones that consist of a single piece and do not contain "holes". For a polygon that has

boundary in the form of h + 1 simple closed curves, the slightly more complicated formula

gives the area.

Ptolemy's theorem

In Euclidean geometry, Ptolemy's theorem

a cyclic quadrilateral (a quadrilateral whose

The theorem is named after

the Greek astronomer and mathematician

Ptolemaeus) Ptolemy used the theorem as an aid to creating

of chords, a trigonometric table that he applied to astronomy.

If the quadrilateral is given with its four vertices

order, then the theorem states that:

where the vertical lines denote the lengths of the line segments between the named

This relation may be verbally expressed as follows:

If a quadrilateral is inscribable in a circle then the product of the measures of its diagonals is equal to the

sum of the products of the measures of the pairs of opposite sides.

Moreover, the converse of Ptolemy's theorem is also true:

In a quadrilateral, if the sum of the products of its two pairs of opposite sides is equal to the product of its

diagonals, then the quadrilateral can be inscribed in a circle.


constructed on a grid of equal-distanced points (i.e., points

coordinates) such that all the polygon's vertices are grid points, Pick's theorem

area A of this polygon in terms of the number i

located in the polygon and the number b of lattice points on the

placed on the polygon's perimeter

= 7 interior points and b = 8 boundary

1 = 7 + 4 1 = 10 (square units)

Note that the theorem as stated above is only valid for simple polygons,

i.e., ones that consist of a single piece and do not contain "holes". For a polygon that has

simple closed curves, the slightly more complicated formula

Ptolemy's theorem

Ptolemy's theorem is a relation between the four sides and two diagonals of

(a quadrilateral whose vertices lie on a common circle).

mathematician Ptolemy (Claudius

Ptolemaeus) Ptolemy used the theorem as an aid to creating his table

tric table that he applied to astronomy.

If the quadrilateral is given with its four vertices A, B, C, and D in

where the vertical lines denote the lengths of the line segments between the named

relation may be verbally expressed as follows:


sum of the products of the measures of the pairs of opposite sides.

olemy's theorem is also true:


diagonals, then the quadrilateral can be inscribed in a circle.


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0

distanced points (i.e., points

Pick's theorem provides a

i of lattice points in the

i.e., ones that consist of a single piece and do not contain "holes". For a polygon that has h holes, with a

simple closed curves, the slightly more complicated formula i + b/2 + h 1

is a relation between the four sides and two diagonals of

where the vertical lines denote the lengths of the line segments between the named vertices.




Varignon's theorem

Varignon's theorem is a statement in

geometry by Pierre Varignon that was first

published in 1731. It deals with the construction of a

particular parallelogram (Varignon parallelogram

from an arbitrary quadrangle.

The midpoints of the sides of an arbitrary

quadrangle form a parallelogram. If the quadrangle

is convex or reentrant, i.e. not a crossing

quadrangle, then the area of the parallelogram is

half as big as the area of the quadrangle

Viviani's theorem

Viviani's theorem, named after Vincenzo Viviani,

distances from any interior point to the sides of anequilateral triangle

length of the triangle's altitude.

Proof

This proof depends on the readily-proved proposition that the area of

a triangle is half its base times its height

side.

Let ABC be an equilateral triangle whose height is

Let P be any point inside the triangle, and

each of A, B, and C, forming three triangles PAB, PBC, and PCA.

Now, the areas of these triangles are

sum of these areas is equal to the area of the enclosing triangle. So we can w

and thus u + s + t = h.


Varignon's theorem

statement in Euclidean

that was first

published in 1731. It deals with the construction of a

Varignon parallelogram)

The midpoints of the sides of an arbitrary

form a parallelogram. If the quadrangle

is convex or reentrant, i.e. not a crossing

quadrangle, then the area of the parallelogram is

half as big as the area of the quadrangle.

Vincenzo Viviani, states that the sum of the

interior point to the sides of anequilateral triangle equals the

proved proposition that the area of

es its heightthat is, half the product of one side with the altitude from that

Let ABC be an equilateral triangle whose height is h and whose side is a.

Let P be any point inside the triangle, and u, s, t the distances of P from the sides. Draw a l

each of A, B, and C, forming three triangles PAB, PBC, and PCA.

Now, the areas of these triangles are , , and . They exactly fill the enclosing triangle, so the

sum of these areas is equal to the area of the enclosing triangle. So we can write:


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1

that is, half the product of one side with the altitude from that

the distances of P from the sides. Draw a line from P to

. They exactly fill the enclosing triangle, so the


Barrow's inequalityBarrow's inequality is an inequality

arbitrary point within a triangle, the vertices of the triangle, and

certain points on the sides of the triangle.

Let P be an arbitrary point inside the

define U, V, and W as the points where the

and APB intersect the sides BC,CA, AB

inequality states that

with equality holding only in the case of an

Euler's theorem in geometry

Euler's theorem states that the distance

expressed as

where R and r denote the circumradius and inradius respectively (the radii of the above two circles).

From the theorem follows the Euler inequality

Weitzenbck's inequality

Weitzenbck's inequality, named after

, , , and area , the following inequality holds:

Equality occurs if and only if the triangle is equilateral.


Barrow's inequality inequality relating the distances between

triangle, the vertices of the triangle, and

certain points on the sides of the triangle.

arbitrary point inside the triangle ABC. From P and ABC,

as the points where the angle bisectors of BPC, CPA,

AB, respectively. Then Barrow's

case of an equilateral triangle

Euler's theorem in geometry

states that the distance d between the circumcentre and incentre

denote the circumradius and inradius respectively (the radii of the above two circles).

Euler inequality:

Weitzenbck's inequality

, named after Roland Weitzenbck, states that for a triangle of

, the following inequality holds:

Equality occurs if and only if the triangle is equilateral.


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an

incentre of a triangle can be

denote the circumradius and inradius respectively (the radii of the above two circles).

Roland Weitzenbck, states that for a triangle of side lengths


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geometry theorems

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width of d

curves of constant width

circle of width w

reuleaux triangle of

n d c o

o o rh

boundary of d

circle of radius w