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found by placing in the circle from the same point, two lines respectively

equal to the sides of the regular hexagon and pentagon.

The centers of the inscribed and ciixumscribed circles of any regular

polygon are coincident.

Besides the circumscription and inscription of triangles and regular

polygons about and in circles, some very important problems are solved

in the constructions respecting the division of the circumferences of

circles into equal parts.

By inscribing an equilateral triangle, a square, a pentagon, a hex-

agon, &c. in a circle, the circumference is divided into three, four, five,

six, &c. equal parts. In Prop. 26, Book iii, it has been shewn that equal

angles at the centers of equal circles, and therefore at the center of the

same circle, subtend equal arcs ; by bisecting the angles at the center,

the arcs which are subtended by them are also bisected, and hence, a

sixth, eighth, tenth, twelfth, &c. part of the circumference of a circle

may be found.

If the right angle be considered as divided into 90 degrees, each degree

into 60 minutes, and each minute into 60 seconds, and so on, according

to the sexagesimal division of a degree; by the aid of the first corollary

to Prop. 32, Book t, may be found the numerical magnitude of an interior

angle of any regular polygon whatever.

Let 6 denote the magnitude of one of the interior angles of a regular

polygon of w sides,

then Â«e is the sum of all the interior angles.

But all the interior angles of any rectilineal figure together with four

right angles, are equal to twice as many right angles as the figure has sides,

that is, if TT be assumed to designate two right angles,

.'. nd + 2'7r = UTT,

and tid = WTT â€” 27r = (n â€” 2) . -tt,

n

the magnitude of an interior angle of a regular polygon of 7i sides.

By taking 7i â€” 3, 4, 5, 6, &c. may be found the magnitude in terms of

two right angles, of an interior angle of any regular polygon whatever.

Pythagoras was the first, as Proclus informs us in his commentary,

who discovered that a multiple of the angles of three regular figures only,

namely, the trigon, the square, and the hexagon, can fill up space round

a point in a plane.

It has been shewn that the interior angle of any regular polygon of n

KOTES TO BOOK IV. 193

sides in. terms of two right angles, is expressed by the equation

n

Let 0s denote the magnitude of the interior angle of a regular figure

of three sides, in which case, w = 3.

3 _ 2 TT

Then 6s = â€” 3 â€” ^"^ = o = ^^Â® tliird of two right angles,

and 603 = 2'7r,

that is, six angles, each equal to the interior angle of an equilateral tri-

angle, are equal to four right angles, and therefore six equilateral triangles

may be placed so as completely to fill up the space round the point at

which they meet in a plane.

In a similar way, it may be shewn that four squares and three hexagons

may be placed so as completely to fill up the space round a point.

Also it will appear from the results deduced, that no other regular

figures besides these three, can be made to fill up the space round a point;

for any multiple of the interior angles of any other regular polygon, will

be found to be in excess above, or in defect from four right angles.

The equilateral triangle or trigon, the square or tetragon, the penta-

gon, and the hexagon, were the only regular polygons known to the

Greeks, capable of being inscribed in circles, besides those which may

be derived from them.

M. Gauss in his Disquisitiones Arithmeticse, has extended the number

by shewing that in general, a regular polygon of 2" + 1 sides is capable

of being inscribed in a circle by means of straight lines and circles, in

those cases in which 2" + 1 is a prime number.

The case in which w = 4, in 2" + 1, was proposed by Mr. Lowry of the

Royal Military College, to be answered in the seventeenth number of

Leybourn's Mathematical Repository, in the following form : â€”

Required a geometrical demonstration of the following method of

constructing a regular polygon of seventeen sides in a circle.

Draw the radius CO at right angles to the diameter AB ; on OC and

0J5, take OQ equal to the half, and OD equal to the eighth part of the

radius ; make BE and DB' each equal to Z)Q, and EG and Fi^ respectively

equal to EQ and FQ\ take OK a mean proportional between OH and

OQ^ and through K, draw iO/ parallel to AB^ meeting the semicircle

described on OG in My draw MN parallel to OC cutting the given circle

in h\ the arc AN is the seventeenth part of the whole circumference.

A demonstration of the truth of this construction has been given by

Mr. Lowry himself, and will be found in the fourth volume of Leybourn's

Repository. The demonstration including the two lemmas occupies

more than eight pages, and is by no means of an elementary character.

QUESTIONS ON BOOK IV.

11. What is the general object of the Fourth Book of Euclid?

2. What consideration renders necessary the first proposition of the

ourth Book of Euclid ?

3. When is a circle said to be inscribed within, and cu'cumscribed

)Out a rectilineal figure ?

I

194 Euclid's elements.

4. When is one rectilineal figure said to be inscribed in, and circum-

scribed about another rectilineal figure ?

5. Modify the construction of Euc. iv. 4, so that the circle may

touch one side of the triangle and the other two sides produced.

6. The sides of a triangle are 5, 6, 7 units respectively, find the radii

of the inscribed and circumscribed circle.

7. Give the constructions by which the centers of circles described

about, and inscribed in triangles are found. In what triangles will they

coincide ?

8. How is it shown that the radius of the circle inscribed in an

equilateral triangle is half the radius described about the same triangle ?

9. The equilateral triangle inscribed in a circle is one-fourth of the

equilateral triangle circumscribed about the same circle.

10. What relation subsists between the square inscribed in, and the

square circumscribed about the same circle ?

1 1 . Enunciate Euc. iii. 22 : and extend this property to any inscribed

polygon having an even number of sides.

12. Trisect a quadrantal arc of a circle, and show that every arc

m

which is an â€” th part of a quadrantal arc may be trisected geometrically :

m and n being whole numbers;

13. If one side of a quadrilateral figure inscribed in a circle be pro-

duced, the exterior angle is equal to the interior and opposite angle of the

figure. Is this property true of any inscribed polygon having an even

number of sides ?

14. In what parallelograms can circles be inscribed ?

15. Give the analysis and synthesis of the problem : to describe

an isosceles triangle, having each of the angles at the base double of

the thii-d angle ?

16. Shew that in the figure Euc. iv. 10, there are two triangles pos-

sessing the required property.

1 7. How is it made to appear that the line BD is the side of a regular

decagon inscribed in the larger circle, and the side of a xe^ulsj: pentagon

inscribed in the smaller circle ? fig. Euc. iv. 10.

18. In the construction of Euc. i.v. 3, Euclid has omitted to shew

that the tangents drawn through the points A and B will meet in some

point M. How may this be shewn ?

19. Shew that if the points of intersection of the circles in Euclid's

figure, Book iv. Prop. 10, be joined with the vertex of the triangle and

with each other, anotlier triangle will be formed equiangular and equal

to the former.

^ 20. Divide a right angle into five equal parts. How may an isosceles

triangle be described upon a given base, having each angle at the base

one-third of the angle at the vertex ?

21. What regular figures may be inscribed in a circle by the help of

Euc. IV. 10 ?

22. Wliat is Euclid's definition of a regular pentagon ? Would the

stellated figure, which is formed by joining the alternate angles of a

regular pentagon, as described in the Fourth Book, satisfy this definition r

23. Shew that each of the interior angles of a regular pentagon in-

scribed in a circle, is equal to three-fifths of two right angles.

24. If two sides not adjacent, of a regular pentagon, be produced to

meet : what is the magnitude of the angle contained at the point where

they meet ?

25. Is there any method more direct than Euclid's for inscribing

a regular pentagon in a circle ?

QUESTIONS ON BOOK IV. 195

26. In what sense is a regular hexagon also a parallelogram ? Would

the same observation apply to all regular figures with an even number of

sides ?

27. Why has Euclid not shewn how to inscribe an equilateral triangle

in a circle, before he requires the use of it in Prop. 16, Book iv. ?

28. An equilateral triangle is inscribed in a circle by joining the first,

third, and fifth angles of the inscribed hexagon.

29. If the sides of a hexagon be produced to meet, the angles formed

by these lines will be equal to four right angles.

30. Shew that the area of an equilateral triangle inscribed in a circle

is one-half of a regular hexagon inscribed in the same circle.

31. If a side of an equilateral triangle be six inches : what is the

radius of the inscribed circle ?

32. Find the area of a regular hexagon inscribed in a circle whose

diameter is twelve inches. What is the difierence between the inscribed

and the circumscribed hexagon ?

33. AVhich is the greater, the difference between the side of the square

and the side of the regular hexagon inscribed in a circle whose radius is

unity ; or the difierence between the side of the equilateral triangle and

the side of the regular pentagon inscribed in the same circle ?

34. The regular hexagon inscribed in a circle, is three-fourths of the

regular circumscribed hexagon.

35. Are the interior angles of an octagon equal to twelve right angles.?

36. What figure is formed by the production of the alternate sides of

a regular octagon ?

37. How many square inches are in the area of a regular octagon

whose side is eight inches ?

38. If an irregular octagon be capable of having a circle described

about it, shew that the sums of the angles taken alternately are equal.

39. Find an algebraical formula for the number of degrees contained

by an interior angle of a regular polygon of n sides.

40. What are the three regular 'figures which can be used in paving

a plane area ? Shew that no other regular figures but these will fiU up

the space round a point in a plane.

41. Into what number of equal parts may a right angle be divided

geometrically ? What connection has the solution of this problem with

the possibility of inscribing regular figures in circles ?

42. Assuming the demonstrations in Euc. iv, shew that any equila-

teral figure of 3.2", 4.2", 5.2'*, or 15.2" sides may be inscribed in a

circle, when n is any of the numbers, 0, 1,2, 3, &c.

43. With a pair of compasses only, shew how to divide the circum-

ference of a given circle into twenty-four equal parts.

44. Shew that if any polygon inscribed in a circle be equilateral, it

must also be equiangular. Is the converse true ?

45. Shew that if the circumference of a circle pass through three

angular points of a regular polygon, it will pass through all of them.

46. Similar polygons are always equiangular : is the converse of this

proposition true ?

47. What are the limits to the Geometrical inscription of regular

figures in circles? What does GeomeiJncG!^ mean when used in this way?

48. What is the difficulty of inscribing geometrically an equilateral

and equiangular undecagon in a circle ? Why is the solution of this pro-

blem said to be beyond the limits of plane geometry ? Why is it so difficult

to prove that the geometrical solution of such problems is impossible?

K2

GEOMETEICAL EXERCISES ON BOOK IV.

PROPOSITION I. THEOREM.

If an equilateral triangle be inscribed in a circle, the square on the side

of the triangle is triple of the square on the radius, or on the side of the

regular hexagon inscribed in the same circle.

Let ABD be an equilateral triangle inscribed in the circle ABD^

of which the center is C.

A

Join JBC, and produce BC to meet the circumference in J5, also

join AJS.

And because ABD is an equilateral triangle inscribed in the cu'cle;

therefore AED is one-thu'd of the whole circumference,

and therefore AE is one-sixth of the circumference,

and consequently ,*the straight line AE is the side of a regular hexagon

(IV. 15.), and is equal to EC.

And because BE is double of EC or AE,

therefore the square on BE is quadi'uple of the square on AE^

but the square on BE is equal to the squares on AB,AE',

therefore the squares on AB, AE are quadruple of the square on AE,

and taking from these equals the square on AE,

therefore the square on AB is triple of the square on AE.

PROPOSITION II. PROBLEM.

To describe a circle which shall touch a straight line given in position, and

pass through two given points.

Analysis. Let AB be the given straight line, and C, D the two

given points.

Suppose the circle required which passes through the points C, D

to touch the line AB in the point E.

A E F B

Join C, D, and produce DC to meet AB in F,

and let the circle be described having the center L,

join also LE, and draw Z^ perpendicular to CD.

Then CD is bisected in H, and LE is perpendicular to AB,

ON BOOK IV. 197

Also, since from tlie point F without the circle, are drawn two

straight lines, one of which FE touches the circle, and the other FD C

cuts it ; the rectangle contained by FC, FB, is equal to the square on

FE. (TIL 36.)

Synthesis. Join (7, D, and produce CD to meet AB in F,

take the point E in FB, such that the square on FE, shall be equal

to the rectangle FD, FC.

Bisect CD in H, and draw JTJT perpendicular to CD',

then ZTA" passes through the center, (ill. 1, Cor. 1.)

At E draw EG perpendicular to FB,

then EG passes through the center, (ill. 1 9.)

consequently L, the point of intersection of these two lines, is

the center of the circle.

It is also manifest, that another circle may be described passing

through C, D, and touching the line AB on the other side of the

point -P; and this circle will be equal to, greater than, or less than the

other circle, according as the angle CFB is equal to, greater than, or

less than the angle CFA.

PROPOSITION III. PROBLEM.

Inscribe a circle in a given sector of a circle.

Analysis. Let CAB be the given sector, and let the required circle

whose center is O, touch the radii in P, Q, and the arc of the sector

mD.

c

ED F

Join OP, OQ, these lines are equal to one another.

Join also CO.

Then in the triangles CPO, CQO, the two sides PC, CO, are equal

to QC, CO, and the base OP is equal to the base OQ ;

therefore the angle PCO is equal to the angle QCO;

and the angle ACB is bisected by CO :

also CO produced will bisect the arc AB in D. (ill. 26.)

If a tangent EDFhe drawn to touch the arc AB in D;

and CA, CB be produced to meet it in ^, i^:

the inscription of the circle in the sector is reduced to the inscrip-

tion of a cii'cle in a triangle, (iv. 4.)

PROPOSITION IV. PROBLEM.

ABCD is a rectangular parallelogram. Bequired to draxo EG, EG

â€¢parallel to AD, DC, so that the rectangle EF may he equal to the figure

EMD, and EB equal to ED.

Analysis. Let EG, EG be drawn, as required, bisecting the rect-

angle ^^CZ).

198 GEOMETRICAL EXERCISES

Draw the diagonal BD cutting EG in ^ffand FG in K,

Then BD also bisects the rectangle ABCD-,

and therefore the area of the triangle KGH is equal to that of the

two triangles EHB, FKD,

A E B

Draw GL perpendicular to BD, and join GB,

also produce FG to M, and EG to N.

If the triangle LGHhe supposed to be equal to the triangle EHB^

by adding HGB to each,

the triangles LGB, GEB are equal, and they are upon the same

base GB, and on the same side of it ;

therefore they are between the same parallels,

that is, if L, E were joined, LE would be parallel to GB ;

and if a semicircle were described on GB as a diameter, it would

pass through the points Ej L; for the angles at E, L are right

angles :

also LE would be a chord parallel to the diameter GB;

therefore the arcs intercepted between the parallels LE, GB are

equal,

and consequently the chord"? EB, LG are also equal ;

but EB is equal to GM, and G3Â£to GN-,

wherefore L G, GM, GN, are equal to one another ;

hence G is the center of the circle inscribed in the triangle BDC.

Synthesis. Draw the diagonal BD.

Find G the center of the circle inscribed in the triangle BDC',

through G draw jE^G^iV^ parallel to BC, and i^Xlf parallel to AB.

Then EG and FG bisect the rectangle ABCD.

Draw GL perpendicular to the diagonal BD.

In the triangles GLH, EHB, the angles GLII, IIEB are equal,

each being a right angle, and the vertical angles LUG, EHB, also the

side LG is equal to the side EB ;

therefore the triangle LUG is equal to the triangle EHB.

Similarly, it may be proved, that the triangle GLK is equal to the

triangle KFD ,

therefore the whole triangle KGH is equal to the two triangles

EHB, EFD',

and consequently EG, FG bisect the rectangle ABCD,

ON BOOK IV. 199

I.

1. In a given circle, place a straight line equal and parallel to a

given straight line not greater than the diameter of the circle.

2. Trisect a given circle by dividing it into three equal sectors.

3. The centers of the circle inscribed in, and circumscribed about

an equilateral triangle coincide ; and the diameter of one is twice the

diameter of the other.

4. If a line be drawn from the vertex of an equilateral triangle,

perpendicular to the base, and intersecting a line drawn from either of

the angles at the base perpendicular to the opposite side ; the distance

from the vertex to the point of intersection, shall be equal to the radius

of the circumscribing circle.

5. If an equilateral triangle be inscribed in a circle, and a straight

line be drawn from the vertical angle to meet the circumference, it

will be equal to the sum or difference of the straight lines drawn from

the extremities of the base to the point where the line meets the cir-

cumference, according as the line does or does not cut the base.

6. The perpendicular from the vertex on the base of an equi-

lateral triangle, is equal to the side of an equilateral triangle inscribed

in a circle whose diameter is the base. Required proof.

7. If an equilateral triangle be inscribed in a circle, and the

adjacent arcs cut off by two of its sides be bisected, the line joining

the points of bisection shall be trisected by the sides.

8. If an equilateral triangle be inscribed in a circle, any of its

sides will cut off one-fourth part of the diameter drawn through the

opposite angle.

9. The perimeter of an equilateral triangle inscribed in a circle is

greater than the perimeter of any other isosceles triangle inscribed in

the same circle.

10. If any two consecutive sides of a hexagon inscribed in a circle

be respectively parallel to their opposite sides, the remaining sides are

parallel to each other.

11. Prove that the area of a regular hexagon is greater than that

of an equilateral triangle of the same perimeter.

12. If two equilateral triangles be inscribed in a circle so as to

have the sides of one parallel to the sides of the other, the figure

common to both will be a regular hexagon, whose area and perimeter

will be equal to the remainder of the area and perimeter of the two

triangles.

13. Determine the distance between the opposite sides of an equi-

lateral and equiangular hexagon inscribed in a circle.

14. Inscribe a regular hexagon in a given equilateral triangle.

15. To inscribe a regular dnodecao^on in a given circle, and shew

that its area is equal to the square of the side of an equilateral triangle

inscribed in the circle.

XL

16. Describe a circle touching three straight lines.

17. Any number of triangles having the same base and the same

vertical angle, will be circumscribed by one circle.

18. Find a point in a tiiangle from which two straight lines

200 GEOMETRICAL EXERCISES

drawn to the extremities of the base shall contain an angle equal to

twice the vertical angle of the triangle. Within what limitations is

this possible ?

19. Given the base of a triangle, and the point from which the

perpendiculars on its three sides are equal; construct the triangle.

To what limitation is the position of this point subject in order that

the triangle may lie on the same side of the base ?

20. From any point B in the radius CA of a given circle whose

center is C, a straight line is di'awn at right angles to CA meeting the

circumference in I)', the circle described round the triangle CBD

touches the given circle in D.

21. If a circle be described about a triangle ABC, and perpen-

diculars be let fall from the angular points A, B, C, on the opposite

sides, and produced to meet the circle in D, JE, F, respectively, the

circumferences UF, FD, DF, are bisected in the points A, B, C.

22. If from the angles of a triangle, lines be di-awn to the points

w^here the inscribed circle touches the sides ; these lines shall intersect

in the same point.

23. The straight line which bisects any angle of a triangle in-

scribed in a circle, cuts the circumference in a point which is equi-

distant from the extremities of the side opposite to the bisected angle,

and from the center of a circle inscribed in the triangle.

24. Let three perpendiculars from the angles of a triangle ABC

on the opposite sides meet in P, a circle described so as to pass through

P and any two of the points A, B, C, is equal to the circumscribing

circle of the triangle.

25. If perpendiculars Aa, Bh, Cc be drawn from the angular

points of a triangle ^PCupon the opposite sides, shew that they will

bisect the angles of the triangle ahc, and thence prove that the peri-

meter of abc will be less than that of any other triangle which can

be inscribed in ABC.

26. Find the least triangle which can be circumscribed about a

given circle.

27. If ABC he a plane triangle, GCF its circumscribing circle,

and GFF a diameter perpendicular to the base AB, then if CF be

joined, the angle GFCis equal to half the difference of the angles at

the base of the triangle.

28. The line joining the centers of the inscribed and circumscribed

circles of a triangle, subtends at any one of the angular points an angle

equal to the semi-difference of the other two angles.

III.

29. The locus of the centers of the circles, which are inscribed

in all right-angled triangles on the same hypotenuse, is the quadrant

described on the hj^otenuse.

30. The center of the circle which touches the two semicircles

described on the sides of a right-angled triangle is the middle point of

the hypotenuse.

31. If a circle be inscribed in a right-angled triangle, the excess

of the sides containing the right angle above the hypotenuse is equal

to the diameter of the inscribed circle.

ON BOOK IV. SOI

32. Having given the hypotenuse of a right-angled triangle, and

the radius of the inscribed circle, to construct the triangle.

33. ABC is a triangle inscribed in a circle, the line joining the

middle points of the arcs AB, AC, will cut off equal portions of the

two contiguous sides measured from the angle A.

IV.

34. Having given the vertical angle of a triangle, and the radii of

the inscribed and circumscribed circles, to construct the triangle.

35. Given the base and vertical angle of a triangle, and also the

radius of the inscribed circle, required to construct it.

36. Given the three angles of a triangle, and the radius of the

inscribed circle, to construct the triangle.

37. If the base and vertical angle of a plane triangle be given,

prove that the locus of the centers of the inscribed circle is a circle,

and find its position and magnitude.

V.

38. In a given triangle inscribe a parallelogram which shall be

equal to one-half the triangle. Is there any limit to the number of

such parallelograms ?

39. In a given triangle to inscribe a triangle, the sides of which

shall be parallel to the sides of a given triangle.

40. If any number of parallelograms be inscribed in a given

parallelogram, the diameters of all the figures shall cut one another

in the same point.

41. A square is inscribed in another, the difi'erence of the areas

is twice the rectangle contained by the segments of the side which

are made at the angular point of the inscribed square.

42. Inscribe an equilateral triangle in a square, (1) "When the

vertex of the triangle is in an angle of the square. (2) W hen the ver-

tex of the triangle is in the point of bisection of a side of the square.

equal to the sides of the regular hexagon and pentagon.

The centers of the inscribed and ciixumscribed circles of any regular

polygon are coincident.

Besides the circumscription and inscription of triangles and regular

polygons about and in circles, some very important problems are solved

in the constructions respecting the division of the circumferences of

circles into equal parts.

By inscribing an equilateral triangle, a square, a pentagon, a hex-

agon, &c. in a circle, the circumference is divided into three, four, five,

six, &c. equal parts. In Prop. 26, Book iii, it has been shewn that equal

angles at the centers of equal circles, and therefore at the center of the

same circle, subtend equal arcs ; by bisecting the angles at the center,

the arcs which are subtended by them are also bisected, and hence, a

sixth, eighth, tenth, twelfth, &c. part of the circumference of a circle

may be found.

If the right angle be considered as divided into 90 degrees, each degree

into 60 minutes, and each minute into 60 seconds, and so on, according

to the sexagesimal division of a degree; by the aid of the first corollary

to Prop. 32, Book t, may be found the numerical magnitude of an interior

angle of any regular polygon whatever.

Let 6 denote the magnitude of one of the interior angles of a regular

polygon of w sides,

then Â«e is the sum of all the interior angles.

But all the interior angles of any rectilineal figure together with four

right angles, are equal to twice as many right angles as the figure has sides,

that is, if TT be assumed to designate two right angles,

.'. nd + 2'7r = UTT,

and tid = WTT â€” 27r = (n â€” 2) . -tt,

n

the magnitude of an interior angle of a regular polygon of 7i sides.

By taking 7i â€” 3, 4, 5, 6, &c. may be found the magnitude in terms of

two right angles, of an interior angle of any regular polygon whatever.

Pythagoras was the first, as Proclus informs us in his commentary,

who discovered that a multiple of the angles of three regular figures only,

namely, the trigon, the square, and the hexagon, can fill up space round

a point in a plane.

It has been shewn that the interior angle of any regular polygon of n

KOTES TO BOOK IV. 193

sides in. terms of two right angles, is expressed by the equation

n

Let 0s denote the magnitude of the interior angle of a regular figure

of three sides, in which case, w = 3.

3 _ 2 TT

Then 6s = â€” 3 â€” ^"^ = o = ^^Â® tliird of two right angles,

and 603 = 2'7r,

that is, six angles, each equal to the interior angle of an equilateral tri-

angle, are equal to four right angles, and therefore six equilateral triangles

may be placed so as completely to fill up the space round the point at

which they meet in a plane.

In a similar way, it may be shewn that four squares and three hexagons

may be placed so as completely to fill up the space round a point.

Also it will appear from the results deduced, that no other regular

figures besides these three, can be made to fill up the space round a point;

for any multiple of the interior angles of any other regular polygon, will

be found to be in excess above, or in defect from four right angles.

The equilateral triangle or trigon, the square or tetragon, the penta-

gon, and the hexagon, were the only regular polygons known to the

Greeks, capable of being inscribed in circles, besides those which may

be derived from them.

M. Gauss in his Disquisitiones Arithmeticse, has extended the number

by shewing that in general, a regular polygon of 2" + 1 sides is capable

of being inscribed in a circle by means of straight lines and circles, in

those cases in which 2" + 1 is a prime number.

The case in which w = 4, in 2" + 1, was proposed by Mr. Lowry of the

Royal Military College, to be answered in the seventeenth number of

Leybourn's Mathematical Repository, in the following form : â€”

Required a geometrical demonstration of the following method of

constructing a regular polygon of seventeen sides in a circle.

Draw the radius CO at right angles to the diameter AB ; on OC and

0J5, take OQ equal to the half, and OD equal to the eighth part of the

radius ; make BE and DB' each equal to Z)Q, and EG and Fi^ respectively

equal to EQ and FQ\ take OK a mean proportional between OH and

OQ^ and through K, draw iO/ parallel to AB^ meeting the semicircle

described on OG in My draw MN parallel to OC cutting the given circle

in h\ the arc AN is the seventeenth part of the whole circumference.

A demonstration of the truth of this construction has been given by

Mr. Lowry himself, and will be found in the fourth volume of Leybourn's

Repository. The demonstration including the two lemmas occupies

more than eight pages, and is by no means of an elementary character.

QUESTIONS ON BOOK IV.

11. What is the general object of the Fourth Book of Euclid?

2. What consideration renders necessary the first proposition of the

ourth Book of Euclid ?

3. When is a circle said to be inscribed within, and cu'cumscribed

)Out a rectilineal figure ?

I

194 Euclid's elements.

4. When is one rectilineal figure said to be inscribed in, and circum-

scribed about another rectilineal figure ?

5. Modify the construction of Euc. iv. 4, so that the circle may

touch one side of the triangle and the other two sides produced.

6. The sides of a triangle are 5, 6, 7 units respectively, find the radii

of the inscribed and circumscribed circle.

7. Give the constructions by which the centers of circles described

about, and inscribed in triangles are found. In what triangles will they

coincide ?

8. How is it shown that the radius of the circle inscribed in an

equilateral triangle is half the radius described about the same triangle ?

9. The equilateral triangle inscribed in a circle is one-fourth of the

equilateral triangle circumscribed about the same circle.

10. What relation subsists between the square inscribed in, and the

square circumscribed about the same circle ?

1 1 . Enunciate Euc. iii. 22 : and extend this property to any inscribed

polygon having an even number of sides.

12. Trisect a quadrantal arc of a circle, and show that every arc

m

which is an â€” th part of a quadrantal arc may be trisected geometrically :

m and n being whole numbers;

13. If one side of a quadrilateral figure inscribed in a circle be pro-

duced, the exterior angle is equal to the interior and opposite angle of the

figure. Is this property true of any inscribed polygon having an even

number of sides ?

14. In what parallelograms can circles be inscribed ?

15. Give the analysis and synthesis of the problem : to describe

an isosceles triangle, having each of the angles at the base double of

the thii-d angle ?

16. Shew that in the figure Euc. iv. 10, there are two triangles pos-

sessing the required property.

1 7. How is it made to appear that the line BD is the side of a regular

decagon inscribed in the larger circle, and the side of a xe^ulsj: pentagon

inscribed in the smaller circle ? fig. Euc. iv. 10.

18. In the construction of Euc. i.v. 3, Euclid has omitted to shew

that the tangents drawn through the points A and B will meet in some

point M. How may this be shewn ?

19. Shew that if the points of intersection of the circles in Euclid's

figure, Book iv. Prop. 10, be joined with the vertex of the triangle and

with each other, anotlier triangle will be formed equiangular and equal

to the former.

^ 20. Divide a right angle into five equal parts. How may an isosceles

triangle be described upon a given base, having each angle at the base

one-third of the angle at the vertex ?

21. What regular figures may be inscribed in a circle by the help of

Euc. IV. 10 ?

22. Wliat is Euclid's definition of a regular pentagon ? Would the

stellated figure, which is formed by joining the alternate angles of a

regular pentagon, as described in the Fourth Book, satisfy this definition r

23. Shew that each of the interior angles of a regular pentagon in-

scribed in a circle, is equal to three-fifths of two right angles.

24. If two sides not adjacent, of a regular pentagon, be produced to

meet : what is the magnitude of the angle contained at the point where

they meet ?

25. Is there any method more direct than Euclid's for inscribing

a regular pentagon in a circle ?

QUESTIONS ON BOOK IV. 195

26. In what sense is a regular hexagon also a parallelogram ? Would

the same observation apply to all regular figures with an even number of

sides ?

27. Why has Euclid not shewn how to inscribe an equilateral triangle

in a circle, before he requires the use of it in Prop. 16, Book iv. ?

28. An equilateral triangle is inscribed in a circle by joining the first,

third, and fifth angles of the inscribed hexagon.

29. If the sides of a hexagon be produced to meet, the angles formed

by these lines will be equal to four right angles.

30. Shew that the area of an equilateral triangle inscribed in a circle

is one-half of a regular hexagon inscribed in the same circle.

31. If a side of an equilateral triangle be six inches : what is the

radius of the inscribed circle ?

32. Find the area of a regular hexagon inscribed in a circle whose

diameter is twelve inches. What is the difierence between the inscribed

and the circumscribed hexagon ?

33. AVhich is the greater, the difference between the side of the square

and the side of the regular hexagon inscribed in a circle whose radius is

unity ; or the difierence between the side of the equilateral triangle and

the side of the regular pentagon inscribed in the same circle ?

34. The regular hexagon inscribed in a circle, is three-fourths of the

regular circumscribed hexagon.

35. Are the interior angles of an octagon equal to twelve right angles.?

36. What figure is formed by the production of the alternate sides of

a regular octagon ?

37. How many square inches are in the area of a regular octagon

whose side is eight inches ?

38. If an irregular octagon be capable of having a circle described

about it, shew that the sums of the angles taken alternately are equal.

39. Find an algebraical formula for the number of degrees contained

by an interior angle of a regular polygon of n sides.

40. What are the three regular 'figures which can be used in paving

a plane area ? Shew that no other regular figures but these will fiU up

the space round a point in a plane.

41. Into what number of equal parts may a right angle be divided

geometrically ? What connection has the solution of this problem with

the possibility of inscribing regular figures in circles ?

42. Assuming the demonstrations in Euc. iv, shew that any equila-

teral figure of 3.2", 4.2", 5.2'*, or 15.2" sides may be inscribed in a

circle, when n is any of the numbers, 0, 1,2, 3, &c.

43. With a pair of compasses only, shew how to divide the circum-

ference of a given circle into twenty-four equal parts.

44. Shew that if any polygon inscribed in a circle be equilateral, it

must also be equiangular. Is the converse true ?

45. Shew that if the circumference of a circle pass through three

angular points of a regular polygon, it will pass through all of them.

46. Similar polygons are always equiangular : is the converse of this

proposition true ?

47. What are the limits to the Geometrical inscription of regular

figures in circles? What does GeomeiJncG!^ mean when used in this way?

48. What is the difficulty of inscribing geometrically an equilateral

and equiangular undecagon in a circle ? Why is the solution of this pro-

blem said to be beyond the limits of plane geometry ? Why is it so difficult

to prove that the geometrical solution of such problems is impossible?

K2

GEOMETEICAL EXERCISES ON BOOK IV.

PROPOSITION I. THEOREM.

If an equilateral triangle be inscribed in a circle, the square on the side

of the triangle is triple of the square on the radius, or on the side of the

regular hexagon inscribed in the same circle.

Let ABD be an equilateral triangle inscribed in the circle ABD^

of which the center is C.

A

Join JBC, and produce BC to meet the circumference in J5, also

join AJS.

And because ABD is an equilateral triangle inscribed in the cu'cle;

therefore AED is one-thu'd of the whole circumference,

and therefore AE is one-sixth of the circumference,

and consequently ,*the straight line AE is the side of a regular hexagon

(IV. 15.), and is equal to EC.

And because BE is double of EC or AE,

therefore the square on BE is quadi'uple of the square on AE^

but the square on BE is equal to the squares on AB,AE',

therefore the squares on AB, AE are quadruple of the square on AE,

and taking from these equals the square on AE,

therefore the square on AB is triple of the square on AE.

PROPOSITION II. PROBLEM.

To describe a circle which shall touch a straight line given in position, and

pass through two given points.

Analysis. Let AB be the given straight line, and C, D the two

given points.

Suppose the circle required which passes through the points C, D

to touch the line AB in the point E.

A E F B

Join C, D, and produce DC to meet AB in F,

and let the circle be described having the center L,

join also LE, and draw Z^ perpendicular to CD.

Then CD is bisected in H, and LE is perpendicular to AB,

ON BOOK IV. 197

Also, since from tlie point F without the circle, are drawn two

straight lines, one of which FE touches the circle, and the other FD C

cuts it ; the rectangle contained by FC, FB, is equal to the square on

FE. (TIL 36.)

Synthesis. Join (7, D, and produce CD to meet AB in F,

take the point E in FB, such that the square on FE, shall be equal

to the rectangle FD, FC.

Bisect CD in H, and draw JTJT perpendicular to CD',

then ZTA" passes through the center, (ill. 1, Cor. 1.)

At E draw EG perpendicular to FB,

then EG passes through the center, (ill. 1 9.)

consequently L, the point of intersection of these two lines, is

the center of the circle.

It is also manifest, that another circle may be described passing

through C, D, and touching the line AB on the other side of the

point -P; and this circle will be equal to, greater than, or less than the

other circle, according as the angle CFB is equal to, greater than, or

less than the angle CFA.

PROPOSITION III. PROBLEM.

Inscribe a circle in a given sector of a circle.

Analysis. Let CAB be the given sector, and let the required circle

whose center is O, touch the radii in P, Q, and the arc of the sector

mD.

c

ED F

Join OP, OQ, these lines are equal to one another.

Join also CO.

Then in the triangles CPO, CQO, the two sides PC, CO, are equal

to QC, CO, and the base OP is equal to the base OQ ;

therefore the angle PCO is equal to the angle QCO;

and the angle ACB is bisected by CO :

also CO produced will bisect the arc AB in D. (ill. 26.)

If a tangent EDFhe drawn to touch the arc AB in D;

and CA, CB be produced to meet it in ^, i^:

the inscription of the circle in the sector is reduced to the inscrip-

tion of a cii'cle in a triangle, (iv. 4.)

PROPOSITION IV. PROBLEM.

ABCD is a rectangular parallelogram. Bequired to draxo EG, EG

â€¢parallel to AD, DC, so that the rectangle EF may he equal to the figure

EMD, and EB equal to ED.

Analysis. Let EG, EG be drawn, as required, bisecting the rect-

angle ^^CZ).

198 GEOMETRICAL EXERCISES

Draw the diagonal BD cutting EG in ^ffand FG in K,

Then BD also bisects the rectangle ABCD-,

and therefore the area of the triangle KGH is equal to that of the

two triangles EHB, FKD,

A E B

Draw GL perpendicular to BD, and join GB,

also produce FG to M, and EG to N.

If the triangle LGHhe supposed to be equal to the triangle EHB^

by adding HGB to each,

the triangles LGB, GEB are equal, and they are upon the same

base GB, and on the same side of it ;

therefore they are between the same parallels,

that is, if L, E were joined, LE would be parallel to GB ;

and if a semicircle were described on GB as a diameter, it would

pass through the points Ej L; for the angles at E, L are right

angles :

also LE would be a chord parallel to the diameter GB;

therefore the arcs intercepted between the parallels LE, GB are

equal,

and consequently the chord"? EB, LG are also equal ;

but EB is equal to GM, and G3Â£to GN-,

wherefore L G, GM, GN, are equal to one another ;

hence G is the center of the circle inscribed in the triangle BDC.

Synthesis. Draw the diagonal BD.

Find G the center of the circle inscribed in the triangle BDC',

through G draw jE^G^iV^ parallel to BC, and i^Xlf parallel to AB.

Then EG and FG bisect the rectangle ABCD.

Draw GL perpendicular to the diagonal BD.

In the triangles GLH, EHB, the angles GLII, IIEB are equal,

each being a right angle, and the vertical angles LUG, EHB, also the

side LG is equal to the side EB ;

therefore the triangle LUG is equal to the triangle EHB.

Similarly, it may be proved, that the triangle GLK is equal to the

triangle KFD ,

therefore the whole triangle KGH is equal to the two triangles

EHB, EFD',

and consequently EG, FG bisect the rectangle ABCD,

ON BOOK IV. 199

I.

1. In a given circle, place a straight line equal and parallel to a

given straight line not greater than the diameter of the circle.

2. Trisect a given circle by dividing it into three equal sectors.

3. The centers of the circle inscribed in, and circumscribed about

an equilateral triangle coincide ; and the diameter of one is twice the

diameter of the other.

4. If a line be drawn from the vertex of an equilateral triangle,

perpendicular to the base, and intersecting a line drawn from either of

the angles at the base perpendicular to the opposite side ; the distance

from the vertex to the point of intersection, shall be equal to the radius

of the circumscribing circle.

5. If an equilateral triangle be inscribed in a circle, and a straight

line be drawn from the vertical angle to meet the circumference, it

will be equal to the sum or difference of the straight lines drawn from

the extremities of the base to the point where the line meets the cir-

cumference, according as the line does or does not cut the base.

6. The perpendicular from the vertex on the base of an equi-

lateral triangle, is equal to the side of an equilateral triangle inscribed

in a circle whose diameter is the base. Required proof.

7. If an equilateral triangle be inscribed in a circle, and the

adjacent arcs cut off by two of its sides be bisected, the line joining

the points of bisection shall be trisected by the sides.

8. If an equilateral triangle be inscribed in a circle, any of its

sides will cut off one-fourth part of the diameter drawn through the

opposite angle.

9. The perimeter of an equilateral triangle inscribed in a circle is

greater than the perimeter of any other isosceles triangle inscribed in

the same circle.

10. If any two consecutive sides of a hexagon inscribed in a circle

be respectively parallel to their opposite sides, the remaining sides are

parallel to each other.

11. Prove that the area of a regular hexagon is greater than that

of an equilateral triangle of the same perimeter.

12. If two equilateral triangles be inscribed in a circle so as to

have the sides of one parallel to the sides of the other, the figure

common to both will be a regular hexagon, whose area and perimeter

will be equal to the remainder of the area and perimeter of the two

triangles.

13. Determine the distance between the opposite sides of an equi-

lateral and equiangular hexagon inscribed in a circle.

14. Inscribe a regular hexagon in a given equilateral triangle.

15. To inscribe a regular dnodecao^on in a given circle, and shew

that its area is equal to the square of the side of an equilateral triangle

inscribed in the circle.

XL

16. Describe a circle touching three straight lines.

17. Any number of triangles having the same base and the same

vertical angle, will be circumscribed by one circle.

18. Find a point in a tiiangle from which two straight lines

200 GEOMETRICAL EXERCISES

drawn to the extremities of the base shall contain an angle equal to

twice the vertical angle of the triangle. Within what limitations is

this possible ?

19. Given the base of a triangle, and the point from which the

perpendiculars on its three sides are equal; construct the triangle.

To what limitation is the position of this point subject in order that

the triangle may lie on the same side of the base ?

20. From any point B in the radius CA of a given circle whose

center is C, a straight line is di'awn at right angles to CA meeting the

circumference in I)', the circle described round the triangle CBD

touches the given circle in D.

21. If a circle be described about a triangle ABC, and perpen-

diculars be let fall from the angular points A, B, C, on the opposite

sides, and produced to meet the circle in D, JE, F, respectively, the

circumferences UF, FD, DF, are bisected in the points A, B, C.

22. If from the angles of a triangle, lines be di-awn to the points

w^here the inscribed circle touches the sides ; these lines shall intersect

in the same point.

23. The straight line which bisects any angle of a triangle in-

scribed in a circle, cuts the circumference in a point which is equi-

distant from the extremities of the side opposite to the bisected angle,

and from the center of a circle inscribed in the triangle.

24. Let three perpendiculars from the angles of a triangle ABC

on the opposite sides meet in P, a circle described so as to pass through

P and any two of the points A, B, C, is equal to the circumscribing

circle of the triangle.

25. If perpendiculars Aa, Bh, Cc be drawn from the angular

points of a triangle ^PCupon the opposite sides, shew that they will

bisect the angles of the triangle ahc, and thence prove that the peri-

meter of abc will be less than that of any other triangle which can

be inscribed in ABC.

26. Find the least triangle which can be circumscribed about a

given circle.

27. If ABC he a plane triangle, GCF its circumscribing circle,

and GFF a diameter perpendicular to the base AB, then if CF be

joined, the angle GFCis equal to half the difference of the angles at

the base of the triangle.

28. The line joining the centers of the inscribed and circumscribed

circles of a triangle, subtends at any one of the angular points an angle

equal to the semi-difference of the other two angles.

III.

29. The locus of the centers of the circles, which are inscribed

in all right-angled triangles on the same hypotenuse, is the quadrant

described on the hj^otenuse.

30. The center of the circle which touches the two semicircles

described on the sides of a right-angled triangle is the middle point of

the hypotenuse.

31. If a circle be inscribed in a right-angled triangle, the excess

of the sides containing the right angle above the hypotenuse is equal

to the diameter of the inscribed circle.

ON BOOK IV. SOI

32. Having given the hypotenuse of a right-angled triangle, and

the radius of the inscribed circle, to construct the triangle.

33. ABC is a triangle inscribed in a circle, the line joining the

middle points of the arcs AB, AC, will cut off equal portions of the

two contiguous sides measured from the angle A.

IV.

34. Having given the vertical angle of a triangle, and the radii of

the inscribed and circumscribed circles, to construct the triangle.

35. Given the base and vertical angle of a triangle, and also the

radius of the inscribed circle, required to construct it.

36. Given the three angles of a triangle, and the radius of the

inscribed circle, to construct the triangle.

37. If the base and vertical angle of a plane triangle be given,

prove that the locus of the centers of the inscribed circle is a circle,

and find its position and magnitude.

V.

38. In a given triangle inscribe a parallelogram which shall be

equal to one-half the triangle. Is there any limit to the number of

such parallelograms ?

39. In a given triangle to inscribe a triangle, the sides of which

shall be parallel to the sides of a given triangle.

40. If any number of parallelograms be inscribed in a given

parallelogram, the diameters of all the figures shall cut one another

in the same point.

41. A square is inscribed in another, the difi'erence of the areas

is twice the rectangle contained by the segments of the side which

are made at the angular point of the inscribed square.

42. Inscribe an equilateral triangle in a square, (1) "When the

vertex of the triangle is in an angle of the square. (2) W hen the ver-

tex of the triangle is in the point of bisection of a side of the square.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38