Euclid. # A text-book of Euclid's Elements : for the use of schools : Books I-VI and XI online

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Join CM, ON, CD meeting AB, AC and the given circle at P, Q

and R.

Then a circle described from centre O with radius OP will touch

AB, AC and the given circle.

For since O is the centre of the Â©MND,

.-. 0M = 0N=30D.

But PM = QN = RD ; Comtr.

.-. OP = OQ=:OR.

.'. a circle described from centre O, with radius OP, will pass through

Q and R.

And since the z ^ at M and N are rt. angles, iii. 18.

.-. the Z ' at P and Q are rt. angles ; i. 29.

.-. the PQR touches AB and AC.

238

EUCLID'S KLEMENTS.

And since R, the point in which the circles meet, is on the Hnc of

centres OD,

. '. the PQR touches the given circle. q. k. v.

Note. There will be two solutions of this problem, since two

circles may be drawn to touch EF, GH and to pass through D.

25. To describe a circle to pass through a given poiiit and touch a

given straight line and a given circle.

Let P be the given point, AB the

given St. line, and DHE the given

circle, of which C is the centre :

it is required to describe a circle to

pass through P, and to touch AB

and the Â©DHE.

Through C draw DCEF perp. to

AB, cutting the circle at the points

D and E, of which E is between C

and AB.

Join DP;

and by describing a circle through

F, E, and P, find a point K in DP (or DP produced) such that the

rect. DE, DF=:therect. DK, DP.

Describe a circle to pass through P, K and touch AB : Ex. 21, p. 235.

This circle shall also touch the given Â© DHE.

For' let G be the point at which this circle touches AB.

Join DG, cutting the given circle DHE at H.

Join HE.

Then the Z DHE is a rt. angle, being in a semicircle. iii. 31.

also the angle at F is a rt. angle; Constr.

.'. the points E, F, G, H are concyclic :

.â™¦. the rect. DE, DF = the rect. DH, DG : m. 36.

but the rect. DE, DF = the rect. DK, DP : Constr.

.'. the rect. DH, DG=therect. DK, DP:

.*. the point H is on the Â© PKG.

Let O be the centre of the PHG.

Join OG, OH, CH.

Then OG and DF are par", since they are both perp. to AB ;

and DG meets them.

.-. thezOGD=:thezGDC. i. 29.

But since OG = OH, and CD = CH,

.â™¦. theZOGH=thezOHG ; and the z CDH=the Z CHD :

.-.thezOHG^thezCHD;

.'. OH and CH are in one st. line.

.*. the PHG touches the given Â© DHE. q. e. f.

THEOREMS AXD EXAMPLES OX BOOK III. 239

Note, (i) Since two circles may be drawn to pass through

P, K and to touch AB, it follows that there will be two solutions

of the present problem.

(ii) Two more solutions may be obtained by joining PE, and

proceeding as before.

The student should examine the nature of the contact between the

circles in each case.

26. Describe a circle to pass through a given point, to touch

a given straight line, and to have its centre on another given straight

line.

27. Describe a circle to pass through a given point, to touch

a given circle, and to have its centre on a given straight line.

28. Describe a circle to pass through two given points, and to

intercept an arc of given length on a given circle.

29. Describe a circle to touch a given circle and a given straight

line at a given point.

30. Describe a circle to touch two given circles and a given

straight line.

V. ON MAXIMA AND MINIMA.

We gather from the Theory of Loci that the position of an

angle, line or figure is capable under suitable conditions of

gradual change ; and it is usually found that change of position

involves a corresponding and gradual change of magnitude.

Under these circumstances we may be required to note if

any situations exist at which the magnitude in question, after

increasing, begins to decrease ; or after decreasing, to increase :

in such situations the Magnitude is said to have reached a

Maximum or a Minimum value; for in the former case it is

greater, and in the latter case less than in adjacent situations

on either side. In the geometry of the circle and straight line

we only meet with such cases of continuous change as admit of

one transition from an increasing to a decreasing state â€” or vice

versS, â€” so that in all the problems with which we have to deal

(where a single circle is involved) there can be only one Maximum

and one Minimum â€” the Maximum being the greatest, and the

Minimum being the least value that the variable magnitude is

capable of taking.

'240 KLChlUÂ« KliKMKNT.N.

Tims a variaMe geometrical magnitude reaches ita maxiinum

or minimum ^â€¢ahle at a turning point, towards which the magni-

tude may mount or descend from either side : it is natural there-

fore to expect a maximum or minimum value to occur when, in

the coui-se of its change, the magnitude assumes a symmetrical

form or position ; and this is usually found to be the case.

This general connection between a symmetrical form or posi-

tion and a maximum or minimum value is not exact enough to

constitute a proof in any particular problem ; but by means of

it a situation is suggested, which on further examination may be

shewn to give the maximum or minimum value sought for.

For example, suppose it is required

to determine the greatest straight line that may he drawn perpen-

dicular to the chord of a segment of a circle and intercepted

heticeen the chord and the arc:

we immediately anticipate that the greatest perpendicular is

that which occupies a symmetrical position in the figure, namely

the perpendicular which passes through the middle point of the

chord ; and on further examination this may be proved to be the

case by means of i. 19, and i. 34.

Again we are able to find at what point a geometrical magni-

tude, varying under certain conditions, assumes its Maximum or

Minimum value, if we can discover a construction for drawing

the magnitude so that it may have an assigned value : for we

may "then examine between what limits the assigned value must

lie in order that the construction may be possible; and the

higher or lower limit will give the Maximum or Minimum

sought for.

It Avas pointed out in the chapter on the Intersection of Loci,

[see page 119] that if under certain conditions existing among

the data, two solutions of a problem are possible, and under other

conditions, no solution exists, there will always be some inter-

mediate condition under which one and only one distinct solution

is possible.

Under these circumstances this single or limiting solution

will always be found to correspond to the maximum or minimum

value of the magnitude to be constructed.

1. For example, suppose it is required

to divide a given straight line so that the rectangle contained hy the

two segments may he a maximum.

We may first attempt to divide the given straight line so that the

rectangle contained by its segments may have a given area â€” that is,

be equal to the square on a given straight line.

THEOREMS AND EXAMPLES ON IJOOK III.

241

Let AB be the given straight line, and K the side of the given

square :

p:^....

n

1 \

A M'

r

M B

it is required to divide the st. line AB at a point M, so that

the rect. AM, MB may be equal to the sq. on K,

Adopting a construction suggested by ii. 14,

describe a semicircle on AB; and at any point X in AB, or AB

produced, draw XY perp. to AB, and equal to K.

Through Y draw YZ par^ to AB, to meet the arc of the semicircle

at P.

Then if the perp. PM is drawn to AB, it may be shewn after the

manner of ii. 14, or by m. 35 that

the rect. AM, MB = the sq. on PM.

= the sq. on K.

So that the rectangle AM, MB increases as K increases.

Now if K is less than the radius CD, then YZ will meet the arc

of the semicircle in two points P, P'; and it follows that AB may be

divided at tivo points, so that the rectangle contained by its segments

maybe equal to the square on K. If K increases, the st. line YZ

will recede from AB, and the points of intersection P, P' will con-

tinually approach one another ; until, when K is equal to the radius

CD, the St. line YZ (now in the position Y'Z') will meet the arc in

tico coincident points, that is, will touch the semicircle at D; and

there will be only one solution of the problem.

If K is greater than CD, the straight line YZ will not meet the

semicircle, and the j)roblem is impossible.

Hence the greatest length that K may have, in order that the con-

struction may be possible, is the radius CD.

.â€¢. the rect. AM, MB is a maximum, when it is equal to the square

on CD;

that is, when PM coincides with DC, and consequently when M

is the middle point of AB.

06s. The special feature to be noticed in this problem is that the

maximum is found at the transitional point between two solutions

and no solution; that is, when the two solutions coincide and become

identicah

U. Â£.

Iti

24iJ KUCIJDH KI,KMKN'l*S.

The following example illustrates the same point.

2. To find at lohat point in a given straight line the angle subtended

by the line joining two given points, which are on the same side of the

given straight line, is a maximum.

Let CD be the given st. line, and A, B the given points on the

same side of CD:

it is required to find at what point in CD the angle subtended by the

St. line AB is a maximum.

First determine at what point in CD, the st. line AB subtends a

given angle.

This is done as follows:â€”

On AB describe a segment of a circle containing an angle equal to

the given angle. iii, 33.

If the arc of this segment intersects CD, tico points in CD are

found at which AB subtends the given angle: but if the arc does not

meet CD, no solution is given.

In accordance with the principles explained above, we expect that

a maximum angle is determined at the limiting position, that is,

when the arc touches CD ; or meets it at two coincident points.

[See page 218.]

This we may prove to be the case.

Describe a circle to pass through A and

B, and to touch the st. line CD.

[Ex. 21, p. 235.]

Let P be the point of contact.

Then shall the Z APB be greater than

any other angle subtended by A B at a point

in CD on the same si^e of AB as P.

For take Q, any other point in CD, on

the same side of AB as P ;

and join AQ, QB.

Since Q is a point in the tangent other

than the point of contact, it must be with-

out the circle,

.â€¢, either BQ or AQ must meet the arc of the segment APB.

Let BQ meet the arc at K : join AK.

Then the Z APB = the z AKB, in the same segment:

but the ext. Z AKB is greater than the int. opp. Z AQB.

.-. the Z APB is greater than AQB.

Similarly the Z APB may be shewn to be greater than any other

angle subtended by AB at a point in CD on the same side of AB:

that is, the Z APB is the greatest of all such angles, q. e. d.

Note. Two circles may be described to pass through A and B,

and to touch CD, the points of contact being on opposite sides of AB:

THEOREMS AND EXAMPLES ON BOOK III. 24.3

hence two points in CD may be found such that the angle subtended

by AB at each of them is greater than the angle subtended at any

other point in CD on the same side of AB.

We add two more examples of considerable importance.

3. In a straight line of indefinite length find a point such that the

sum of its distances from two given points, on the same side of the given

line, shall be a minimum.

Let CD be the given st. line of

indefinite length, and A, B the given

points on the same side of CD :

it is required to find a point P in

CD such that the sum of AP, PB is

a minimum.

Draw AF perp. to CD ;

and produce AF to E, making FE

equal to AF.

Join EB, cutting CD at P.

Join AP, PB.

Then of all lines drawn from A Â£

and B to a point in CD,

the sum of AP, PB shall be the least.

For, let Gl be any other point in CD.

Join AQ, BQ, EQ.

Now in the aÂ« AFP, EFP,

( AF=EF, Co7istr.

Because <and FP is common;

(and the zAFP = the Z EFP, being rt. angles,

.-. AP=EP. 1.4.

Similarly it may be shewn that

AQ=:Ea

Now in the A EQB, the two sides EQ, QB are together greater

than EB;

hence, AQ, QB are together greater than EB,

that is, greater than AP, PB.

Similarly the sum of the st. lines drawn from A and B to any other

point in CD may be shewn to be greater than AP, PB.

.-. the sum of AP, PB is a minimum.

Q. E.D.

Note. It follows from the above proof that

the Z APF = tbe Z EPF i. 4.

= the Z BPD. I. 15.

Thus the sum of AP, PB is a minimum, when these lines are

equally inclined to CD.

16-2

244

EUCLID

4. Given tico intersecting straight lines AB, AC, and a point P

between them; shew that of all straight lines which pass through P

and are terminated by AB, AC, that which is bisected at P cuts off the

triangle of minimum area.

Let EF be the st. line, terminated

by AB, AC, wfiich is bisected at P:

then the a FAE shall be of mini-

mum area.

For let HK be any other st. line

passing through P :

through E draw EM par' to AC.

Then in the a" HPF, MPE,

( the z HPF = the z MPE, i. 15.

Because^ and the z HFP = the z MEP, i. 29.

( andFP=EP; Hyp.

:. the A HPF = the A MPE. i. 26, Cor.

But the A MPE is less than the a KPE;

.-. the A HPF is less than the a KPE:

to each add the fig. AHPE;

then the a FAE is less than the a HAK.

Similarly it may be shewn that the a FAE is less than any other

triangle formed by drawing a st. line through P:

that is, the a FAE is a minimum.

Examples.

1. Two sides of a triangle are given in length ; how must they

be placed in order that the area of the triangle may be a maximum ?

2 Of all triangles of given base and area, the isosceles is that

which has the least perimeter.

3. Given the base and vertical angle of a triangle ; construct it

so that its area may be a maximum.

4. Find a point in a given straight line such that the tangents

drawn from it to a given circle contain the greatest angle possible.

5. A straight rod slips between two straight rulers placed at

right angles to one another; in what position is the triangle

intercepted between the rulers and rod a maximum ?

THEOREMS AND EXAMPLES ON BOOK III. 245

6. Divide a given straight line into two parts, so that the sum of

the squares on the segments may

(i) be equal to a given square,

(ii) may be a minimum.

7. Through a point of intersection of two circles draw a straight

line terminated by the circumferences,

(i) so that it may be of given length,

(ii) so that it may be a maximum.

8. Two tangents to a circle cut one another at right angles ;

find the point on the intercepted arc such that the sum of the

perpendiculars drawn from it to the tangents may be a minimum.

9. Straight lines are drawn from two given points to meet

one another on the circumference of a given circle : prove that their

sum is a minimum when they make equal angles with the tangent

at the point of intersection.

10. Of all triangles of given vertical angle and altitude, the

isosceles is that which has the least area.

11. Two straight lines CA, CB of indefinite length are drawn

from the centre of a circle to meet the circumference at A and B ;

then of all tangents that may be drawn to the circle at points on the

arc AB, that whose intercept is bisected at the point of contact cuts

off the triangle of minimum area.

12. Given two intersecting tangents to a circle, draw a tangent to

the convex arc so that the triangle formed by it and the given tan-

gents may be of maximum area.

13. Of all triangles of given base and area, the isosceles is that

which has the greatest vertical angle.

14. Find a point on the circu]?iference of a circle at which the

straight line joining two given points (of which both are within,

or both without the circle) subtends the greatest angle.

15. A bridge consists of three arches, whose spans are 49 ft.,

32 ft. and 49 ft. respectively : shew that the point on either bank

of the river at which the middle arch subtends the greatest angle

is 63 feet distant from the bridge.

16. From a given point P without a circle whose centre is C,

draw a straight line to cut the circumference at A and B, so that the

triangle ACB may be of maximum area.

17. Shew that the greatest rectangle which can be inscribed

in a circle is a square.

18. A and B are two fixed points without a circle : find a point

P on the circumference such that the sum of the squares on AP, PB

may be a minimum. [See p. 147, Ex. 24.]

24(> JiUCLlD'a ELEMENTJS.

19. A segment of a circle ig described on the chord AB : find a

point P on its arc so that the sum of AP, BP may be a maximum.

20. 0/ all triangles that can be inscribed in a circle that which

has the greatest perimeter is equilateral.

21. Of all triangles that can be inscribed in a given circle tliat

which has the greatest area is equilateral.

22. Of all triangles that can be inscribed in a given triangle that

which has the least perimeter is the triangle formed bij joining the feet

of the perpendiculars drawn from the vertices on opposite sides.

23. Of all rectangles of given area, the square has the least peri-

meter.

24. Describe the triangle of maximum area, having its angles

equal to those of a given triangle, and its sides passing through three

given points.

VI. HARDER MISCELLANEOUS EXAMPLES.

1. AB is a diameter of a given circle; and AC, BD, two chords

on the same side of AB, intersect at E : shew that the circle which

passes through D, E, C cuts the given circle orthogonally.

2. Two circles whose centres are C and D intersect at A and B,

and a straight hne PAQ is drawn through A and terminated by the

circumferences : prove that

(i) the angle PBQr^the angle CAD

(ii) the angle BPC = the angle BQD.

3. Two chords A B, CD of a circle whose centre is O intersect at

right angles at P : shew that

(i) PA2+ PB2+ PC2+ PD2=4 (radiusp.

(ii) AB2 + CD2 + 40P2 =8(radius)2.

4. Two parallel tangents to a circle intercept on any third

tangent a portion which is so divided at its point of contact that the

rectangle contained by its two parts is equal to the square on the

radius.

5. Two equal circles move between two straight lines placed

at right angles, so that each straight line is touched by one circle,

and the two circles touch one another : find the locus of the point

of contact.

6. AB is a given diameter of a circle, and CD is any parallel

chord: if any point X in AB is joined to the extremities of CD,

shew that

XC2 + XD2=XA24-XB2.

THEOREMS AND EXAMPLES ON BOOK III. 247

7. PQ is a fixed chord in a circle, and PX, QY any two parallel

chords through A and B : shew that XY touches a fixed concentric

circle.

8. Two equal circles intersect at A and B ; and from C any point

on the circumference of one of them a perpendicular is drawn to AB,

meeting the other circle at O and O' : shew that either O or O' is the

orthocentre of the triangle ABC. Distinguish between the two cases.

9. Three equal circles pass through the same point A, and their

other points of intersection are B, C, D : shew that of the four

points A, B, C, D, each is the orthocentre of the triangle formed

by joining the other three.

10. From a given point without a circle draw a straight line

to the concave circumference so as to be bisected by the convex

circumference. When is this problem impossible ?

11. Draw a straight line cutting two concentric circles so that

the chord intercepted by the circumference of the greater circle may

be double of the chord intercepted by the less.

12. ABC is a triangle inscribed in a circle, and A', B', C are the

middle points of the arcs subtended by the sides (remote from the

opposite vertices) : find the relation between the angles of the two

triangles ABC, A'B'C ; and prove that the pedal triangle of A'B'C is

equiangular to the triangle ABC.

13. The opposite sides of a quadrilateral inscribed in a circle are

produced to meet: shew that the bisectors of the two angles so

formed are perpendicular to one another.

14. If a quadrilateral can have one circle inscribed in it, and

another circumscribed about it ; shew that the straight lines joining

the opposite points of contact of the inscribed circle are perpendicular

to one anotherl

15. Given the base of a triangle and the sum of the remaining

sides ; find the locus of the foot of the perpendicular from one

extremity of the base on the bisector of the exterior vertical angle.

16. Two circles touch each other at C, and straight lines are

drawn through C at right angles to one another, meeting the

circles at P, P' and Q, Gl' respectively: if the straight line which

joins the centres is terminated by the circumferences at A and A',

shew that

P'P2 + Q'Q2^A'A2.

17. Two circles cut one another orthogonally at A and B ; P

is any point on the arc of one circle intercepted by the other, and

PA, PB are produced to meet the circumference of the second circle

at C and D : shew that CD is a diameter.

248 kuclid's klements.

18. ABC is a triangle, and from any point P perpendiculars

PD, PE, PF are drawn to the sides: if S^ So, S., are the centroM

of the circles circumscribed about the triangles DPE, EPF, FPD,

shew that the triangle SjSgSg is ecjuiangular to the triangle ABC,

and that the sides of the one are respectively half of the sides of the

other.

19. Two tangents PA, PB are drawn from an external point P to

a given circle, and C is the middle point of the chord of contact

AB: if XY is any chord through P, shew that AB bisects the angle

XCY.

20. Given the sum â– of two straight lines and the rectangle con-

tained by them (equal to a given square) : find the lines.

21. Given the sum of the squares on two straight lines and the

rectangle contained by them : find the lines.

22. Given the sum of two straight lines and the sum of the

squares on them : find the lines.

23. Given the difference between two straight lines, and the rect-

angle contained by them : find the lines.

24. Given the difference between two straight lines and the differ-

ence of their squares : find the lines.

25. ABC is a triangle, and the internal and external bisectors of

the angle A meet BC, and BC produced, at P and P': if O is the

middle point of PP', shew that OA is a tangent to the circle circum-

scribed about the triangle ABC.

26. ABC is a triangle, and from P, any point on the circum-

ference of the circle circumscribed about it, perjjendiculars are drawn

to the sides BC, CA, AB meeting the circle again in A', B', C ;

prove that

(i) the triangle A'B'C is identically equal to the triangle ABC.

(ii) AA', BB', CC are parallel.

27. Two equal circles intersect at fixed points A and B, and from

any point in AB a perpendicular is drawn to meet the circumferences

on the same side of AB at P and Q : shew that PQ is of constant

length.

28. The straight lines which join the vertices of a triangle to the

centre of its circumscribed circle, are perpendicular respectively to the

sides of the pedal triangle.

29. P is any point on the circumference of a circle circumscribed

about a triangle ABC ; and perpendiculars PD, PE are drawn from P

to the sides BC, CA. I'md the locus of the centre of the circle circum-

scribed about the triangle PDE.

THEOREMS AND EXAMPLES ON BOOK III. 249

30. P is any point on the circumference of a circle circumscribed

about a iriangle ABC : shew that the angle between Simson's Line

for the point P and the side BC, is equal to the angle between AP

and the diameter of the circumscribed circle.

31. Shew that the orthocentres of the four triangles formed by two

pairs of intersecting straight lines are coUinear.

32. Shew that the circles circumscribed about the four triangles

formed by two pairs of intersecting straight lines meet in a point.

On the Construction of Triangles.

83. Given the vertical angle, one of the sides containing it, and

the length of the perpendicular from the vertex on the base : construct

the triangle.

34. Given the feet of the perpendiculars drawn from the vertices

on the opposite sides : construct the triangle.

35. Given the base, the altitude, and the radius of the circum-

scribed circle : construct the triangle.

36. Given the base, the vertical angle, and the sum of the squares

on the sides containing the vertical angle : construct the triangle.

37. Given the base, the altitude and the sum of the squares on

the sides containing the vertical angle : construct the triangle.

38. Given the base, the vertical angle, and the difference of the

squares on the sides containing the vertical angle : construct the tri-

angle.

39. Given the vertical angle, and the lengths of the two medians

drawn from the extremities of the base : construct the triangle.

40. Given the base, the vertical angle, and the difference of the

angles at the base : construct the triangle.

41. Given the base, and the position of the bisector of the vertical

angle : construct the triangle.

and R.

Then a circle described from centre O with radius OP will touch

AB, AC and the given circle.

For since O is the centre of the Â©MND,

.-. 0M = 0N=30D.

But PM = QN = RD ; Comtr.

.-. OP = OQ=:OR.

.'. a circle described from centre O, with radius OP, will pass through

Q and R.

And since the z ^ at M and N are rt. angles, iii. 18.

.-. the Z ' at P and Q are rt. angles ; i. 29.

.-. the PQR touches AB and AC.

238

EUCLID'S KLEMENTS.

And since R, the point in which the circles meet, is on the Hnc of

centres OD,

. '. the PQR touches the given circle. q. k. v.

Note. There will be two solutions of this problem, since two

circles may be drawn to touch EF, GH and to pass through D.

25. To describe a circle to pass through a given poiiit and touch a

given straight line and a given circle.

Let P be the given point, AB the

given St. line, and DHE the given

circle, of which C is the centre :

it is required to describe a circle to

pass through P, and to touch AB

and the Â©DHE.

Through C draw DCEF perp. to

AB, cutting the circle at the points

D and E, of which E is between C

and AB.

Join DP;

and by describing a circle through

F, E, and P, find a point K in DP (or DP produced) such that the

rect. DE, DF=:therect. DK, DP.

Describe a circle to pass through P, K and touch AB : Ex. 21, p. 235.

This circle shall also touch the given Â© DHE.

For' let G be the point at which this circle touches AB.

Join DG, cutting the given circle DHE at H.

Join HE.

Then the Z DHE is a rt. angle, being in a semicircle. iii. 31.

also the angle at F is a rt. angle; Constr.

.'. the points E, F, G, H are concyclic :

.â™¦. the rect. DE, DF = the rect. DH, DG : m. 36.

but the rect. DE, DF = the rect. DK, DP : Constr.

.'. the rect. DH, DG=therect. DK, DP:

.*. the point H is on the Â© PKG.

Let O be the centre of the PHG.

Join OG, OH, CH.

Then OG and DF are par", since they are both perp. to AB ;

and DG meets them.

.-. thezOGD=:thezGDC. i. 29.

But since OG = OH, and CD = CH,

.â™¦. theZOGH=thezOHG ; and the z CDH=the Z CHD :

.-.thezOHG^thezCHD;

.'. OH and CH are in one st. line.

.*. the PHG touches the given Â© DHE. q. e. f.

THEOREMS AXD EXAMPLES OX BOOK III. 239

Note, (i) Since two circles may be drawn to pass through

P, K and to touch AB, it follows that there will be two solutions

of the present problem.

(ii) Two more solutions may be obtained by joining PE, and

proceeding as before.

The student should examine the nature of the contact between the

circles in each case.

26. Describe a circle to pass through a given point, to touch

a given straight line, and to have its centre on another given straight

line.

27. Describe a circle to pass through a given point, to touch

a given circle, and to have its centre on a given straight line.

28. Describe a circle to pass through two given points, and to

intercept an arc of given length on a given circle.

29. Describe a circle to touch a given circle and a given straight

line at a given point.

30. Describe a circle to touch two given circles and a given

straight line.

V. ON MAXIMA AND MINIMA.

We gather from the Theory of Loci that the position of an

angle, line or figure is capable under suitable conditions of

gradual change ; and it is usually found that change of position

involves a corresponding and gradual change of magnitude.

Under these circumstances we may be required to note if

any situations exist at which the magnitude in question, after

increasing, begins to decrease ; or after decreasing, to increase :

in such situations the Magnitude is said to have reached a

Maximum or a Minimum value; for in the former case it is

greater, and in the latter case less than in adjacent situations

on either side. In the geometry of the circle and straight line

we only meet with such cases of continuous change as admit of

one transition from an increasing to a decreasing state â€” or vice

versS, â€” so that in all the problems with which we have to deal

(where a single circle is involved) there can be only one Maximum

and one Minimum â€” the Maximum being the greatest, and the

Minimum being the least value that the variable magnitude is

capable of taking.

'240 KLChlUÂ« KliKMKNT.N.

Tims a variaMe geometrical magnitude reaches ita maxiinum

or minimum ^â€¢ahle at a turning point, towards which the magni-

tude may mount or descend from either side : it is natural there-

fore to expect a maximum or minimum value to occur when, in

the coui-se of its change, the magnitude assumes a symmetrical

form or position ; and this is usually found to be the case.

This general connection between a symmetrical form or posi-

tion and a maximum or minimum value is not exact enough to

constitute a proof in any particular problem ; but by means of

it a situation is suggested, which on further examination may be

shewn to give the maximum or minimum value sought for.

For example, suppose it is required

to determine the greatest straight line that may he drawn perpen-

dicular to the chord of a segment of a circle and intercepted

heticeen the chord and the arc:

we immediately anticipate that the greatest perpendicular is

that which occupies a symmetrical position in the figure, namely

the perpendicular which passes through the middle point of the

chord ; and on further examination this may be proved to be the

case by means of i. 19, and i. 34.

Again we are able to find at what point a geometrical magni-

tude, varying under certain conditions, assumes its Maximum or

Minimum value, if we can discover a construction for drawing

the magnitude so that it may have an assigned value : for we

may "then examine between what limits the assigned value must

lie in order that the construction may be possible; and the

higher or lower limit will give the Maximum or Minimum

sought for.

It Avas pointed out in the chapter on the Intersection of Loci,

[see page 119] that if under certain conditions existing among

the data, two solutions of a problem are possible, and under other

conditions, no solution exists, there will always be some inter-

mediate condition under which one and only one distinct solution

is possible.

Under these circumstances this single or limiting solution

will always be found to correspond to the maximum or minimum

value of the magnitude to be constructed.

1. For example, suppose it is required

to divide a given straight line so that the rectangle contained hy the

two segments may he a maximum.

We may first attempt to divide the given straight line so that the

rectangle contained by its segments may have a given area â€” that is,

be equal to the square on a given straight line.

THEOREMS AND EXAMPLES ON IJOOK III.

241

Let AB be the given straight line, and K the side of the given

square :

p:^....

n

1 \

A M'

r

M B

it is required to divide the st. line AB at a point M, so that

the rect. AM, MB may be equal to the sq. on K,

Adopting a construction suggested by ii. 14,

describe a semicircle on AB; and at any point X in AB, or AB

produced, draw XY perp. to AB, and equal to K.

Through Y draw YZ par^ to AB, to meet the arc of the semicircle

at P.

Then if the perp. PM is drawn to AB, it may be shewn after the

manner of ii. 14, or by m. 35 that

the rect. AM, MB = the sq. on PM.

= the sq. on K.

So that the rectangle AM, MB increases as K increases.

Now if K is less than the radius CD, then YZ will meet the arc

of the semicircle in two points P, P'; and it follows that AB may be

divided at tivo points, so that the rectangle contained by its segments

maybe equal to the square on K. If K increases, the st. line YZ

will recede from AB, and the points of intersection P, P' will con-

tinually approach one another ; until, when K is equal to the radius

CD, the St. line YZ (now in the position Y'Z') will meet the arc in

tico coincident points, that is, will touch the semicircle at D; and

there will be only one solution of the problem.

If K is greater than CD, the straight line YZ will not meet the

semicircle, and the j)roblem is impossible.

Hence the greatest length that K may have, in order that the con-

struction may be possible, is the radius CD.

.â€¢. the rect. AM, MB is a maximum, when it is equal to the square

on CD;

that is, when PM coincides with DC, and consequently when M

is the middle point of AB.

06s. The special feature to be noticed in this problem is that the

maximum is found at the transitional point between two solutions

and no solution; that is, when the two solutions coincide and become

identicah

U. Â£.

Iti

24iJ KUCIJDH KI,KMKN'l*S.

The following example illustrates the same point.

2. To find at lohat point in a given straight line the angle subtended

by the line joining two given points, which are on the same side of the

given straight line, is a maximum.

Let CD be the given st. line, and A, B the given points on the

same side of CD:

it is required to find at what point in CD the angle subtended by the

St. line AB is a maximum.

First determine at what point in CD, the st. line AB subtends a

given angle.

This is done as follows:â€”

On AB describe a segment of a circle containing an angle equal to

the given angle. iii, 33.

If the arc of this segment intersects CD, tico points in CD are

found at which AB subtends the given angle: but if the arc does not

meet CD, no solution is given.

In accordance with the principles explained above, we expect that

a maximum angle is determined at the limiting position, that is,

when the arc touches CD ; or meets it at two coincident points.

[See page 218.]

This we may prove to be the case.

Describe a circle to pass through A and

B, and to touch the st. line CD.

[Ex. 21, p. 235.]

Let P be the point of contact.

Then shall the Z APB be greater than

any other angle subtended by A B at a point

in CD on the same si^e of AB as P.

For take Q, any other point in CD, on

the same side of AB as P ;

and join AQ, QB.

Since Q is a point in the tangent other

than the point of contact, it must be with-

out the circle,

.â€¢, either BQ or AQ must meet the arc of the segment APB.

Let BQ meet the arc at K : join AK.

Then the Z APB = the z AKB, in the same segment:

but the ext. Z AKB is greater than the int. opp. Z AQB.

.-. the Z APB is greater than AQB.

Similarly the Z APB may be shewn to be greater than any other

angle subtended by AB at a point in CD on the same side of AB:

that is, the Z APB is the greatest of all such angles, q. e. d.

Note. Two circles may be described to pass through A and B,

and to touch CD, the points of contact being on opposite sides of AB:

THEOREMS AND EXAMPLES ON BOOK III. 24.3

hence two points in CD may be found such that the angle subtended

by AB at each of them is greater than the angle subtended at any

other point in CD on the same side of AB.

We add two more examples of considerable importance.

3. In a straight line of indefinite length find a point such that the

sum of its distances from two given points, on the same side of the given

line, shall be a minimum.

Let CD be the given st. line of

indefinite length, and A, B the given

points on the same side of CD :

it is required to find a point P in

CD such that the sum of AP, PB is

a minimum.

Draw AF perp. to CD ;

and produce AF to E, making FE

equal to AF.

Join EB, cutting CD at P.

Join AP, PB.

Then of all lines drawn from A Â£

and B to a point in CD,

the sum of AP, PB shall be the least.

For, let Gl be any other point in CD.

Join AQ, BQ, EQ.

Now in the aÂ« AFP, EFP,

( AF=EF, Co7istr.

Because <and FP is common;

(and the zAFP = the Z EFP, being rt. angles,

.-. AP=EP. 1.4.

Similarly it may be shewn that

AQ=:Ea

Now in the A EQB, the two sides EQ, QB are together greater

than EB;

hence, AQ, QB are together greater than EB,

that is, greater than AP, PB.

Similarly the sum of the st. lines drawn from A and B to any other

point in CD may be shewn to be greater than AP, PB.

.-. the sum of AP, PB is a minimum.

Q. E.D.

Note. It follows from the above proof that

the Z APF = tbe Z EPF i. 4.

= the Z BPD. I. 15.

Thus the sum of AP, PB is a minimum, when these lines are

equally inclined to CD.

16-2

244

EUCLID

4. Given tico intersecting straight lines AB, AC, and a point P

between them; shew that of all straight lines which pass through P

and are terminated by AB, AC, that which is bisected at P cuts off the

triangle of minimum area.

Let EF be the st. line, terminated

by AB, AC, wfiich is bisected at P:

then the a FAE shall be of mini-

mum area.

For let HK be any other st. line

passing through P :

through E draw EM par' to AC.

Then in the a" HPF, MPE,

( the z HPF = the z MPE, i. 15.

Because^ and the z HFP = the z MEP, i. 29.

( andFP=EP; Hyp.

:. the A HPF = the A MPE. i. 26, Cor.

But the A MPE is less than the a KPE;

.-. the A HPF is less than the a KPE:

to each add the fig. AHPE;

then the a FAE is less than the a HAK.

Similarly it may be shewn that the a FAE is less than any other

triangle formed by drawing a st. line through P:

that is, the a FAE is a minimum.

Examples.

1. Two sides of a triangle are given in length ; how must they

be placed in order that the area of the triangle may be a maximum ?

2 Of all triangles of given base and area, the isosceles is that

which has the least perimeter.

3. Given the base and vertical angle of a triangle ; construct it

so that its area may be a maximum.

4. Find a point in a given straight line such that the tangents

drawn from it to a given circle contain the greatest angle possible.

5. A straight rod slips between two straight rulers placed at

right angles to one another; in what position is the triangle

intercepted between the rulers and rod a maximum ?

THEOREMS AND EXAMPLES ON BOOK III. 245

6. Divide a given straight line into two parts, so that the sum of

the squares on the segments may

(i) be equal to a given square,

(ii) may be a minimum.

7. Through a point of intersection of two circles draw a straight

line terminated by the circumferences,

(i) so that it may be of given length,

(ii) so that it may be a maximum.

8. Two tangents to a circle cut one another at right angles ;

find the point on the intercepted arc such that the sum of the

perpendiculars drawn from it to the tangents may be a minimum.

9. Straight lines are drawn from two given points to meet

one another on the circumference of a given circle : prove that their

sum is a minimum when they make equal angles with the tangent

at the point of intersection.

10. Of all triangles of given vertical angle and altitude, the

isosceles is that which has the least area.

11. Two straight lines CA, CB of indefinite length are drawn

from the centre of a circle to meet the circumference at A and B ;

then of all tangents that may be drawn to the circle at points on the

arc AB, that whose intercept is bisected at the point of contact cuts

off the triangle of minimum area.

12. Given two intersecting tangents to a circle, draw a tangent to

the convex arc so that the triangle formed by it and the given tan-

gents may be of maximum area.

13. Of all triangles of given base and area, the isosceles is that

which has the greatest vertical angle.

14. Find a point on the circu]?iference of a circle at which the

straight line joining two given points (of which both are within,

or both without the circle) subtends the greatest angle.

15. A bridge consists of three arches, whose spans are 49 ft.,

32 ft. and 49 ft. respectively : shew that the point on either bank

of the river at which the middle arch subtends the greatest angle

is 63 feet distant from the bridge.

16. From a given point P without a circle whose centre is C,

draw a straight line to cut the circumference at A and B, so that the

triangle ACB may be of maximum area.

17. Shew that the greatest rectangle which can be inscribed

in a circle is a square.

18. A and B are two fixed points without a circle : find a point

P on the circumference such that the sum of the squares on AP, PB

may be a minimum. [See p. 147, Ex. 24.]

24(> JiUCLlD'a ELEMENTJS.

19. A segment of a circle ig described on the chord AB : find a

point P on its arc so that the sum of AP, BP may be a maximum.

20. 0/ all triangles that can be inscribed in a circle that which

has the greatest perimeter is equilateral.

21. Of all triangles that can be inscribed in a given circle tliat

which has the greatest area is equilateral.

22. Of all triangles that can be inscribed in a given triangle that

which has the least perimeter is the triangle formed bij joining the feet

of the perpendiculars drawn from the vertices on opposite sides.

23. Of all rectangles of given area, the square has the least peri-

meter.

24. Describe the triangle of maximum area, having its angles

equal to those of a given triangle, and its sides passing through three

given points.

VI. HARDER MISCELLANEOUS EXAMPLES.

1. AB is a diameter of a given circle; and AC, BD, two chords

on the same side of AB, intersect at E : shew that the circle which

passes through D, E, C cuts the given circle orthogonally.

2. Two circles whose centres are C and D intersect at A and B,

and a straight hne PAQ is drawn through A and terminated by the

circumferences : prove that

(i) the angle PBQr^the angle CAD

(ii) the angle BPC = the angle BQD.

3. Two chords A B, CD of a circle whose centre is O intersect at

right angles at P : shew that

(i) PA2+ PB2+ PC2+ PD2=4 (radiusp.

(ii) AB2 + CD2 + 40P2 =8(radius)2.

4. Two parallel tangents to a circle intercept on any third

tangent a portion which is so divided at its point of contact that the

rectangle contained by its two parts is equal to the square on the

radius.

5. Two equal circles move between two straight lines placed

at right angles, so that each straight line is touched by one circle,

and the two circles touch one another : find the locus of the point

of contact.

6. AB is a given diameter of a circle, and CD is any parallel

chord: if any point X in AB is joined to the extremities of CD,

shew that

XC2 + XD2=XA24-XB2.

THEOREMS AND EXAMPLES ON BOOK III. 247

7. PQ is a fixed chord in a circle, and PX, QY any two parallel

chords through A and B : shew that XY touches a fixed concentric

circle.

8. Two equal circles intersect at A and B ; and from C any point

on the circumference of one of them a perpendicular is drawn to AB,

meeting the other circle at O and O' : shew that either O or O' is the

orthocentre of the triangle ABC. Distinguish between the two cases.

9. Three equal circles pass through the same point A, and their

other points of intersection are B, C, D : shew that of the four

points A, B, C, D, each is the orthocentre of the triangle formed

by joining the other three.

10. From a given point without a circle draw a straight line

to the concave circumference so as to be bisected by the convex

circumference. When is this problem impossible ?

11. Draw a straight line cutting two concentric circles so that

the chord intercepted by the circumference of the greater circle may

be double of the chord intercepted by the less.

12. ABC is a triangle inscribed in a circle, and A', B', C are the

middle points of the arcs subtended by the sides (remote from the

opposite vertices) : find the relation between the angles of the two

triangles ABC, A'B'C ; and prove that the pedal triangle of A'B'C is

equiangular to the triangle ABC.

13. The opposite sides of a quadrilateral inscribed in a circle are

produced to meet: shew that the bisectors of the two angles so

formed are perpendicular to one another.

14. If a quadrilateral can have one circle inscribed in it, and

another circumscribed about it ; shew that the straight lines joining

the opposite points of contact of the inscribed circle are perpendicular

to one anotherl

15. Given the base of a triangle and the sum of the remaining

sides ; find the locus of the foot of the perpendicular from one

extremity of the base on the bisector of the exterior vertical angle.

16. Two circles touch each other at C, and straight lines are

drawn through C at right angles to one another, meeting the

circles at P, P' and Q, Gl' respectively: if the straight line which

joins the centres is terminated by the circumferences at A and A',

shew that

P'P2 + Q'Q2^A'A2.

17. Two circles cut one another orthogonally at A and B ; P

is any point on the arc of one circle intercepted by the other, and

PA, PB are produced to meet the circumference of the second circle

at C and D : shew that CD is a diameter.

248 kuclid's klements.

18. ABC is a triangle, and from any point P perpendiculars

PD, PE, PF are drawn to the sides: if S^ So, S., are the centroM

of the circles circumscribed about the triangles DPE, EPF, FPD,

shew that the triangle SjSgSg is ecjuiangular to the triangle ABC,

and that the sides of the one are respectively half of the sides of the

other.

19. Two tangents PA, PB are drawn from an external point P to

a given circle, and C is the middle point of the chord of contact

AB: if XY is any chord through P, shew that AB bisects the angle

XCY.

20. Given the sum â– of two straight lines and the rectangle con-

tained by them (equal to a given square) : find the lines.

21. Given the sum of the squares on two straight lines and the

rectangle contained by them : find the lines.

22. Given the sum of two straight lines and the sum of the

squares on them : find the lines.

23. Given the difference between two straight lines, and the rect-

angle contained by them : find the lines.

24. Given the difference between two straight lines and the differ-

ence of their squares : find the lines.

25. ABC is a triangle, and the internal and external bisectors of

the angle A meet BC, and BC produced, at P and P': if O is the

middle point of PP', shew that OA is a tangent to the circle circum-

scribed about the triangle ABC.

26. ABC is a triangle, and from P, any point on the circum-

ference of the circle circumscribed about it, perjjendiculars are drawn

to the sides BC, CA, AB meeting the circle again in A', B', C ;

prove that

(i) the triangle A'B'C is identically equal to the triangle ABC.

(ii) AA', BB', CC are parallel.

27. Two equal circles intersect at fixed points A and B, and from

any point in AB a perpendicular is drawn to meet the circumferences

on the same side of AB at P and Q : shew that PQ is of constant

length.

28. The straight lines which join the vertices of a triangle to the

centre of its circumscribed circle, are perpendicular respectively to the

sides of the pedal triangle.

29. P is any point on the circumference of a circle circumscribed

about a triangle ABC ; and perpendiculars PD, PE are drawn from P

to the sides BC, CA. I'md the locus of the centre of the circle circum-

scribed about the triangle PDE.

THEOREMS AND EXAMPLES ON BOOK III. 249

30. P is any point on the circumference of a circle circumscribed

about a iriangle ABC : shew that the angle between Simson's Line

for the point P and the side BC, is equal to the angle between AP

and the diameter of the circumscribed circle.

31. Shew that the orthocentres of the four triangles formed by two

pairs of intersecting straight lines are coUinear.

32. Shew that the circles circumscribed about the four triangles

formed by two pairs of intersecting straight lines meet in a point.

On the Construction of Triangles.

83. Given the vertical angle, one of the sides containing it, and

the length of the perpendicular from the vertex on the base : construct

the triangle.

34. Given the feet of the perpendiculars drawn from the vertices

on the opposite sides : construct the triangle.

35. Given the base, the altitude, and the radius of the circum-

scribed circle : construct the triangle.

36. Given the base, the vertical angle, and the sum of the squares

on the sides containing the vertical angle : construct the triangle.

37. Given the base, the altitude and the sum of the squares on

the sides containing the vertical angle : construct the triangle.

38. Given the base, the vertical angle, and the difference of the

squares on the sides containing the vertical angle : construct the tri-

angle.

39. Given the vertical angle, and the lengths of the two medians

drawn from the extremities of the base : construct the triangle.

40. Given the base, the vertical angle, and the difference of the

angles at the base : construct the triangle.

41. Given the base, and the position of the bisector of the vertical

angle : construct the triangle.

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