CHAPTER I
3. Let the vectors joining O to the points P, Q, R, S be denoted by p, q, r, s. Then the vector from O to the centre of mass of the triangle PQR is given by (p + q + r)/3 and (2) the vector joining O to the centre of mass of the tetrahedron by 3/4·1/4·(p+p+r)+s/4=(p+p+r+s)/4; this expression is independent of the order in which the vertices are taken.
4. A, A', ··· , C' are the final points of the vectors (p + q)/2, (r + s)/2, ··· , (q + r)/2 and all the three lines AA', BB', CC' have the same mid point, the final point of the vector (p + q + r + s )/4 , which is the centre of mass of the tetrahedron.
1. The answer is given by the length of the vector product of a unit vector lying along l and any vector joining P to a point A (b, d, f) in l:
2. The shortest distance h between l and l', two straight lines in space, is perpendicular to both l and I', i.e., it is parallel to the vector product of two arbitrary vectors lying along l and l', respectively. Also, the shortest distance between l and l' is obtained by projecting a line joining any two
3. The left hand side may be interpreted as the volume of a tetrahedron.
4. The length of the vector product of the vectors (wa , wb, wg) and (x, y, z)•:
6. It is sufficient to prove the statement for the case where the origin is inside the polygon, as the sum of the determinants is unaltered by translation of the co-ordinate system. If the origin is inside the polygon, all the determinants have the same sign and yield the areas of the triangles OP1P2, OP2P3, ··· ,OPnP1.
2. If we write -d = d
´ -1, -e
= e ´
-1, -f =
f ´ -1,
the three equations may be regarded as three homogeneous
equations in x, y, -1; the necessary condition for the
existence of a solution is therefore D= 0. If D=0
and, for example, 
. then the third equation is a consequence
of the first two, and the first two equations in x and y
have a solution, as their determinant does not vanish.
3. The lines intersect if the three equations
for t, t have a solution (2). The condition is
5. Subtract the last row of the determinant from the first three.
1. 
2. a + c = 2b,
3. (a) Introduce the three vectors
Then D = [xy]z. Now for any two vectors a and b
(b) If, and only if, the vectors represented by the columns of D are mutually orthogonal.
4. 
6. It is sufficient to show that there is one point (x0, y0, z0) which remains on the same ray through the origin, i.e., that there are four quantities x0, y0, z0, l (the first three of which do not all vanish) and that the equations
are satisfied. Now we have only to choose l so that the determinant of these three homogeneous equations in x0, y0, z0 vanishes; this gives an equation of the third degree in l, which can always be satisfied, as an equation of the third degree always has a real root.
7. 
9. By 1. in Exercise 1.1 and the rule for the multiplication of determinants, the square of the determinant is equal to 1.
CHAPTER 2
1. cf. Exercise 2, 
3. 
2. We may take the origin at the vertex of the cone; its equation is then of the form u = j(y/x)x.
4. 
6. Cf.
1. xy.
2. Use Taylor's theorem, expressing f(2h, e-1/2h) and f(0, 0) in terms of f and its first and second derivatives at (h, e-1/h); add and divide by h².
1. Use Taylor's theorem to express the co-ordinates of a point on the curve in terms of f, g, h and their first and second derivatives at t0; then apply Exercise 3, 1.2
3. If y is the centre of the sphere, the
expression 
must be as stationary as possible, i.e., 
must
vanish (the dots denoting differentiation with respect to s).
Using the relations 
we obtain the equations 
Hence
we have
5. Cf. 3. and also Lagrange's identity.
7. From the definitions of x1, x2, x3 we
have 
Obviously, 
In order to determine 
we
calculate their components with respect to a rectangular
coordinate system Ox1,
Ox2,
Ox3.
From the relations
we obtain by differentiation
whence 
is perpendicular both to x1
and x3
, and therefore
We define the sign of t so
as to give 
· This implies that t
is positive or negative according to whether the screw defined by
the motion of the osculating plane in the direction of increasing
s is right-handed or .left- handed. In order to prove the
second formula, note that
8. Use 6. and 3.:
9. 
hence x3
is a constant vector h,
say; 
where h
is a fixed vector, i.e., the curve lies in a fixed plane.
10. (b) If the curve is given by x = f(t),.y = g(t), z= h(t), the surface has the parametric equations
then express 
in terms of the derivatives with respect
to t and s.
1. (a) As R is
closed, there is a point B in R the distance of
which from A is less than that of any other point in R.
Let n be the normal to AB at B. Then no
point C in R lies on the same side of n as A;
in fact, otherwise not only B and C, but the entire
segment BC would belong to R, and on this segment
there would be points nearer to A than is B. Hence
the parallel to n through A cannot meet R.
(b) There is a sequence of points P1,
P2 ··· , not in R, converging
to P. Let l1, l2
··· be straight lines passing through P1,
P2 ···, respectively , and dividing the
plane into two half-planes, one of which contains no point of R
(cf. (a)). From these straight lines, we can choose a
sub-sequence for which the directions also converge. The limiting
straight line is then a line of support through P.
(c) If A were not in R, then there would
exist, by the proof of (a), a line of support n separating
A from B.
(d) Let G be the centre of
mass of R and g any line of support, which we take
as x-axis. Then, the y-co-ordinates of all
points in R have the same sign. By the definition
of the centre of mass, the y-co-ordinate of G
also has this sign, i.e., G and R are on the same
side of the arbitrary line of support. Now apply (c).
(g) The curvature is equal to dj/ds, where j is the angle which the tangent makes with the x-axis, s the arc length; j is a continuous function of s. Hence j increases monotonically from j(0) to j(0) + 2p, i.e., j cannot have the same value for two different points of the curve. If the curve were cut at three points s1, s2, s3 by a straight line l (ax + by = c), then the function
would have three zeros; in this case F'(s) would also have at least three zeros, i.e., there would be three tangents parallel to l. In addition, two of these would certainly have the same sense, i.e., they would have the same value of j, which contradicts the statement above.
2. (a)
The set consisting of the points which lie in all convex regions
containing S has the properties (1), (2), (3).
(b) If P is in E, there can be no straight line l
separating P from S, because otherwise one could
take, for example, a large square Q with one side on l
and containing S; Q would then be a convex
region containing S but not P. If P is not
in E, there is at least one convex region Q
containing S but not P; then there
is a straight line separating Q from P, and
therefore, as Q contains S, also separating S
from P.
(c) Cf. 1(d).
1. (a) No. (b) No. (c) Yes (Volume I, 9.2.4.
CHAPTER III
1. 
2. 1
3. Use the fact that the tangents at the origin are given by y = 0 and ax + by = 0. 2c/a, 2(a³g + ab²e-b³c)/a(a² + b²)3/2.
4. Write equation in form 
6. x(y + z) = ay.
8. (a) Double point. (6) Two branches touching one another. (c) Corner. (d) Cusp,
9. Differentiate the equation F = 0 twice with respect to x and use the fact that Fy = 0.
10. a = 1, b = -½.
Then any circle passing through A and B is given
by K' + lK"
= 0. The conditions that the circle K should be
orthogonal to K' and K" are 
From
these conditions the corresponding relation expressing the
orthogonality of K and K' + lK" readily follows.
13. 
2. 
3. Take O as the origin and invert; then the curvilinear triangle is transformed into an ordinary triangle with the same angles.
5. (b) If we denote the left-hand side of the equation defining t1 and t2 by F(x, y, t), two curves t1 = const and t2 = const are given implicitly by the equations F(x, y, t1) = 1 and F(x, y, t2) = 1, respectively, whence the condition that these should be orthogonal is
but this relation is an immediate consequence of F(x,
y, t1) — F(x,
y, t2) = 0.
(e) The coefficients of the quadratic equation defining t1
and t2 are respectively equal to t1,
t2 and -( t1 + t2).
We thus obtain two linear equations in x² and y²,
whence
6. (a) Let .F(t) be the left
hand side of the equation defining t. F is a
continuous function of t in -¥
< t < c, for which F(-¥)=+0, F(c - 0) =
+¥, whence F = 1I at
one point at least of that interval. Similar conclusions apply to
the other intervals.
(b) cf. 5(b)
(c) cf. 5(c): 
with similar formulae for y and z.
7. (b) Let x = r
cos 9, y = r
sin q. Then, the straight
line q = const is transformed
into the conic 
and the circle r = const
into the conic 
9. 
1. (b) A circle on the sphere is given by a linear
equation in x, y, z.
(d) 
2. 
3. 
use the
transformation formula for Jacobians.
4. Introduce co-ordinates x,y, z
such that P becomes the origin, the tangent plane at P
the xy-plane, and t the x-axis. The
equation of S then takes the form z = f(x,
y), where f(0, 0) = fx(0,
0) = fy(0, 0) = 0. A plane S through t is given by the
equation z = ay. We
now introduce 
and x as co-ordinates in S; then the intersection of S and S is given implicitly by the
equation
The curvature of the curve of intersection at the point x = 0, r = 0 is therefore given by
Thus, the centre of curvature of this section has the co-ordinates
i.e., it lies on the circle
5. Take the tangent plane at P as the xy-plane.
Then, the equation of S may be taken to be z = f(x,
y). A normal plane is given by the equation x
= a.y. Take 
and z
as co-ordinates in the plane; then the curve of intersection is
given by
and its curvature at r = 0 by
the final point of the vector of length 
along the line t then has the
co-ordinates
i.e., it lies on the conic
6. (a) By differentiating the two equations with respect to a parameter t of the curve, we obtain
We can find from these relations the ratio x': y': z', i.e., the direction of the tangent. If (x, h, z) are current co-ordinates, the equations of the tangent are
(b) By differentiating the equations of the curve a second time and using the result of (a), we obtain
where l is a proportionality factor. Eliminating l, we have
This linear equation in x", y", z" remains valid if we substitute x', y', z ' for x", y", z". Hence it is still satisfied if we replace x", y", z" by some linear combination lx' + m x", ly'+ m y", lz'+ m x", respectively. Now, if (x, h, z) is in the osculating plane, x - x, h - y, z - z are just such a linear combination ( 6. 2.5).
Hence the equation of the osculating plane is
1. Let P(x, y, z) be a point on the tube-surface S and S be the sphere of the family which has the point P in common with S. Then S and S have the same tangent plane at P, i.e., the same values of x, y, z, zx, zy at that point. It is therefore sufficient to prove that the relation is true for any sphere of unit radius which has its centre in the xy-plane, i.e., for
2. 
4. We may introduce t as a parameter on the curve, so that the latter is given by x = x(t), y = y(t), z = z(t) and the tangent at the point with
CHAPTER I
3. Let the vectors joining O to the points P, Q, R, S be denoted by p, q, r, s. Then the vector from O to the centre of mass of the triangle PQR is given by (p + q + r)/3 and (2) the vector joining O to the centre of mass of the tetrahedron by 3/4·1/4·(p+p+r)+s/4=(p+p+r+s)/4; this expression is independent of the order in which the vertices are taken.
4. A, A', ··· , C' are the final points of the vectors (p + q)/2, (r + s)/2, ··· , (q + r)/2 and all the three lines AA', BB', CC' have the same mid point, the final point of the vector (p + q + r + s )/4 , which is the centre of mass of the tetrahedron.
1. The distance is given by the length of the vector product of a unit vector lying along l and any vector joining P to a point A (b, d, f) in l:
2. The shortest distance h between l and l', two straight lines in space, is perpendicular to both l and I', i.e., it is parallel to the vector product of two arbitrary vectors lying along l and l', respectively. Also, the shortest distance between l and l' is obtained by projecting a line joining any two
3. The left hand side may be interpreted as the volume of a tetrahedron.
4. The length of the vector product of the vectors (wa , wb, wg) and (x, y, z)•:
6. It is sufficient to prove the statement for the case where the origin is inside the polygon, as the sum of the determinants is unaltered by translation of the co-ordinate system. If the origin is inside the polygon, all the determinants have the same sign and yield the areas of the triangles OP1P2, OP2P3, ··· ,OPnP1.
2. If we write -d = d
´ -1, -e
= e ´
-1, -f =
f ´ -1,
the three equations may be regarded as three homogeneous
equations in x, y, -1; the necessary condition for the
existence of a solution is therefore D= 0. If D=0
and, for example, . then the third equation is a consequence
of the first two, and the first two equations in x and y
have a solution, as their determinant does not vanish.
3. The lines intersect if the three equations
for t, t have a solution (2). The condition is
5. Subtract the last row of the determinant from the first three.
1.
2. a + c = 2b,
3. (a) Introduce the three vectors
Then D = [xy]z. Now for any two vectors a and b
(b) If, and only if, the vectors represented by the columns of D are mutually orthogonal.
4.
6. It is sufficient to show that there is one point (x0, y0, z0) which remains on the same ray through the origin, i.e., that there are four quantities x0, y0, z0, l (the first three of which do not all vanish) and that the equations
are satisfied. Now we have only to choose l so that the determinant of these three homogeneous equations in x0, y0, z0 vanishes; this gives an equation of the third degree in l, which can always be satisfied, as an equation of the third degree always has a real root.
7.
9. By 1. in Exercise 1.1 and the rule for the multiplication of determinants, the square of the determinant is equal to 1.
CHAPTER 2
1. cf. Exercise 2,
3.
2. We may take the origin at the vertex of the cone; its equation is then of the form u = j(y/x)x.
4.
6. Cf.
1. xy.
2. Use Taylor's theorem, expressing f(2h, e-1/2h) and f(0, 0) in terms of f and its first and second derivatives at (h, e-1/h); add and divide by h².
1. Use Taylor's theorem to express the co-ordinates of a point on the curve in terms of f, g, h and their first and second derivatives at t0; then apply Exercise 3, 1.2
3. If y is the centre of the sphere, the
expression must be as stationary as possible, i.e., must
vanish (the dots denoting differentiation with respect to s).
Using the relations we obtain the equations Hence
we have
5. Cf. 3. and also Lagrange's identity.
7. From the definitions of x1, x2, x3 we
have Obviously, In order to determine we
calculate their components with respect to a rectangular
coordinate system Ox1,
Ox2,
Ox3.
From the relations
we obtain by differentiation
whence is perpendicular both to x1
and x3
, and therefore
We define the sign of t so
as to give · This implies that t
is positive or negative according to whether the screw defined by
the motion of the osculating plane in the direction of increasing
s is right-handed or .left- handed. In order to prove the
second formula, note that
8. Use 6. and 3.:
9. hence x3
is a constant vector h,
say; where h
is a fixed vector, i.e., the curve lies in a fixed plane.
10. (b) If the curve is given by x = f(t),.y = g(t), z= h(t), the surface has the parametric equations
then express in terms of the derivatives with respect
to t and s.
1. (a) As R is
closed, there is a point B in R the distance of
which from A is less than that of any other point in R.
Let n be the normal to AB at B. Then no
point C in R lies on the same side of n as A;
in fact, otherwise not only B and C, but the entire
segment BC would belong to R, and on this segment
there would be points nearer to A than is B. Hence
the parallel to n through A cannot meet R.
(b) There is a sequence of points P1,
P2 ··· , not in R, converging
to P. Let l1, l2
··· be straight lines passing through P1,
P2 ···, respectively , and dividing the
plane into two half-planes, one of which contains no point of R
(cf. (a)). From these straight lines, we can choose a
sub-sequence for which the directions also converge. The limiting
straight line is then a line of support through P.
(c) If A were not in R, then there would
exist, by the proof of (a), a line of support n separating
A from B.
(d) Let G be the centre of
mass of R and g any line of support, which we take
as x-axis. Then, the y-co-ordinates of all
points in R have the same sign. By the definition
of the centre of mass, the y-co-ordinate of G
also has this sign, i.e., G and R are on the same
side of the arbitrary line of support. Now apply (c).
(g) The curvature is equal to dj/ds, where j is the angle which the tangent makes with the x-axis, s the arc length; j is a continuous function of s. Hence j increases monotonically from j(0) to j(0) + 2p, i.e., j cannot have the same value for two different points of the curve. If the curve were cut at three points s1, s2, s3 by a straight line l (ax + by = c), then the function
would have three zeros; in this case F'(s) would also have at least three zeros, i.e., there would be three tangents parallel to l. In addition, two of these would certainly have the same sense, i.e., they would have the same value of j, which contradicts the statement above.
2. (a)
The set consisting of the points which lie in all convex regions
containing S has the properties (1), (2), (3).
(b) If P is in E, there can be no straight line l
separating P from S, because otherwise one could
take, for example, a large square Q with one side on l
and containing S; Q would then be a convex
region containing S but not P. If P is not
in E, there is at least one convex region Q
containing S but not P; then there
is a straight line separating Q from P, and
therefore, as Q contains S, also separating S
from P.
(c) Cf. 1(d).
1. (a) No. (b) No. (c) Yes (Volume I, 9.2.4.
CHAPTER III
1.
2. 1
3. Use the fact that the tangents at the origin are given by y = 0 and ax + by = 0. 2c/a, 2(a³g + ab²e-b³c)/a(a² + b²)3/2.
4. Write equation in form
6. x(y + z) = ay.
8. (a) Double point. (6) Two branches touching one another. (c) Corner. (d) Cusp,
9. Differentiate the equation F = 0 twice with respect to x and use the fact that Fy = 0.
10. a = 1, b = -½.
Then any circle passing through A and B is given
by K' + lK"
= 0. The conditions that the circle K should be
orthogonal to K' and K" are From
these conditions the corresponding relation expressing the
orthogonality of K and K' + lK" readily follows.
13.
2.
3. Take O as the origin and invert; then the curvilinear triangle is transformed into an ordinary triangle with the same angles.
5. (b) If we denote the left-hand side of the equation defining t1 and t2 by F(x, y, t), two curves t1 = const and t2 = const are given implicitly by the equations F(x, y, t1) = 1 and F(x, y, t2) = 1, respectively, whence the condition that these should be orthogonal is
but this relation is an immediate consequence of F(x,
y, t1) — F(x,
y, t2) = 0.
(e) The coefficients of the quadratic equation defining t1
and t2 are respectively equal to t1,
t2 and -( t1 + t2).
We thus obtain two linear equations in x² and y²,
whence
6. (a) Let .F(t) be the left
hand side of the equation defining t. F is a
continuous function of t in -¥
< t < c, for which F(-¥)=+0, F(c - 0) =
+¥, whence F = 1I at
one point at least of that interval. Similar conclusions apply to
the other intervals.
(b) cf. 5(b)
(c) cf. 5(c):
with similar formulae for y and z.
7. (b) Let x = r
cos 9, y = r
sin q. Then, the straight
line q = const is transformed
into the conic and the circle r = const
into the conic
9.
1. (b) A circle on the sphere is given by a linear
equation in x, y, z.
(d)
2.
3. use the
transformation formula for Jacobians.
4. Introduce co-ordinates x,y, z
such that P becomes the origin, the tangent plane at P
the xy-plane, and t the x-axis. The
equation of S then takes the form z = f(x,
y), where f(0, 0) = fx(0,
0) = fy(0, 0) = 0. A plane S through t is given by the
equation z = ay. We
now introduce and x as co-ordinates in S; then the intersection of S and S is given implicitly by the
equation
The curvature of the curve of intersection at the point x = 0, r = 0 is therefore given by
Thus, the centre of curvature of this section has the co-ordinates
i.e., it lies on the circle
5. Take the tangent plane at P as the xy-plane.
Then, the equation of S may be taken to be z = f(x,
y). A normal plane is given by the equation x
= a.y. Take and z
as co-ordinates in the plane; then the curve of intersection is
given by
and its curvature at r = 0 by
the final point of the vector of length along the line t then has the
co-ordinates
i.e., it lies on the conic
6. (a) By differentiating the two equations with respect to a parameter t of the curve, we obtain
We can find from these relations the ratio x': y': z', i.e., the direction of the tangent. If (x, h, z) are current co-ordinates, the equations of the tangent are
(b) By differentiating the equations of the curve a second time and using the result of (a), we obtain
where l is a proportionality factor. Eliminating l, we have
This linear equation in x", y", z" remains valid if we substitute x', y', z ' for x", y", z". Hence it is still satisfied if we replace x", y", z" by some linear combination lx' + m x", ly'+ m y", lz'+ m x", respectively. Now, if (x, h, z) is in the osculating plane, x - x, h - y, z - z are just such a linear combination ( 6. 2.5).
Hence the equation of the osculating plane is
1. Let P(x, y, z) be a point on the tube-surface S and S be the sphere of the family which has the point P in common with S. Then S and S have the same tangent plane at P, i.e., the same values of x, y, z, zx, zy at that point. It is therefore sufficient to prove that the relation is true for any sphere of unit radius which has its centre in the xy-plane, i.e., for
2.
4. We may introduce t as a parameter on the curve, so that the latter is given by x = x(t), y = y(t), z = z(t) and the tangent at the point with the parameter t lies in the two planes corresponding to t; this yields the relations
We thus obtain by differentiating the equations of the straight lines with respect to t
We then have with the relation
three homogeneous equations in x, y, z and the determinant must vanish.
5. We have for the envelope the two equations
These two equations give a family of straight lines with
parameter t; if there exists a curve having these lines
as tangents, it equations obtained by differentiating once again.
(a) 
(b) the curve is given by 
7. Use inversion. Since S1, S2, S3 pass through the origin, they are transformed into planes; we must hen merely find the envelope of the spheres touching three planes, i.e. a certain circular cone, which we reinvert:
8. 
1.
2. 
3. Maxima for x = 0, y = ± 1, minimum for x = y = 0.
4. The maximum value is the same as for the expression ax²+2bxy+cy², subject to the subsidiary condition ax²+2fxy+cy²=1.
5. cf. 4., (a) 
(b) the function has an improper
maximum equal to 1.05, when y/x = 0-64.
6. Saddle points: 
Minima: 
7. The ellipse obviously touches the circle,
i.e., the two equations must give a double
root in x, whence the condition for
contact is 
8. Introduce the angles between a, b and c, d as variables: The cyclic quadrilateral.
9. 
10. Cf. the similar proof for triangles. A minimum point O does exist. First show that if O is not one of the vertices, then it can only be the point of intersection of the diagonals. Use the fact that the final points of four unit vectors with zero vector sum form a rectangle. Then prove that the sum of the distances from the vertices is less for the point of intersection of the diagonals than for any of the vertices.
11. 
together with the subsidiary condition
12. The vertices are given by 
12. The vertices are given by 
15. The largest axis is given by the maximum of 
with
the subsidiary condition that (x, y, z) lies on the
ellipsoid, whence we have the three equations
Multiplying these by (x, y, z), respectively, and adding, we have
1. On the other hand, we may regard the equations as three linear homogeneous equations in x, y, z, the determinant of which must vanish..
1. 
where 
On the other hand, the subsidiary condition gives
