Curiosa mathematica, Part 1

Book II.

Reading edition

Certain universally-true Propositions, not provable from genuine Axioms, but provable if the following Axiom be accepted.

[N.B. This Axiom cannot claim to be more than a 'Quasi-Axiom,' i. e. one whose self-evident character is disputable.]

Axioms.

Diagram with inscribed square within a circle, divided by two perpendicular diameters into eight triangles, here appearing within the text to support a specific proposition about the inscribed tetragon.

In any Circle, the inscribed equilateral Tetragon is greater than any one of the Segments which lie outside it.

Note.—If, in any Circle, 2 Diameters be drawn at right angles to each other, and their extremities joined, the joining Lines will, by Euc. I. 4 be equal to each other. Hence the Figure, thus formed, will be an inscribed equilateral Tetragon. (It will also be equiangular; but that is of no importance for our present purpose.)

PROP. I. Theorem.

An isosceles Triangle, whose vertical angle is one-eighth of a right angle, has its base less than either of its sides.

A quarter-circle arc from K to B, with multiple lines radiating from corner A to labelled points K, J, H, G, F, E, D, C, B along the arc and sides, illustrating how angles or chord lengths vary progressively around a quadrant.

Let be an isosceles Triangle, whose vertical angle at is one-eighth of a right angle.

It shall be proved that is less than .

If we deny this, we must assert that is not-less than .

Let this be our Hypothesis.

Construct 7 more Triangles , &c., equal to . With centre , and distance , describe quadrant passing through , , &c. (See Note.) And join , , , , , , .

Note.—The Reader is requested to imagine chords drawn to the arcs , , &c.

Then Triangle is one-fourth of an equilateral Tetragon inscribed in the Circle.

Hence, 4 times this Triangle is greater than Segment .

[II. Ax. 1.

Because angle is greater than angle , and that angle is less than angle ;

angle is greater than angle ;

is greater than .

[Euc. I. 19.

Similarly, is greater than .

Hence, on our Hypothesis, and are both of them greater than .

Also, is greater than ;

Triangle is greater than Triangle ;

[I. Prop. 2.

to each of these add Triangle ;

Figure is greater than twice Triangle .

Again, is greater than ;

Triangle is greater than Triangle ;

[I. Prop. 2.

Triangles , are together greater than Figure ;

to each of these add Figure ;

Figure is greater than twice Figure , i.e. greater than 4 times Triangle .

Again, is not-less than ;

Triangle is not-less than Triangle ;

[I. Prop. 2.

Triangles , , , are together not-less than Figure ; i.e. they are together greater than 4 times Triangle ;

, a fortiori, Segment is greater than 4 times Triangle .

But this is absurd, since it has been already proved less than 4 times this Triangle.

Hence our Hypothesis, that is not-less than , is false; i.e. is less than .

Therefore an isosceles Triangle &c.

Q.E.D.

Corollary.

Hence, by Book I, Prop. III, it is possible to describe, on a given base, an isosceles Triangle having each base-angle equal to one-eighth of a right angle.

PROP. II. Theorem.

The angles of any Triangle are together not-less than one-eighth of a right angle.

Let one-eighth of a right angle be represented by '.'

Two nested triangles sharing base AB, outer triangle ADB with apex D, inner triangle ACB with apex C just below, with apexes D and C positioned nearer together.

Let be a Triangle; it shall be proved that its 'amount' is not-less than .

If we deny this, we must assert that its 'amount' is less than .

Let this be our Hypothesis.

Hence each of its angles is less than .

On describe an isosceles Triangle having each base-angle equal to ;

[II. Prop. 1. Cor.

hence , , must lie within this Triangle;

i.e. Triangle must lie within it;

angle is less than angle ;

[Euc. I. 21.

i.e. less than ;

but angle is not-less than angle ;

[I. Prop. 3. Cor.

i.e. not-less than ; which is absurd.

Hence our Hypothesis, that 'amount' of Triangle is less than , is false; i.e. it is not less than .

Therefore the angles of any Triangle &c.

Q.E.D.

Corollary.

The 'possible region' does not extend below one-eighth of a right angle.

PROP. III. Theorem.

There is a Triangle whose angles are together not-less than two right angles.

If we deny this, we must assert that any 'amount' is less than .

Let this be our First Hypothesis.

It shall be proved that, if we assert this, we must also assert that any 'amount' is not-less-than , where '' represents one-eighth of a right angle.

If we deny this, we must assert that there is an 'amount' less than .

Let this be our Second Hypothesis.

Now we know that the 'possible region' does not extend below ;

[II. Prop. 2. Cor.

i.e. it has an 'inferior limit.'

Let a 'possible amount' be selected, more than half-way from to this 'inferior limit,' and call it '.' Then it is evident that will lie below the 'inferior limit,' and will therefore be an 'impossible amount.'

Let a Triangle be taken, whose 'amount' is ;

any one of its angles, which is not-greater than either of the others, is not-greater than ; i.e. is less than .

Call this angle '.'

Now one, at least, of the remaining angles must be acute.

[Euc. I. 17.

Call this ''; and call the third angle .

Let 2 such Triangles, and , be taken; and let them be so placed that their -vertices coincide and their -sides lie in one straight line; and join .

Symmetric diagram with baseline AA', apex D above, and central point B on the base. Inner points C and C' form a small diamond shape, with multiple lines radiating from D and B, illustrating a proposition about symmetric or mirror triangles.

On describe an isosceles Triangle , having each base-angle equal to .

[II. Prop. 1. Cor.

Now each of the angles , , is less than .

Lines , , will fall within angles , ;

i.e. Figure will fall within Triangle .

Join , DC′.

Now 'amounts' of Triangles , , together = , and those of the other 4 Triangles are, by our First Hypothesis, together less than ;

'amounts' of all 6 Triangles are together less than ;

but these make up 'amount' of Triangle , plus angles at , , , which together = ;

'amount' of Triangle , plus , is less than .

Now we know that is not-greater than ;

, adding these inequalities, 'amount' of Triangle , plus , is less than ;

this 'amount,' alone, is less than ; which is absurd, since the latter lies below the 'inferior limit,' and is therefore an 'impossible amount';

one of our two Hypotheses must be false;

i.e. either there is an 'amount' not-less than , or else any 'amount' is not-less than .

Suppose we maintain our First Hypothesis: then we must abandon our Second; i.e. we must admit that any 'amount' is not-less than .

It shall be proved that, in this case, we must also admit that any 'amount' is not-less than .

For, if we deny this, we must assert that there is an 'amount' less than .

Let this be our Third Hypothesis.

Then, in the above proof, and may be replaced by and , and a similar absurdity will follow.

Hence either our First or our Third Hypothesis must be false;

i.e. either there is an 'amount' not-less than , or else any amount is not-less than .

A similar proof will hold for ; and then for .

Hence, either there is an 'amount' not-less than , or else any amount is not-less than .

But .

Hence the second clause of this alternative contains the first.

Hence the first clause must be true.

That is, there is a Triangle &c.

Q.E.D.

Corollary.

The 'possible region' does not lie wholly below two right angles.

PROP. IV. Theorem.

There is a Triangle whose angles are together equal to two right angles.

For the 'possible region' does not lie wholly above ;

[I. Prop. 6. Cor.

neither does it lie wholly below ;

[II. Prop. 3. Cor.

it includes .

[I. 4. Cor. 2.

That is, there is a Triangle &c.

Q.E.D.

PROP. V. Theorem.

There is a quadrilateral Figure which is 'rectangular,' that is, which has all its angles right angles.

Two parallel horizontal lines DE and AB, with vertical lines and diagonals connecting labelled points D, H, C, L, E above and A, G, K, B below, via intermediate points F and J, illustrating a proposition about parallel lines and transversals.

Let be a Triangle whose 'amount' = .

[II. Prop. 4.

At make angle equal to angle , and angle equal to angle ;

hence angles , , , together = ;

, , are in a straight Line.

[Euc. I. 14.

Bisect at ; from draw perpendicular to ; from cut off equal to ; and join .

, in Triangles , , , , are respectively equal to , , and angle to angle , [Euc. I. 4.

the Triangles are equal in all respects;

angle is a right angle;

and angle ;

angles , , together = angles , ; i. e. together = ;

, are in a straight line;

[Euc. I. 14.

is a common perpendicular to Lines , .

Similarly, by bisecting at , it may be proved that is a common perpendicular.

Hence Figure is rectangular.

Therefore there is a quadrilateral Figure &c.

Q.E.D.

Definition.

A rectangular quadrilateral Figure may be called a 'Rectangle.'

PROP. VI. Theorem.

The opposite sides of a Rectangle are equal.

A rectangle ABCD with two crossing diagonals inside, plus point C' above D and D' on the right side, illustrating a proposition about diagonals, rectangles, and points displaced from the corners.

Let be a Rectangle; and let it be reversed so that , , may change places.

Then will lie along , and along .

Now, if were not equal to , , , would not change places, but would take new positions, as ;

hence exterior angle would be greater than interior opposite angle ;

[Euc. I. 18.

but they are also equal, being right angles; which is absurd;

Similarly it may be proved that .

Therefore the opposite sides &c.

Q.E.D.

PROP. VII. Theorem.

There is a Pair of Lines, each of which is 'equidistant' from the other, that is, is such that all Points on it are equally distant from the other Line.

Let be a Rectangle; and let a vertical Line be supposed, first to coincide with , and then to move along , continuing always at right angles to it, till it reaches some intermediate position .

Rectangle ABCD with interior point D' connected by lines to corners D and C above, and to point A' on the base, illustrating a proposition about an internal point and its distances to the sides or corners.

Now, if its top be not now on , it must have either dropped below it or risen above it.

First, let it be supposed to have dropped below it: and join , .

Then, if the Figure be reversed, and applied to the same base, it is evident that and will exchange places;

angle , i.e. it is less than ;

Similarly angle is less than ;

but angle is less than .

angles at are together less than ; which is absurd;

[Euc. I. 13. Cor.

top of vertical Line has not dropped below .

Similarly it may be proved that it has not risen above .

Hence it moves along ; i. e. it describes a straight Line, and will evidently continue to do so, however far the vertical Line move, either way, along .

Therefore there is a Pair of Lines &c.

Q.E.D.

Corollaries.

If a Pair of Lines have a common perpendicular: each of them is equidistant from the other.

Corollary 2.

It is possible to form a Rectangle of any given width and height.

For, in the above Pair of horizontal equidistant Lines, 2 common perpendiculars may be drawn, at a given width apart; and the Figure, so formed, will be a Rectangle; and its sides will be a Pair of vertical equidistant Lines, which may be treated in the same way.

PROP. VIII. Theorem.

The angles of any Triangle are together equal to two right angles.

Rectangle with points F, C, E along the top and A, D, B along the base. Point C connects by lines to A, D, and B, forming triangles within, illustrating a proposition about a point on the top side and its distances to the base.

Let be a Triangle, so placed that each base-angle is acute; from draw perpendicular to ; and make Rectangles , .

[II. Prop. 7. Cor. 2.

has its opposite sides equal,

[II. Prop. 6.

, in Triangles , , the sides of the one are respectively equal to the sides of the other;

;

[Euc. I. 8.

Similarly angle ;

the angles of the Triangle together = the angles , , ; i. e. together = .

Therefore the angles of any Triangle &c.

Q.E.D.

PROP. IX. Theorem.

A Pair of Lines, which are equally inclined to a certain transversal, are so to any transversal.

Two parallel horizontal lines AB and CD, with a small triangle formed between them by points E, G above and F, H below, illustrating a proposition about transversals or angles between parallel lines.

Let , , be equally inclined to transversal ; and let be any other transversal.

Join .

Now 'amounts' of Triangles , , together = .

[II. Prop. 8.

But these make up angles of Figure ;

angles of Figure together = ;

but angles , together = ;

[I. Prop. 1.

angles , , together = ;

, , are equally inclined to .

[I. Prop. 1.

Therefore a Pair of Lines &c.

Q.E.D.

Axioms (continued).

If two homogeneous magnitudes be both of them finite: the lesser may be so multiplied, by a finite number, as to exceed the greater.

PROP. X. Theorem.

If a Pair of Lines make, with a certain transversal, two interior angles, on the same side of it, which are together less than two right angles, the defect being a finite angle: these Lines are intersectional on that side of the transversal.

A diagram showing multiple lines radiating from central point E toward labelled points Z, K, V, Y, T, S, R, N, H, M, B, D, F, L, C, G, A,a dense fan of rays and intersecting lines illustrating a proposition about angles, parallels, or convergence.

Let , make with the interior angles , together less than .

Make angle equal to angle ; produce to ; from draw at right angles to , and at right angles to .

Hence is also at right angles to ;

[II. Prop. 9.

i. e. , have a common perpendicular;

each is equidistant from the other;

[II. Prop. 7, Cor. 1.

also = the common distance between them.

Now angle is the defect, from , of the sum of the 2 interior angles , ;

hence, by hypothesis, it is finite;

it may be so multiplied, by a finite number, as to exceed angle .

[II. Ax. 2.

Call this finite number '.'

In , produced if necessary, take -times ; from draw at right angles to ; turn Triangle about into position ; then, about , into position , and so on, till there are such Triangles altogether; and let its final position be .

Then angle is -times angle ; i. e. it is greater than angle .

Let cut broken-Line at ; and join .

Then is greater than , which is greater than , which is greater than ;

[Euc. I. 20, 17, 19.

is greater than ;

but is the same fraction of which is of ;

is greater than ; i. e. the distance of , from , is greater than the common distance between and .

, are intersectional towards , .

Therefore, if a Pair of Lines &c.

Q.E.D.