unity, or some other known quantity. For, hence the other parts of the figure may be found, by the known proportions of the like sides, or parts, and so an equation be obtained. For examples, take the following problems.

PROBLEM I.

In a right-angled triangle, having given the base (3), and the sum of the hypothenuse and perpendicular (9); to find both these two sides.

Let ABC represent the proposed triangle right-angled at B. Put the base AB = 3 = b, and the sum AC + BC of the hypothenuse and perpendicular = 9 = s; also, let x denote the hypothenuse AC, and y the perpendicular BC.

C

N. B. In this solution, and the following ones, the notation is made by using as many unknown letters, x and y, as there are unknown sides of the triangle, a separate letter for each; in preference to using only one unknown letter for one side, and expressing the other unknown side in terms of that letter and the given sum or difference of the sides; though this latter way would render the solution shorter and sooner; because the former way gives occasion for more and better practice in reducing equations; which is the very end and reason for which these problems are given at all.

PROBLEM 11.

In a right-angled triangle, having given the hypothenuse (5), and the sum of the base and perpendicular (7); to find both these two sides.

Let ABC represent the proposed triangle right-angled at B. Put the given hypothenuse AC = 5 = a, and the sum AB + BC of the base and perpendicular = 7 = s; also let a denote the base AB, and y the perpendicular BC.

PROBLEM III.

In a rectangle, having given the diagonal (10), and the perimeter, or sum of all the four sides (28); to find each of the sides severally.

Let ABCD be the proposed rectangle; and put the diagonal AC = 10=d, and half the perimeter AB + BC or AD + DC = 14 = a; also put one side AB=x, and the other BC= y.

Hence, by right-angled triangles,

And by the question

Then by transposing y in the 2d, gives
This value substituted in the 1st, gives
Transposing a2, gives

And dividing by 2, gives

By completing the square, it is

And extracting the root, gives

And transposing a, gives

x2 + y2 = d2,
x + y = a,

x = α

a2

2y2

y 2

y,

D

C

2ay+2y= d2,

2ay = d2

- a2,

ay = {d2 — }a2,

B

y2

ay + 4a2 = }d2.

or 6, the values of x and y.

PROBLEM IV.

Having given the base and perpendicular of any triangle; to find the side of a square inscribed in the same.

Let ABC represent the given triangle, and EFGH its inscribed square. Put the base AB = b, the perpendicular CD = a, and the side of the square GF or GH = DI=x; then will CI = CD — DI = a — x.

G

DEB

Then, because the like lines in the similar triangles ABC, GFC, are proportional (by theorem 84, Geom.), AB: CD:: GF: CI, that is, b:a::x:a-x. Hence, by multiplying extremes and means, ab bx = ax, and transposing bæ, gives ab= ax + bx; then dividing by a+b, gives x =

side of the inscribed square; which therefore is of the same magnitude, whatever the species or the angles of the triangles may be.

PROBLEM V.

In an equilateral triangle, having given the lengths of the three perpendiculars, drawn from a certain point within, on the three sides; to determine the sides. Let ABC represent the equilateral triangle, and DE, DF, and DG, the given perpendiculars from the point D. Draw the lines DA, DB, DC, to the three angular points; and let fall the perpendicular CH on the base AB. Put the three given perpendiculars, DE = a, DF=b, DG = c, and put x = AH or BH, half the side of the equilateral triangle. Then is AC or BC= 2x, and by right-angled triangles the perpendicular CH = VAC AH' = √ 4x3-22 = √3a2 = 2√/3.

A

H G

Now, since the area or space of a rectangle, is expressed by the product of the base and height (cor. 2, th. 81, Geom.), and since a triangle is equal to half a rectangle of equal base and height (cor. 1, th. 26), it follows that,

But the three last triangles make up, or are equal to, the whole former or

great triangle;

that is, x2√3 = ax + bx + cx: hence, dividing by x, gives

x √3 = a + b + c, and dividing by √3, gives

x=

a+b+c, half the side of the triangle sought.

Also, since the whole perpendicular CH is = x √3, it is therefore = a + b+c. That is, the whole perpendicular CH, is just equal to the sum of all the three smaller perpendiculars DE + DF + DG taken together, wherever the point D is situated.

PROBLEM VI.

In a right-angled triangle, having given the base (3), and the difference between the hypothenuse and perpendicular (1); to find both these two sides.

PROBLEM VII.

In a right-angled triangle, having given the hypothenuse (5), and the difference between the base and perpendicular (1); to determine both these two sides.

PROBLEM VIII,

Having given the area, or measure of the space, of a rectangle, inscribed in a given triangle; to determine the sides of the rectangle.

PROBLEM IX.

In a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle; to determine the sides of the triangle.

PROBLEM X.

In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base; to find the sides of the triangle.

PROBLEM XI.

In a triangle, having given the two sides about the vertical angle, with the line bisecting that angle, and terminating in the base; to find the base.

PROBLEM XII.

To determine a right-angled triangle; having given the lengths of two lines drawn from the acute angles, to the middle of the opposite sides.

PROBLEM XIII.

To determine a right angled-triangle; having given the perimeter, and the

radius of its inscribed circle.

PROBLEM XIV.

To determine a triangle; having given the base, the perpendicular, and the ratio of the two sides.

PROBLEM XV.

To determine a right-angled triangle; having given the hypothenuse, and the side of the inscribed square.

PROBLEM XVI.

To determine the radii of three equal circles, described in a given circle, to touch each other and also the circumference of the given circle.

PROBLEM XVII.

In a right-angled triangle, having given the perimeter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypothenuse; to determine the triangle, that is, its sides.

PROBLEM XVIII.

To determine a right-angled triangle; having given the hypothenuse, and the difference of two lines drawn from the two acute angles to the centre of the inscribed circle.

PROBLEM XIX.

To determine a triangle; having given the base, the perpendicular, and the difference of the two other sides.

PROBLEM XX.

To determine a triangle; having given the base, the perpendicular, and the rectangle or product of the two sides.

PROBLEM XXI.

To determine a triangle; having given the lengths of three lines drawn from the three angles, to the middle of the opposite sides.

PROBLEM XXII.

In a triangle, having given all the three sides; to find the radius of the inscribed circle.

PROBLEM XXIII.

To determine a right-angled triangle; having given the side of the inscribed square, and the radius of the inscribed circle.

PROBLEM XXIV.

To determine a triangle, and the radius of the inscribed circle; having given the lengths of three lines drawn from the three angles, to the centre of that circle.

PROBLEM XXV.

To determine a right-angled triangle; having given the hypothenuse, and the radius of the inscribed circle.

PROBLEM XXVI.

To determine a triangle; having given the base, the line bisecting the vertical angle, and the diameter of the circumscribing circle.

PROBLEMS ON MAXIMA AND MINIMA.

TO BE SOLVED GEOMETRICALLY.

1. Divide a right line into two parts so that their rectangle shall be a maximum.

2. Find a point in a given straight line, from which if two straight lines be drawn to two given points on the same side of the given line, and in the same plane with it, their sum shall be a maximum.

3. Let ABC be a right-angled triangle of which AB is the hypothenuse. Draw through the angular point, C, a right line such, that the sum of two perpendiculars let fall upon it from A and B, respectively, shall be a minimum.

4. Through a given point within a circle, which is not the centre, to draw the least chord.

5. Through either of the points of intersection of two given circles that cut each other, to draw the greatest of all straight lines, passing through that point, and terminated both ways by the two circumferences.

6. Two semicircles whose radii are in a known ratio, lie on contrary sides of the same right line, the circumference of one terminating in the centre of the other. Draw the greatest right line perpendicular to the common diametral line, and terminated both ways by the two curves.

7. Through a given point in a given circle, out of the centre, draw a chord which shall cut off the least segment.

8. To find a point in the circumference of a given circle, at which any given straight line drawn from the centre, but less than the radius of the circle, shall subtend the greatest angle.

9. Given the base and the ratio of the sides, to determine the triangle when its area is a maximum

10. In a given triangle to inscribe the greatest rectangle.

11. To divide a given right line into two parts, such that the sum of the squares of the two parts may be a minimum.

12. In a given plane triangle to inscribe another, having its angular points in the three sides of the given one, and its perimeter a minimum.

13. Given the hypothenuse of a right-angled triangle, to construct it when the sum of one leg and the diameter of the inscribed circle is a maximum.