Now, to square the circle, or, in other words, to get exactly equal in superficial area to the circle X, I will show how to find it. From the point G draw a straight line—say G m—perpendicular to ED, making G m equal GD. Produce GA to a point n, making G n equal to 2AG - GD, and join n m. The square on n m will be the required square. (I have indicated this square by dotted lines.) For example:—If AO = 4, then AG = 5, and GD = 1'25; therefore {2 AG - GD} = {10 - 1'25} = 8'75 = Gn: and Gm = 1'25; therefore, Gn2 + Gm2 = 3-1/8 (AB2); that is, {8'752 + 1'252} = 3-1/8 (52), or, {76'5625 + 1'5625} = {3'125 × 25}; and this equation=Area of the Circle X; and area of the square on n m :: and it follows, that the area of every circle, is equal to the area of a square on the hypotenuse of a right-angled triangle, of which the sides that contain the right angle are in the ratio of 7 to 1, and the sum of these two sides equal to the diameter of the circle. In many ways I have proved this fact, by practical or constructive geometry.
[Plate IV.]
Duplication of the Cube.—In his "Young Geometrician; or, Practical Geometry without Compasses," 1865, Mr. Oliver Byrne's 40th Problem is as follows:—
Let AB be the side of a given cube BD. It is required to find AC, the side of another cube CE, so that the solid contents of the cube CE are double the solid contents of the cube BD.
Ancient and modern mathematicians (says Mr. Byrne) have in vain attempted to solve this problem geometrically, that is, by the ruler and compasses only.
Let AB = BG = GR = RQ = QP = QO = OR = VZ. The length of the shortest side of the lesser set square; a line of any other given length may be applied. Draw OP and VR parallel to it; then apply the set squares in close contact, the edge OV of OVT passing through the point O, while the points of V and Z of ZSV fall exactly on the lines RV, RZ. Then draw the line ZBC, cutting FA produced in C; then the cube on AC is double the cube on AB.
[Plate V.]
Trisection of an Angle.—In his work entitled Young Geometrician, 1865, Mr. Oliver Byrne gives as the 39th Problem: To divide a given angle BAC into three equal angles:—
The line A m is made = p q, the least side of the lesser triangular ruler; by (II) p m is drawn parallel, and m n perpendicular to AB. Then both rulers are kept in motion, and at the same time in close contact, as represented in the figure, until p falls on the line p m, and n on the line m n; r nA passing through the angular point A.