If P and Q are the common points of this circle and line, then OP and OQ are the bounding radii for the two values of θ, supposing these angles to be traced in the usual way by a revolving radius starting from the position OX.

The method of arranging the partial products illustrated in the accompanying example should offer no difficulty, if in explaining it we take the steps given below.

Let the quotient be A ÷ B where the possible errors in A and B respectively are a and b.

In a recent issue ofThe Mathematical Gazette (May 1910), Mr C. S. Jackson points out that certain algebraic theorems appear more difficult to some beginners than would have been anticipated, and he instances as a case in point the proof of the proposition that if , each ratio is equal to . He almost apologises (I think quite unnecessarily) for making the suggestion that some concrete illustrations of the proposition may render its simplicity more manifest, and thus prepare the way for an enunciation of the result in its most general form. The illustration Mr Jackson takes of the above proposition is:

“A regiment of 1000 men contains 130 Yorkshiremen; another of 1200 men contains 156.

The following is a slightly simplified version of a well-known proof of this theorem:—

The Simson's Line of P, with respect to δABC, bisects PO, where O is the orthocentre of the triangle.

The position-ratio of any point N on the line SX is defined as the ratio .