Mathematical Questions and Solutions in Continuation of the Mathematical Columns of the Educational Times Volume 52 (Paperback)


This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1890 Excerpt: ...Arithmetica Universalis, says, --"The Ancients taught how to find two mean proportionals by the cissoid; but nobody that I know of hath given a good manual description of this curve. Let AG be the diameter and F the centre of a circle to which the cissoid belongs. At the point F let the perpendicular FD be erected and produced in infinitum, and let FG be produced to P so that FP may be equal to the diameter of the circle. Let the rectangular ruler PED be moved so that the leg EP may always pass through the point P and the other leg ED must be always equal to the diameter AG or FP with its end D always moving in the line FD; and the middle point C of this leg will describe the cissoid." (Wilder's edition.) Because DP, PF = DP, DE; therefore right-angled triangles PFD, PED are congruent; therefore EF is parallel to PD, and PF, ED equally inclined to PD; therefore GC is parallel to PD and EF. Let it meet FD in H, the tangent at A in R, and the parallel to ED through F in Q. Then FQ = EC = FG; therefore Q lies on the circle. Also, because CD is equal and parallel to FQ, therefore HC = HQ; but HR = HG, therefore CR = QG; therefore C lies on a cissoid with the circle AQG for generating circle. See Gregory's Exs. in lliff. Calc, p. 130 (18461). 9490. (Professor Schoute.)--Two non-intersecting lines are the directors of a congruency (1, 1). Show that the locus of the axes of the complexes of the first order passing through the congruency is a ruled surface of the third order, the double line of which is the shortest distance of the two directors, while its simple line, that is no generator, is the line at infinity common to all the planes parallel to both the directors. Solution by the Proposer. As the axis of a linear complex is the line perpendicular to ...

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1890 Excerpt: ...Arithmetica Universalis, says, --"The Ancients taught how to find two mean proportionals by the cissoid; but nobody that I know of hath given a good manual description of this curve. Let AG be the diameter and F the centre of a circle to which the cissoid belongs. At the point F let the perpendicular FD be erected and produced in infinitum, and let FG be produced to P so that FP may be equal to the diameter of the circle. Let the rectangular ruler PED be moved so that the leg EP may always pass through the point P and the other leg ED must be always equal to the diameter AG or FP with its end D always moving in the line FD; and the middle point C of this leg will describe the cissoid." (Wilder's edition.) Because DP, PF = DP, DE; therefore right-angled triangles PFD, PED are congruent; therefore EF is parallel to PD, and PF, ED equally inclined to PD; therefore GC is parallel to PD and EF. Let it meet FD in H, the tangent at A in R, and the parallel to ED through F in Q. Then FQ = EC = FG; therefore Q lies on the circle. Also, because CD is equal and parallel to FQ, therefore HC = HQ; but HR = HG, therefore CR = QG; therefore C lies on a cissoid with the circle AQG for generating circle. See Gregory's Exs. in lliff. Calc, p. 130 (18461). 9490. (Professor Schoute.)--Two non-intersecting lines are the directors of a congruency (1, 1). Show that the locus of the axes of the complexes of the first order passing through the congruency is a ruled surface of the third order, the double line of which is the shortest distance of the two directors, while its simple line, that is no generator, is the line at infinity common to all the planes parallel to both the directors. Solution by the Proposer. As the axis of a linear complex is the line perpendicular to ...

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Product Details

General

Imprint

Rarebooksclub.com

Country of origin

United States

Release date

March 2012

Availability

Supplier out of stock. If you add this item to your wish list we will let you know when it becomes available.

First published

March 2012

Authors

Dimensions

246 x 189 x 2mm (L x W x T)

Format

Paperback - Trade

Pages

32

ISBN-13

978-1-130-69578-6

Barcode

9781130695786

Categories

LSN

1-130-69578-6



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