## Mathematical Questions and Solutions |

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axis base becomes centre chord circle circumscribing common condition conic constant construct coordinates corresponding cubic curve described determined diameter draw drawn easily eliminating ellipse equal equation Evolute expressions fixed formula four geometrically given given point gives groups hence inscribed integer intersection inverse joining locus means meet middle points minor nine-point obtain once opposite origin pairs parallel particular passes perpendicular points of intersection polar positive primitive projective Proposed prove quadrilateral Question Radial radius readily reference represents respectively right angles satisfy segments self-conjugate shown sides similar Solution sphere straight line substituting subtend suppose tangent tangential equation theorem triangle triangle ABC values vertical whence

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Página 22 - D, E, F are the feet of the perpendiculars from A, B, C, on the opposite sides: shew that OA, OB, OC are respectively perpendicular to EF, FD, DE.

Página viii - The opposite angles of a quadrilateral inscribed in a circle are together equal to two right angles, with converse.

Página x - Find the locus of the points of contact of tangents drawn from a given point to a given set of concentric circles.

Página 38 - Let aa', lb', (Fig. 3) be two lines of the system, intersecting perpendicularly in M ; and let F be one of the points determined in Art. 1, and O the centre of the circle which passes through the feet of the perpendiculars from F. Then since AFBM is a rectangle, OF2 + OM2 = OA2 + OB2, therefore OM is given and the locus of M is a circle, which proves (/3).

Página 80 - All that has been here said may be applied to the hyperbola by changing the sign of d2. 5. The first pedal of the cycloid may be simply obtained by the method of Art. 4. For let the circle begin to roll on the axis of x at the origin, and consider the tangent at a point corresponding to a revolution <p of the circle. If its equation be written Ix + my = F (I, m), we must have...

Página vi - ... 1505. [If P, Q, 1, 2, 3, 4 be points on a conic, then the four points PI, Q2; 1"~, Ql ; P3, y I ; I'l, Q3 lie on a conic passing through the points P and Q.

Página 19 - AOK = 9, the condition is expressed by the equation in Art. 4, which may be constructed as in Art. 1. 1505 (Proposed by Professor CAYLEY.)— If P, Q, 1, 2, 3, 4 be points on a conic; then the four points PI, Q2; P2, Ql ; P3, Q4 ; P4, Q3 lie on a conic passing through the points P and Q. Solution by WILLIAM SPOTTISWOODE, MA, FRS If the points P, Q, 4, be taken for the vertices of the triangle of reference, the equation of the conic passing through P, Q, 1, 2, 3, 4, will be = 0 (1). The equations...

Página 22 - A = 4, (A) represents the director of the ellipse which touches the sides of the triangle at their middle points.

Página ix - B") be the sibi-conjugate points (or foci) of the pairs (A, A') and (B, B'), or of the pairs (A, B') and (A', B), then (A", B") lie on a conic passing through the four points of intersection of the two given conies.

Página ix - ... of equations of this kind are solvable algebraically when they are of an odd degree and have only one real root and one general coefficient, viz., the absolute term 2a ....................... 55 1545. A conic is inscribed in a triangle so that the normals at the points of contact meet in a point i prove that the locus of the centre of the conic is a cubic passing through the angular points of the triangle, the centre of gravity, the intersection of the perpendiculars, the centres of the inscribed...