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1923 Paper 3 Q301
D: 1500.0 B: 1500.0

Two triangles \(ABC, A'B'C'\) are such that lines through \(A,B,C\) parallel respectively to \(B'C', C'A', A'B'\) are concurrent. Shew that the same is true of lines through \(A', B', C'\) parallel to \(BC, CA, AB\).

1923 Paper 3 Q302
D: 1500.0 B: 1500.0

Given four points and one line, shew that there is in general one and only one conic through the four points which has an axis parallel to the given line; and give a geometrical construction for any number of points upon it. Indicate the case in which there are an infinite number of conics satisfying the given conditions.

1923 Paper 3 Q303
D: 1500.0 B: 1500.0

Shew that there are in general two triangles whose sides pass through three given points and whose vertices lie on a given conic.

1923 Paper 3 Q304
D: 1500.0 B: 1500.0

Two circles intersect orthogonally in two fixed points. Shew that their common tangent envelopes an ellipse of eccentricity \(1/\sqrt{2}\).

1923 Paper 3 Q305
D: 1500.0 B: 1500.0

Two of the normals from a point \(P\) to a given parabola make equal angles with a given straight line. Prove that the locus of \(P\) is a parabola.

1923 Paper 3 Q306
D: 1500.0 B: 1500.0

Four suits of cards, each suit consisting of thirteen cards numbered from 1 to 13, are dealt to four persons. Find the chance that each person's cards contain all the numbers from 1 to 13.

1923 Paper 3 Q307
D: 1500.0 B: 1500.0

Eliminate \(x, y, z\) from the equations \[ ax^2+by^2+cz^2 = ax+by+cz = yz+zx+xy=0 \] and reduce the result to a symmetrical form.

1923 Paper 3 Q308
D: 1500.0 B: 1500.0

Prove that, if \(u_n = (\alpha+\beta)u_{n-1} - \alpha\beta u_{n-2}\) and \(u_2=\alpha\beta u_1\), then \[ \frac{u_n}{u_1} = \frac{\alpha\beta}{\beta-\alpha} \{\alpha^{n-2} - \alpha^{-1}\beta^{n-1} - \beta^{n-2} + \beta^{-1}\alpha^{n-1} \}. \]

1923 Paper 3 Q309
D: 1500.0 B: 1500.0

If \(\theta=t^n e^{-(x^2+y^2)/4t}\), find what value of \(n\) will make \[ \frac{\partial^2\theta}{\partial x^2} + \frac{\partial^2\theta}{\partial y^2} = \frac{\partial\theta}{\partial t}. \]

1923 Paper 3 Q310
D: 1500.0 B: 1500.0

By finding the fourth differential coefficient of \((\sin^2 x)/x^2\), or otherwise, shew that as \(x\) tends to zero the limit of \[ \frac{15}{x^5} - \frac{2x^4-18x^2+15}{x^6}\cos 2x + \frac{8x^2-24}{x^5}\sin 2x \] is \(\frac{4}{15}\).