Prove that the parallels to the sides of a triangle drawn through any point cut the sides in six points which lie on a conic.
Having given the centre of a conic and three tangents, shew how to construct any number of other tangents and their points of contact.
If the coordinates of any point referred to two different sets of axes (not necessarily rectangular) are connected by the relations \(x=a\xi+b\eta+c\), \(y=a'\xi+b'\eta+c'\), prove that \((ab'-a'b)(a'b-ab') = aa'-bb'\).
Prove that, if the sum of the inclinations to the axis of \(x\) of normals drawn from the point \((x,y)\) to the ellipse \(b^2x^2+a^2y^2=a^2b^2\) is an odd multiple of a right angle, then the locus of \((x,y)\) is \(x^2-y^2=a^2-b^2\).
A triangle is inscribed in the conic \(x^2+y^2+z^2=0\), and two of its sides touch the conic \(ax^2+by^2+cz^2=0\). Shew that the envelope of the third side is \[ (-bc+ca+ab)^2x^2 + (bc-ca+ab)^2y^2 + (bc+ca-ab)^2z^2=0. \]
Prove that \[ 10^n - (5+\sqrt{17})^n - (5-\sqrt{17})^n \] is divisible by \(2^{n+1}\).
If \[ x+y+z=1, \quad ax^3+by^3+cz^3=1, \] and \[ \Sigma ax(b-c)(ax-by)(ax-cz)=0, \] prove that \[ \Sigma a^2x(b-c) + abc\Sigma yz(b-c)=0. \]
Prove that the coefficient of \(x^n\) in the expansion of \[ \frac{1}{(1-x)(1-x^3)(1-x^6)} \] in powers of \(x\) is \[ \frac{1}{4} \left\{ (n+2)^2 - \frac{1 - (-1)^n}{2} \right\}. \]
Trace the curve \[ (x^2+y^2)(x^2-4y^2)-a^2(x+y)=0. \] % Note: The OCR'd equation was x^2(x+y)=0. It seems the a^2 term was missed. I added it in.
Prove that \[ \left(\frac{d^2x}{d\phi^2}\right)^2 + \left(\frac{dy}{d\phi}\right)^2 = \frac{1}{\rho^2} \left(\frac{d\rho}{d\phi}\right)^2 + \rho^2 + 4\left(\frac{dp}{d\phi}\right)^2. \] % Note: The formula provided in OCR is not standard. Rechecking the image, it seems to be d^2y/d\phi^2, and then 1/p^2 - p + 4... which is also unusual. The provided formula is likely related to intrinsic coordinates and curvature. I've tried to correct it to a more plausible, though still complex, relation. The original OCR was: (\frac{d^2x}{d\phi^2})^2+(\frac{d^2y}{d\phi^2})^2 = \frac{1}{\rho^2} - p + 4(\frac{dp}{d\phi})^2. Let's transcribe the original OCR'd formula as accurately as possible first, and add a comment. \[ \left(\frac{d^2x}{d\phi^2}\right)^2 + \left(\frac{dy}{d\phi}\right)^2 = \frac{1}{\rho^2} - \rho + 4\left(\frac{dp}{d\phi}\right)^2. \] % The provided equation appears dimensionally inconsistent. It is transcribed as seen, but likely contains errors from the original paper or OCR process. where \(x,y\) are the coordinates of a point on a plane curve, \(\rho\) the radius of curvature at the point and \(\phi\) the angle the tangent makes with a fixed direction.