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If $\sqrt{2}+\sqrt{3}$ were rational, squaring would make $\sqrt{6}$ rational, a contradiction. Alsoso it is irrational but algebraic, satisfying $x^2+2x-1=0$.

For lattice vertices $u,v,w$, the triangle area isa half-integer. Triangulating a convex lattice polygon from one vertex makes its area rational.

A regular octagon of side length $s$ has area $2(1+\sqrt{2})s^2$. Lattice endpoints make $s^2$ a positive integer, so this area would be irrational, contradicting the preceding result. Hence no such octagon exists.

Put $R=R_A+R_B$, $c=\sqrt{R^2-b^2}/R$, and $s=b/R$. At impact the unit vector from A to B is $n=(-c,s)$. The impulse is normal, so tangential velocity is unchanged, while the normal components undergo a one-dimensional elastic collision. Therefore

Newton's equation and its energy integral areThusLet $a=-V^{\prime \prime }(x_{+}^{\ast })>0$. Near the maximum,and the nearby turning point has distance $ϵ\sim \sqrt{2E^{\ast }\delta /a}$ from $x_{+}^{\ast }$. The singular part of the period is an $\mathrm{arcosh}$ integral:HencePhysically, the limiting orbit approaches the unstable equilibrium at the barrier top with vanishing speed and spends arbitrarily long there.

The series $\sum a_n$ converges when its partial sums tend to a finite real limit. Put $m=n+1$. Thenby strict concavity of the square root. Taylor expansion at $m$ givesso $m^2{a}_n\to 1/8$. Limit comparison with $\sum n^{-2}$ proves convergence.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

For axis unit vector $\hat{\omega }$,For a body, replace the sum by $I=\int \rho (x)r_{\perp }^{2}dV$.

For the hollow ball,Energy conservation and $v=R_{+}\Omega$ giveThe factor $\beta$ increases with $\mu$, so if ${\mu }_A>{\mu }_B$, ball B accelerates more and arrives first. Moving the cavity off centre removes material with a larger lever arm and lowers the moment of inertia about the centre of mass; despite the resulting wobble, ball C therefore rolls down faster than A.

Binet's equation givesConsequently