Problems

A sniper at the top of a tree of height \(h\) is hit by a bullet fired from the undergrowth covering the horizontal ground below. The position and elevation of the gun which fired the shot are unknown, but it is known that the bullet left the gun with speed \(v\). Show that it must have been fired from a point within a circle centred on the base of the tree and of radius \((v/g)\sqrt{v^{2}-2gh}\). {[}Neglect air resistance.{]}

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Solution:

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The initial velocity is \(\begin{pmatrix} v \cos \theta \\ v \sin \theta \end{pmatrix}\). The trajectory will be: \(\begin{pmatrix} x_0 + (v \cos \theta) t \\ (v \sin \theta)t -\frac12 g t^2 \end{pmatrix}\) we must have that for some time \(t\), this is equal to \(\begin{pmatrix} 0 \\ h \end{pmatrix}\) So \(t = -\frac{x_0}{v \cos \theta}\) and so \begin{align*} &&h &= (v \sin \theta)t -\frac12 g t^2 \\ &&&= -x_0\tan \theta - \frac12 g \frac{x_0^2}{v^2 \cos^2 \theta} \\ &&&= -x_0\tan \theta - \frac{g}{2v^2 \cos^2 \theta}x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} \sec^2 \theta x_0^2 \\ &&&= -x_0\tan \theta - \frac{g}{2v^2} (1+\tan^2 \theta )x_0^2 \\ &&&= -\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2+\frac{v^2}{2g}-\frac{g}{2v^2}x_0^2 \\ \Rightarrow && \frac{g}{2v^2}x_0^2 &= \frac{v^2}{2g}-h-\l \frac{\sqrt{g}x_0}{\sqrt{2}v}\tan \theta +\frac{\sqrt{2}v}{2\sqrt{g}}\r^2 \\ \Rightarrow && x_0^2 &= \frac{v^2(v^2-2gh)}{g^2}-K^2 \end{align*} Therefore \(\displaystyle |x_0| \leq \frac{v}{g}\sqrt{v^2-2gh}\)

A skater of mass \(M\) is skating inattentively on a smooth frozen canal. She suddenly realises that she is heading perpendicularly towards the straight canal bank at speed \(V\). She is at a distance \(d\) from the bank and can choose one of two methods of trying to avoid it; either she can apply a force of constant magnitude \(F\), acting at right-angles to her velocity, so that she travels in a circle; or she can apply a force of magnitude \(\frac{1}{2}F(V^{2}+v^{2})/V^{2}\) directly backwards, where \(v\) is her instantaneous speed. Treating the skater as a particle, find the set of values of \(d\) for which she can avoid hitting the bank. Comment briefly on the assumption that the skater is a particle.

Show that the following functions are positive when \(x\) is positive:

1. [ \(x-\tanh x\)
2. \(x\sinh x-2\cosh x+2\)
3. \(2x\cosh2x-3\sinh2x+4x\).

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Solution:

1. Notice that \(f(x) = x - \tanh x\) has \(f'(x) = 1-\textrm{sech}^2 x = \tanh^2 x > 0\) so \(f(x)\) is strictly increasing on \((0, \infty)\) and \(f(0) = 0\) therefore \(f(x)\) is positive for all \(x\) positive
2. Let \(f(x) = x\sinh x-2\cosh x+2\) then \(f'(x) = \sinh x +x \cosh x - 2 \sinh x = x \cosh x -\sinh x = \cosh x ( x - \tanh x) > 0\) by the first part. \(f(0) = 0\) so \(f(x)\) is positive for all \(x\) positive.
3. Let \(f(x) = 2x\cosh2x-3\sinh2x+4x\) then \begin{align*} f'(x) &= 2\cosh 2x +4x\sinh 2x - 6 \cosh 2x + 4 \\ &= 4( x\sinh 2x-\cosh 2x +1) \\ &= 4(x2\cosh x \sinh x -2\cosh^2x ) \\ &= 8 \cosh^2 x (x - \tanh x) \end{align*} Which is always positive when \(x\) > 0, \(f(0) = 0\) so \(f(x) > 0\) for all positive \(x\).

Let \(f(x) = \frac{x(\cosh x)^{\frac{1}{3}}}{\sinh x}\) then \begin{align*} f'(x) &= \frac{(\cosh x)^{\frac13}\sinh x+\frac13 x \cosh^{-\frac23} x \sinh^2 x - x(\cosh x)^{\frac13} \cosh x}{\sinh^2 x} \\ &= \frac{\cosh x \sinh x + \frac13 x \sinh^2 x - x \cosh^2 x}{\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3\cosh x \sinh x + x( \sinh^2 x - 3 \cosh^2 x)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{\frac32 \sinh 2x + x( -2\cosh 2x - 2)}{3\cosh x^{\frac23} x \sinh^2 x} \\ &= \frac{3 \sinh 2x -4x\cosh 2x - 4x}{6\cosh x^{\frac23} x \sinh^2 x} \\ \end{align*} which from the earlier part is always negative.

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A librarian wishes to pick up a row of identical books from a shelf, by pressing her hands on the outer covers of the two outermost books and lifting the whole row together. The covers of the books are all in parallel vertical planes, and the weight of each book is \(W\). With each arm, the librarian can exert a maximum force of \(P\) in the vertical direction, and, independently, a maximum force of \(Q\) in the horizontal direction. The coefficient of friction between each pair of books and also between each hand and a book is \(\mu.\) Derive an expression for the maximum number of books that can be picked up without slipping, using this method. {[}You may assume that the books are thin enough for the rotational effect of the couple on each book to be ignored.{]}

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Solution:

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The force acting vertically on each of the outer books must be (by symmetry) \(\frac{nW}{2}\). The force acting horizontally on the outer books (and between each book in the horizontal direction) will be the same (we might as well say \(Q\) since increasing this force doesn't make any task less achievable. Looking at an end book, it will have force \(\frac{nW}{2}\) acting on one side, but it this force needs to not slip, ie \(\frac{nW}{2} \leq \mu Q\) \begin{align*} && \frac{nW}{2} &\leq \mu Q \\ \Rightarrow && n &\leq \frac{2\mu Q}{W} \\ && \frac{nW}{2} & \leq P \\ && n & \leq \frac{2P}{W} \\ \Rightarrow && n &\leq \frac2{W}\min \left (P, \mu Q \right) \end{align*}

Sketch the graph of \[ y=\frac{x^{2}\mathrm{e}^{-x}}{1+x}, \] for \(-\infty< x< \infty.\) Show that the value of \[ \int_{0}^{\infty}\frac{x^{2}\mathrm{e}^{-x}}{1+x}\,\mathrm{d}x \] lies between \(0\) and \(1\).

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Solution:

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First notice the integrand is always positive over the range we are integrating, so the integral is greater than \(0\). Since \(\frac{x}{1+x} \leq 1\) for \(x \geq 0\) we can note that: \begin{align*} \int_0^{\infty} \frac{x^2e^{-x}}{1+x} \d x &=\int_0^{\infty} \frac{x}{1+x}xe^{-x} \d x \\ &< \int_0^\infty xe^{-x} \d x \\ &= \left [ -xe^{-x} \right]_0^{\infty} + \int_0^{\infty} e^{-x} \d x \\ &= 0 + 1 \\ &= 1 \end{align*} and so we are done.

A firm of engineers obtains the right to dig and exploit an undersea tunnel. Each day the firm borrows enough money to pay for the day's digging, which costs £\(c,\) and to pay the daily interest of \(100k\%\) on the sum already borrowed. The tunnel takes \(T\) days to build, and, once finished, earns £\(d\) a day, all of which goes to pay the daily interest and repay the debt until it is fully paid. The financial transactions take place at the end of each day's work. Show that \(S_{n},\) the total amount borrowed by the end of day \(n\), is given by \[ S_{n}=\frac{c[(1+k)^{n}-1]}{k} \] for \(n\leqslant T\). Given that \(S_{T+m}>0,\) where \(m>0,\) express \(S_{T+m}\) in terms of \(c,d,k,T\) and \(m.\) Show that, if \(d/c>(1+k)^{T}-1,\) the firm will eventually pay off the debt.

A smooth billiard ball moving on a smooth horizontal table strikes another identical ball which is at rest. The coefficient of restitution between the balls is \(e(<1)\). Show that after the collision the angle between the velocities of the balls is less than \(\frac{1}{2}\pi.\) Show also that the maximum angle of deflection of the first ball is \[ \sin^{-1}\left(\frac{1+e}{3-e}\right). \]

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Solution:

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Set up the coordinate frame so that the \(x\)-direction is the line of centres of the spheres. Then if the initial velocities are \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{0}{0}\). Then the final velocities must be: \(\displaystyle \binom{v_{x1}}{u_y}\) and \(\displaystyle \binom{v_{x2}}{0}\) where \(mu_x = mv_{x1}+mv_{x2}\) by conservation of energy and \(\frac{v_{x1}-v_{x2}}{u_x} = -e\). \begin{align*} && \begin{cases} v_{x1}+v_{x2} &= u_x \\ v_{x1}-v_{x2} &= -eu_x \\ \end{cases} \\ \Rightarrow && 2v_{x1} &= (1-e)u_x \\ \Rightarrow && v_{x1} &= \frac{(1-e)}{2} u_x \\ && v_{x2} &= \frac{1+e}{2} u_x \end{align*} Notice that since \(0 < e < 1\) we must have \(v_{x1} > 0\) and so the ball on the left is still continuing in the positive direction, therefore the angle will be less than \(\frac12 \pi\). The angle the first ball is deflected through is the angle between: \(\displaystyle \binom{u_x}{u_y}\) and \(\displaystyle \binom{\frac{1-e}{2}u_x}{u_y}\). We can scale the velocities so \(u_y = 1\). So we are interested in the angle between \(\displaystyle \binom{x}{1}\) and \(\displaystyle \binom{\frac{1-e}{2}x}{1}\). To maximise \(\theta\) we can maximise \(\tan \theta\), so: \begin{align*} && \tan \theta &= \frac{\frac{2}{(1-e)x-\frac{1}{x}}}{1+\frac{2}{(1-e)x^2}} \\ &&&= \frac{2x-(1-e)x}{(1-e)x^2+2} \\ &&&= \frac{(1+e)x}{(1-e)x^2+2} \\ \\ \frac{\d}{\d t}: &&&= \frac{(1+e)((1-e)x^2+2)-2(1+e)(1-e)x^2}{\sim} \\ &&&= \frac{2(1+e)-(1+e)(1-e)x^2}{\sim}\\ \frac{\d}{\d t} = 0: &&0 &= 2(1+e)-(1+e)(1-e)x^2 \\ \Rightarrow && x &= \pm \sqrt{\frac{2}{1-e}} \\ \\ \Rightarrow && \tan \theta &= \frac{\pm(1+e)\sqrt{\frac{2}{1-e}}}{2+2} \\ &&&= \pm \frac{\sqrt{2}(1+e)}{4\sqrt{1-e}} \\ \Rightarrow && \cot^2 \theta &= \frac{8(1-e)}{(1+e)^2} \\ \Rightarrow && \cosec^2 \theta &= \frac{8(1-e)}{(1+e)^2} + 1 \\ &&&= \frac{8-8e+1+2e+e^2}{(1+e)^2} \\ &&&= \frac{9-6e+e^2}{(1+e)^2} \\ &&&= \frac{(3-e)^2}{(1+e)^2} \\ \Rightarrow && \theta &= \sin^{-1} \left ( \frac{1+e}{3-e}\right) \end{align*}

Let \(y=\mathrm{f}(x)\), \((0\leqslant x\leqslant a)\), be a continuous curve lying in the first quadrant and passing through the origin. Suppose that, for each non-negative value of \(y\) with \(0\leqslant y\leqslant\mathrm{f}(a)\), there is exactly one value of \(x\) such that \(\mathrm{f}(x)=y\); thus we may write \(x=\mathrm{g}(y)\), for a suitable function \(\mathrm{g}.\) For \(0\leqslant s\leqslant a,\) \(0\leqslant t\leqslant \mathrm{f}(a)\), define \[ \mathrm{F}(s)=\int_{0}^{s}\mathrm{f}(x)\,\mathrm{d}x,\qquad\mathrm{G}(t)=\int_{0}^{t}\mathrm{g}(y)\,\mathrm{d}y. \] By a geometrical argument, show that \[ \mathrm{F}(s)+\mathrm{G}(t)\geqslant st.\tag{*} \] When does equality occur in \((*)\)? Suppose that \(y=\sin x\) and that the ranges of \(x,y,s,t\) are restricted to \(0\leqslant x\leqslant s\leqslant\frac{1}{2}\pi,\) \(0\leqslant y\leqslant t\leqslant1\). By considering \(s\) such that the equality holds in \((*)\), show that \[ \int_{0}^{t}\sin^{-1}y\,\mathrm{d}y=t\sin^{-1}t-\left(1-\cos(\sin^{-1}t)\right). \] Check this result by differentiating both sides with respect to \(t\).

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Solution:

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The blue area is \(F(s)\) the red area is \(G(t)\), the dashed rectangle (which is a subset of the red and blue areas) has area \(st\) therefore \(F(s) + G(t) \geq st\). Equality holds if \(f(s) = t\). \begin{align*} && \int_0^t \sin^{-1} y \d y + \int_0^{\sin^{-1} t} \sin x \d x &= t \sin^{-1} t \\ \Rightarrow && \int_0^t \sin^{-1} y \d y &= t \sin^{-1} t - \left [ -\cos (x) \right]_0^{\sin^{-1} t} \\ &&&= t \sin^{-1} t - (1- \cos (\sin^{-1} t)) \end{align*} Let \(y = t \sin^{-1} t - (1- \cos (\sin^{-1} t))\) then, \begin{align*} \frac{\d y}{\d t} &= \sin^{-1} t +t \frac{\d}{\d t} \l \sin^{-1} (t) \r - \sin ( \sin^{-1} t) \frac{\d}{\d t} \l \sin^{-1} (t) \r \\ &= \sin^{-1} t \end{align*} as required

Prove that:

1. if \(a+2b+3c=7x\), then \[ a^{2}+b^{2}+c^{2}=\left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2}; \]
2. if \(2a+3b+3c=11x\), then \[ a^{2}+b^{2}+c^{2}=\left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2}. \]

A definite integral can be evaluated approximately by means of the Trapezium rule: \[ \int_{x_{0}}^{x_{N}}\mathrm{f}(x)\,\mathrm{d}x\approx\tfrac{1}{2}h\left\{ \mathrm{f}\left(x_{0}\right)+2\mathrm{f}\left(x_{1}\right)+\ldots+2\mathrm{f}\left(x_{N-1}\right)+\mathrm{f}\left(x_{N}\right)\right\} , \] where the interval length \(h\) is given by \(Nh=x_{N}-x_{0}\), and \(x_{r}=x_{0}+rh\). Justify briefly this approximation. Use the Trapezium rule with intervals of unit length to evaluate approximately the integral \[ \int_{1}^{n}\ln x\,\mathrm{d}x, \] where \(n(>2)\) is an integer. Deduce that \(n!\approx\mathrm{g}(n)\), where \[ \mathrm{g}(n)=n^{n+\frac{1}{2}}\mathrm{e}^{1-n}, \] and show by means of a sketch, or otherwise, that \[ n!<\mathrm{g}(n). \] By using the Trapezium rule on the above integral with intervals of width \(k^{-1}\), where \(k\) is a positive integer, show that \[ \left(kn\right)!\approx k!n^{kn+\frac{1}{2}}\left(\frac{\mathrm{e}}{k}\right)^{k\left(1-n\right)}. \] Determine whether this approximation or \(\mathrm{g}(kn)\) is closer to \(\left(kn\right)!\).

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Solution:

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We can approximate the integral by \(N\) trapeziums, each with height \(x_{i+1}-x_{i} = \frac{x_N-x_0}{N} = \frac{h}{N}\). The will have area \(\frac{(f(x_i)+f(x_{i+1}))h}{2}\) and summing all these areas we will get: \[\frac12 h \l f(x_0) + f(x_1) + f(x_1)+f(x_2) + \cdots + f(x_{N-1})+f(x_N) \r = \frac12 h \l f(x_0) +2 f(x_1) + + \cdots +2f(x_{N-1})+f(x_N) \r\] But this is approximately the integral \(\displaystyle \int_{x_0}^{x_N} f(x) \d x\) \begin{align*} && \int_1^n \ln x \d x &= [x \ln x]_1^n - \int_1^n x \cdot \frac{1}{x} \d x \\ &&&= n \ln n - n+1 \\ &&&\approx \frac12 \l \ln 1 + 2\sum_{k=2}^{n-1} \ln k + \ln n \r \\ &&&= \ln (n!) - \frac12 \ln n \\ \Rightarrow && \ln (n!) &\approx n \ln n + \frac12 \ln n - n + 1 \\ \Rightarrow && n! &\approx \exp(n \ln n + \frac12 \ln n - n + 1) \\ &&&=n^{n+\frac12}e^{1-n} \end{align*} Since \(\ln x\) is a concave function, we should expect all the trapeziums to all lie under the curve, therefore this is always an underestimate for the integral, ie \(n! < g(n)\)

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\begin{align*} && \int_1^n \ln x \d x &= n \ln n - n+1 \\ &&&\approx \frac12 k^{-1} \l \ln 1 + 2\sum_{r=1}^{k(n-1)-1} \ln \l 1+\frac{r}{k} \r + \ln n \r \\ &&&=\frac{1}{2k} \l 2\sum_{r=1}^{k(n-1)-1} \l \ln(k+r) - \ln k)\r + \ln n\r \\ &&&=\frac1{k} \l \ln ((k+k(n-1)-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&=\frac1{k} \l \ln ((kn-1)!) - \ln(k!) - (k(n-1)-1) \ln k+\frac12 \ln n \r \\ &&&=\frac1{k} \l \ln ((kn)! ) -\ln k -\ln n - \ln(k!) - (k(n-1)-1) \ln k+\frac12\ln n \r \\ &&&= \frac1{k} \l \ln ((kn)! ) - \ln(k!) - (k(n-1)) \ln k - \frac12 \ln n\r \\ \Rightarrow && \ln ((kn)!) &\approx kn \ln n - kn + k + \ln(k!) + (k(n-1)) \ln k + \frac12 \ln n\\ \Rightarrow && (kn)! &\approx n^{kn+\frac12}e^{-k(n-1)}k!k^{k(n-1)} \\ &&&= n^{kn+\frac12} k! \l \frac{e}{k} \r^{k(1-n)} \end{align*} I would expect this approximation to be a better approximation for \((kn)!\) since it is created using a finer mesh.

A train of length \(l_{1}\) and a lorry of length \(l_{2}\) are heading for a level crossing at speeds \(u_{1}\) and \(u_{2}\) respectively. Initially the front of the train and the front of the lorry are at distances \(d_{1}\) and \(d_{2}\) from the crossing. Find conditions on \(u_{1}\) and \(u_{2}\) under which a collision will occur. On a diagram with \(u_{1}\) and \(u_{2}\) measured along the \(x\) and \(y\) axes respectively, shade in the region which represents collision. Hence show that if \(u_{1}\) and \(u_{2}\) are two independent random variables, both uniformly distributed on \((0,V)\), then the probability of a collision in the case when initially the back of the train is nearer to the crossing than the front of the lorry is \[ \frac{l_{1}l_{2}+l_{2}d_{1}+l_{1}d_{2}}{2d_{2}\left(l_{2}+d_{2}\right)}. \] Find the probability of a collision in each of the other two possible cases.

Prove that \[ \tan^{-1}t=t-\frac{t^{3}}{3}+\frac{t^{5}}{5}-\cdots+\frac{(-1)^{n}t^{2n+1}}{2n+1}+(-1)^{n+1}\int_{0}^{t}\frac{x^{2n+2}}{1+x^{2}}\,\mathrm{d}x. \] Hence show that, if \(0\leqslant t\leqslant1,\) then \[ \frac{t^{2n+3}}{2(2n+3)}\leqslant\left|\tan^{-1}t-\sum_{r=0}^{n}\frac{(-1)^{r}t^{2r+1}}{2r+1}\right|\leqslant\frac{t^{2n+3}}{2n+3}. \] Show that, as \(n\rightarrow\infty,\) \[ 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}\rightarrow\pi, \] but that the error in approximating \(\pi\) by \({\displaystyle 4\sum_{r=0}^{n}\frac{(-1)^{r}}{(2r+1)}}\) is at least \(10^{-2}\) if \(n\) is less than or equal to \(98\).

Show that, if the lengths of the diagonals of a parallelogram are specified, then the parallogram has maximum area when the diagonals are perpendicular. Show also that the area of a parallelogram is less than or equal to half the square of the length of its longer diagonal. The set \(A\) of points \((x,y)\) is given by \begin{alignat*}{1} \left|a_{1}x+b_{1}y-c_{1}\right| & \leqslant\delta,\\ \left|a_{2}x+b_{2}y-c_{2}\right| & \leqslant\delta, \end{alignat*} with \(a_{1}b_{2}\neq a_{2}b_{1}.\) Sketch this set and show that it is possible to find \((x_{1},y_{1}),(x_{2},y_{2})\in A\) with \[ (x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}\geqslant\frac{8\delta^{2}}{\left|a_{1}b_{2}-a_{2}b_{1}\right|}. \]

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Solution: In a parallelogram the diagonals meet at their mid points. Fixing one diagonal, we can look at the two triangles formed by the other diagonal. Suppose the angle between them is \(\theta\). Then the area of the triangles will be \(\frac12 \frac{l_1}{2} \frac{l_2}2 \sin \theta+\frac12 \frac{l_1}{2} \frac{l_2}2 \sin (\pi -\theta) = \frac{l_1l_2}{4} \sin \theta\). This will be true on both sides. Therefore we can maximise this area by setting \(\theta = \frac{\pi}{2}\).

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Consider the (darker) shaded area. This is our set \(A\). The area of the set is indifferent to a parallel shift in the lines, so without loss of generality, we can consider \(c_1 = 0, c_2 = 0\), so our lines meet at the origin. Now also consider the linear transformation \(\begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \end{pmatrix}^{-1}\) which takes the coordinate axes to these lines. This will take the square \([-\delta, \delta] \times [-\delta, \delta]\) which has area \(4\delta^2\) to our square, which will have area \(\frac{4 \delta^2}{ |a_1b_2 - a_2b_1|}\). If we consider the length of the two diagonals of this area, \(l_1, l_2\) we know that \(\frac{l_1l_2}2 \sin \theta = \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\), if we consider the larger of \(l_1\) and \(l_2\) (wlog \(l_1\)) we must have that \(\frac{l_1^2}{2} \geq \frac{4 \delta^2}{|a_1b_2 - a_2b_1|}\) and so points on opposite ends of the diagonal will satisfy the inequality in the question.