Problems

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Solution: \begin{align*} && \cos(A \pm B) &= \cos A \cos B \mp \sin A \sin B \\ A = a, B = 3a&& \cos 4a + \cos 2a &= 2\cos 3a \cos a \\ \Rightarrow && \cos a \cos 3a &= \tfrac12(\cos 4a + \cos 2a) \\ \\ && \sin(A \pm B) &= \sin A \cos B \pm \cos A \sin B \\ && \sin 4a + \sin(- 2a) &= 2 \sin a \cos 3a \\ \Rightarrow && \sin a \cos 3a &= \tfrac12 (\sin 4a - \sin 2a) \end{align*}

1. \(\,\) \begin{align*} && 1 &= 4 \cos x \cos 2x \cos 3x \\ &&&= 2(\cos 4x +\cos 2x)\cos 2x \\ c = \cos 2x:&&&= 2(2c^2-1+c)c \\ \Rightarrow && 0 &= 4c^3+2c^2-2c-1 \\ &&&= (2c+1)(2c^2-1) \\ \Rightarrow && \cos 2x &= -\frac12 \\ \Rightarrow && x &= \frac{\pi}{3}, \frac{2\pi}{3} \\ && \cos 2x &= \pm \frac1{\sqrt2} \\ \Rightarrow && x&= \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \end{align*}
2. \(\,\) \begin{align*} && \tan x &= \tan 2x \tan 3x \tan 4x \\ \Rightarrow &&1 &= \frac{\cos x\sin 2x \sin 3x \sin 4x}{\sin x \cos 2x \cos 3x \cos 4x} \\ &&&= \frac{\sin 2x \sin 4x (\sin4 x + \sin 2x)}{\cos 2x \cos 4x (\sin 4x - \sin 2x)} \\ &&&= \frac{(\cos 2x - \cos 6x) (\sin4 x + \sin 2x)}{(\cos 6x + \cos 2x) (\sin 4x - \sin 2x)} \\ \Rightarrow && 0 &= 2\cos 6x \sin 4x - 2\cos 2x \sin 2 x\\ &&&= \sin 4 x (2 \cos 6x - 1) \\ \Rightarrow && \sin 4x &= 0 \\ \text{ or }&& \cos 6x &= \frac12 \end{align*} \(\sin 4x = 0 \Rightarrow x = 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi\) \(\cos 6x = \frac12 \Rightarrow x = \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\). We should verify these work, since not all of them will, especially where \(\sin 4x = 0\), so our final answer is \(x = 0, \pi, \frac{\pi}{18}, \frac{5\pi}{18},\frac{7\pi}{18}, \frac{11\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}\)

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Solution:

1. \(\,\) \begin{align*} && \dot{x} &=12 \cos^2 t \sin t \\ && \dot{y} &= 12 \cos t - 12 \sin^2 t \cos t \\ && \frac{\d y}{\d x} &= \frac{12 \cos t - 12 \sin^2 t \cos t}{12 \cos^2 t \sin t} \\ &&&= \frac{1 - \sin^2 t}{\cos t \sin t} \\ &&&= \cot t \\ \\ && \frac{y - (12\sin\phi - 4\sin^3\phi)}{x - (-4 \cos^3 \phi)} &= - \tan \phi \\ && y &= -\tan \phi x -4 \cos^3 \phi \tan \phi + 12 \sin \phi -4\sin^3 \phi \\ &&&= -\tan \phi x -4 \cos^2 \phi \sin \phi + 12 \sin \phi -4\sin^3 \phi \\ &&&= -\tan \phi x - 4\sin \phi+12 \sin \phi \\ &&y&= -\tan \phi x + 8 \sin \phi \end{align*} Note that when \(x = 8\cos^3 \phi\) we have \(y =-8 \cos^2 \phi \sin \phi + 8 \sin \phi = 8 \sin^3 \phi\). So the point lies on the curve. Notice also that \((8\cos^3 \phi, 8 \sin^ 3\phi)\) is a parametrisation of \(x^{2/3} + y^{2/3} = 4\) and so we can use parametric differentiation to see the gradient is \(\frac{24\sin^2 \phi \cos \phi}{-24\cos^2 \phi\sin\phi} = - \tan \phi\) so it also has the same gradient as required.
   

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2. \(\,\) \begin{align*} && \dot{x} &= -\sin t + \sin t + t \cos t \\ &&&= t \cos t \\ && \dot{y} &= \cos t - \cos t + t \sin t \\ &&&= t \sin t \\ && \frac{\d y}{\d x} &= \frac{t \sin t}{t \cos t} = \tan t \\ \\ && \frac{y - (\sin \phi - \phi \cos \phi)}{x - (\cos \phi + \phi \sin \phi)} &= -\cot \phi \\ \Rightarrow && y &= -\cot \phi x + (\cos \phi + \phi \sin \phi) \cot \phi + \sin \phi - \phi \cos \phi \\ &&&= -\cot \phi x + \cos \phi \cot \phi + \phi \cos \phi + \sin \phi - \phi \cos \phi \\ &&&= -\cot \phi x + \frac{\cos^2 \phi + \sin^2 \phi}{\sin \phi} \\ &&&= -\cot \phi x + \cosec \phi \end{align*} The distance to the origin is \(\displaystyle \frac{|\cosec \phi|}{\sqrt{1 + \cot^2 \phi}} = 1\) so this normal is a tangent to \(x^2 + y^2 = 1\)
   

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This is an interesting question because many years ago this question of finding involutes and envelopes of questions would be considered extremely standard. (Particularing finding the involute of a circle). (It also seems to make sense mechanically imagine unwinding (or winding) a piece of string from a circle!)

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Solution: \begin{questionparts} \item \(\,\) \begin{align*} && I_n &= \int_0^{\beta} (\sec x + \tan x)^n \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} \left ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( (\sec x + \tan x)^{2}+1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( \sec^2 x + \tan^2 x + 2\sec x \tan x + 1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( 2\sec x \tan x +2\sec^2 x \right) \, \d x \\ &&& = \left [\frac1n(\sec x + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[(\sec \beta + \tan \beta)^n - 1] \end{align*} Notice that by AM-GM \(\tfrac12( ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}) \geq (\sec x + \tan x)^{n}\) with equality not holding most of the time. Integrating we obtain our result. \item \(\,\) \begin{align*} && J_n &= \int_0^{\beta} (\sec x \cos \beta + \tan x )^n \d x \\ && \tfrac12( J_{n+1} + J_{n-1}) &= \tfrac12 \int_0^{\beta} \left ( (\sec x \cos \beta + \tan x )^{n+1} +(\sec x \cos \beta + \tan x )^{n-1}\right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( (\sec x \cos \beta + \tan x )^{2} + \right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec^2 x \cos^2 \beta + \tan^2 x+ 2\sec x \tan x \cos \beta +1 \right ) \d x \\ && &= \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\tfrac12(\cos^2 \beta +1)\sec^2 x \right ) \d x \\ && &< \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\sec^2 x \right ) \d x \\ &&&= \left [\frac1n (\sec x \cos \beta + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[ (1 + \tan \beta)^n - \cos^n \beta] \end{align*} But notice we can use the same AM-GM argument from before to show that \(J_n < \tfrac12( J_{n+1} + J_{n-1}) < \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]\)

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Solution:

1. \(\,\) \begin{align*} && \cos \angle MOB &= \frac{\mathbf{m} \cdot \mathbf{b}}{|\mathbf{m}||\mathbf{b}|} \\ &&&= \frac{\cos \theta + 1}{2\sqrt{\frac14(\mathbf{a}+\mathbf{b})\cdot(\mathbf{a}+\mathbf{b})}} \\ &&&= \frac{\cos \theta + 1}{\sqrt{1+1+2\cos \theta}} \\ &&&= \frac{1 + \cos \theta}{\sqrt{2(1+\cos \theta})} \\ &&&= \frac1{\sqrt{2}} \sqrt{1+\cos \theta} \\ &&&= \cos \tfrac{\theta}{2} \end{align*} Since \(0 < \theta < \pi\) we must have \(\angle MOB = \tfrac{\theta}{2}\) ie it is the angle bisector.
2. The plane through the origin perpendicular to \(\mathbf{c}\) has \(\mathbf{x} \cdot \mathbf{c} = 0\), so \begin{align*} && \mathbf{a}_1 \cdot \mathbf{c} &= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot \mathbf{c} \\ &&&= \mathbf{a} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{c} \\ &&&= 0 \\ \\ && |\mathbf{a}_1|^2 &= \mathbf{a}_1 \cdot \mathbf{a}_1 \\ &&&= (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \\ &&&= 1 - 2(\mathbf{a} \cdot \mathbf{c})^2 + \mathbf{a} \cdot \mathbf{c} \\ &&&= (1-\cos^2 \alpha) \\ &&&= \sin^2 \alpha \\ \Rightarrow && |\mathbf{a}_1| &= \sin \alpha \\ \Rightarrow && |\mathbf{b}_1| &= \sin \beta \tag{changing a and b} \\ \\ && \cos \phi &= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1}{|\mathbf{a}_1||\mathbf{b}_1|} \\ &&&= \frac{(\mathbf{a} - (\mathbf{a} \cdot \mathbf{c}) \mathbf{c}) \cdot (\mathbf{b} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{c})}{\sin \alpha \sin \beta} \\ &&&= \frac{\mathbf{a} \cdot \mathbf{b} - 2(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})+(\mathbf{a} \cdot \mathbf{c}) \cdot (\mathbf{b} \cdot \mathbf{c})}{\sin \alpha \sin \beta} \\ &&&= \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} \end{align*}
3. Note that \(\mathbf{m}_1 = \tfrac12(\mathbf{a}_1 + \mathbf{b}_1)\) either by expanding or by noting that projection is linear \begin{align*} && \cos \angle M_1OB_1 &= \frac{\mathbf{m}_1 \cdot \mathbf{b}_1}{|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{(\mathbf{a}_1 + \mathbf{b}_1) \cdot \mathbf{b}_1}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{\mathbf{a}_1 \cdot \mathbf{b}_1 + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{|\mathbf{a}_1 || \mathbf{b}_1| \cos \phi + |\mathbf{b}_1|^2}{2|\mathbf{m}_1||\mathbf{b}_1|} \\ &&&= \frac{|\mathbf{a}_1 |\cos \phi + |\mathbf{b}_1|}{2|\mathbf{m}_1|} \\ &&&= \frac{\sin \alpha \cos \phi + \sin \beta}{2\sin \frac{\theta}{2}} \\ &&&= \frac{\sin \alpha \frac{\cos \theta - \cos \alpha \cos \beta}{\sin \alpha \sin \beta} + \sin \beta}{2\sin \frac{\theta}{2}} \\ &&&= \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} \\ \Rightarrow && \cos \angle M_1OA_1 &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \end{align*} \(M_1\) is a bisector iff these two cosines are equal, ie \begin{align*} && \cos \angle M_1OB_1 &= \cos \angle M_1OA_1 \\ \Leftrightarrow && \frac{\cos \theta - \cos \alpha \cos \beta+ \sin^2 \beta}{2\sin \frac{\theta}{2} \sin \beta} &= \frac{\cos \theta - \cos \beta \cos \alpha+ \sin^2 \alpha}{2\sin \frac{\theta}{2} \sin \alpha} \\ \Leftrightarrow && \cos \theta (\sin \alpha - \sin \beta) &= \cos \alpha \cos \beta(\sin \alpha - \sin \beta) + \sin \alpha \sin \beta (\sin \alpha - \sin \beta) \\ \Leftrightarrow &&0 &= (\sin \alpha - \sin \beta)( \cos \theta - (\cos \alpha \cos \beta + \sin \alpha \sin \beta)) \\ &&&= (\sin \alpha - \sin \beta) (\cos \theta - \cos (\alpha - \beta)) \end{align*} From which the result immediately follows

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Solution:

1. The curves meet when they have the same radius for a given \(\theta\) ie \begin{align*} && a + 2 \cos \theta &= 2 + \cos 2 \theta \\ &&&= 2 + 2\cos^2 \theta - 1 \\ \Rightarrow && 0 &= 2 \cos ^2 \theta - 2 \cos \theta + 1 - a \end{align*} The curves touch if this has a repeated root, ie \(0 = \Delta = 4 - 8(1-a) \Rightarrow a = \frac12\). The second way the curves can touch is if there is a single root, but it's at an extreme value of \(\cos \theta = \pm 1\) ie \(0 = 2 - 2\cdot(\pm1) + 1 - a \Rightarrow a = 3 \pm 2 = 1, 5\)
2. Suppose \(a = \frac12\) then the curves touch when \(0 = 2\cos^2 \theta - 2 \cos \theta + \frac12 = (2 \cos \theta-1 )(\cos \theta -\frac12) \Rightarrow \theta = \pm \frac{\pi}{3}\)
   

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3. \(a = 1\)
   

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   \(a = 5\)
   

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Solution: \begin{align*} \int_0^{\frac{1}{4}\pi} \frac{1}{1 + \sin x} dx &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{1 - \sin^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \frac{1-\sin x}{\cos^2 x} dx \\ &= \int_0^{\frac{1}{4}\pi} \sec^2 x - \sec x \tan x dx \\ &= \left [\tan x-\sec x \right]_0^{\frac{1}{4}\pi} \\ &= 2 - \frac{1}{\sqrt{2}} \end{align*} \begin{align*} \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1}{1 + \sec x} \d x &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{1-\sec x}{1 - \sec^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \frac{\sec x-1}{\tan^2 x} \d x \\ &= \int_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \cot x \cosec x-\cot^2 x\d x \\ &= \left [ -\cosec x +x+\cot x\right]_{\frac{1}{4}\pi}^{\frac{1}{3}\pi} \\ &= \l -\frac{2}{\sqrt3}+\frac{\pi}{3}+\frac{1}{\sqrt{3}}\r - \l-\sqrt{2}+\frac{\pi}{4}+1 \r \\ &= \frac{\pi}{12}-\frac{1}{\sqrt{3}}+\sqrt{2}-1 \end{align*} \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{(1-\sin^2 x)^2} \d x \\ &= \int_0^{\frac{1}{3}\pi} \frac{1-2\sin x+\sin^2x}{\cos^4 x} \d x \\ \end{align*} Splitting this up into: \begin{align*} \int_0^{\frac{1}{3}\pi} \frac{-2\sin x}{\cos^4 x} \d x &= -\frac23 \left [ \frac{1}{\cos^3 x}\right]_0^{\frac{1}{3}\pi} \\ &= -\frac{16}3+\frac23 \\ &= -\frac{14}3 \end{align*} and \begin{align*} && \int_0^{\frac{1}{3}\pi} \frac{1+\sin^2x}{\cos^4 x} \d x &= \int_0^{\frac{1}{3}\pi} (\sec^2 x + \tan^2 x) \sec^2 x \d x \\ &&&= \int_0^{\frac{1}{3}\pi} (1+ 2\tan^2 x) \sec^2 x \d x \\ u = \tan x, \d u = \sec^2 x \d x&&&= \int_0^{\sqrt{3}}(1+2u^2) \d u \\ &&&= \left [u + \frac23 u^3 \right]_0^{\sqrt{3}} \\ &&&= \sqrt{3} + 2\sqrt{3} \\ &&&= 3\sqrt{3} \end{align*} And so our complete integral is: \[ \int_0^{\frac{1}{3}\pi} \frac{1}{(1 + \sin x)^2} \d x = 3\sqrt{3} - \frac{14}3\]

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Solution:

Note that the sine rule gives us \begin{align*} && \frac{x}{\sin \alpha} &= \frac{AB}{\sin (180-3\alpha)} \\ \Rightarrow && AB &= x \frac{\sin 3\alpha}{\sin \alpha} \\ &&&= x \frac{\sin \alpha \cos 2\alpha + \cos \alpha \sin 2\alpha}{\sin \alpha} \\ &&&= x \left ( \frac{\sin \alpha (1-2\sin^2\alpha) + 2(1-\sin^2 \alpha)\sin \alpha}{\sin \alpha} \right) \\ &&&= x (3 - 4\sin^2 \alpha) \end{align*} Note that \(AD = (\tfrac32 - 2 \sin^2\alpha)x\) and \(AE = (3-4\sin^2\alpha-\cos2\alpha)x\) so \begin{align*} DE &= (3-4\sin^2\alpha-\cos2\alpha)x - (\tfrac32 - 2 \sin^2\alpha)x \\ &= (\tfrac32 - 2\sin^2 \alpha - (1-2\sin^2 \alpha))x \\ &= \tfrac12x \end{align*} Since \(DE = \frac12x\) if we drop the perpendicular from \(F\) to \(EC\) we have a \(30-60-90\) triangle. Therefore \(\angle FCE = 30^\circ\). Notice that \(\angle CEB = 90^{\circ} - 2\alpha\) and \(\angle ACB = 180^\circ - 3\alpha\), therefore \begin{align*} \angle ACF &= \angle ACB - \angle FCE - \angle ECB \\ &= (180^\circ - 3\alpha) - 30^\circ - (90^{\circ} - 2\alpha) \\ &= 60^\circ - \alpha \\ &= \frac13 \angle ACB \end{align*}

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Solution: \begin{align*} && \tan \left ( \tfrac14 \pi -\tfrac12 x \right) &\equiv \frac{\tan \tfrac{\pi}{4} - \tan \tfrac12 x}{1 + \tan \tfrac{\pi}{4} \tan \tfrac12 x} \\ &&&= \frac{1-\tan \tfrac12 x}{1+\tan \tfrac12 x} \\ \\ && \sec x - \tan x &= \frac{1+t^2}{1-t^2} - \frac{2t}{1-t^2} \\ &&&= \frac{(1-t)^2}{(1-t)(1+t)} \\ &&&= \frac{1-t}{1+t} \end{align*} Therefore both sides are equal to the same thing.

1. \(\tan \tfrac18 \pi = \tan(\tfrac14 \pi - \tfrac12 \tfrac14\pi) = \sec \tfrac14 \pi - \tan \tfrac14 \pi = \sqrt{2} - 1\) \begin{align*} && \tan \tfrac{11}{24} \pi &= \tan (\tfrac13 \pi +\tfrac18 \pi) \\ &&&= \frac{\tan \tfrac13 \pi +\tan \tfrac18 \pi}{1-\tan \tfrac13 \pi \tan \tfrac18 \pi} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2}-1)} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 +\sqrt{3}-\sqrt{6}} \\ \end{align*}
2. \(\,\) \begin{align*} && (\sqrt{3}-\sqrt{6}+1)(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) &= (2\sqrt{3}+\sqrt{6}+3+3\sqrt{2}) + \\ &&&\quad+(-2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-6) + \\ &&&\quad+(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{3}+\sqrt{2}-1 \end{align*}
3. \(\,\) \begin{align*} && \tan \tfrac{1}{48} \pi &= \tan (\tfrac14\pi - \tfrac{11}{48} \pi) \\ &&&= \sec \tfrac{11}{24} \pi - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1+\tan^2 \tfrac{11}{24}\pi} - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1 + (2+\sqrt{2}+\sqrt{3}+\sqrt{6})^2} - (2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{16+10\sqrt{2} + 8\sqrt{3}+6\sqrt{6}} - 2 - \sqrt2 - \sqrt3-\sqrt6 \end{align*}