Problems

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Solution:

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2. \begin{enumerate}
3. since \(\frac{\ln x}{x}\) is decreasing on \((e, \infty)\) we must have that \(\frac{\ln 3}{3} > \frac{\ln \pi}{\pi} \Rightarrow e^\pi > \pi^3\)
4. similarly, since \(\frac{\ln x}{x}\) is increasing on \((0, e)\) we must have that \(\frac{\ln \sqrt{5}}{\sqrt{5}} < \frac{\ln 9/4}{9/4} \Rightarrow \left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
5. Since \(2^4 = 4^2\) notice also that:
   

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   from the graph we must have the green area between \(1\) and \(2\) mapping to the (higher) green area between \(3\) and \(4\). Therefore \((x+2)^x > x^{x+2}\) for \(1 < x < 2\)
6. \begin{align*} && 9^{\sqrt 2} & \stackrel{?}{>} \sqrt{2}^9 \\ \Leftrightarrow && (3^2)^{\sqrt2} &\stackrel{?}{>} (\sqrt{2}^3)^3 \\ \Leftrightarrow && 3^{2 \sqrt2} &\stackrel{?}{>} (2\sqrt2)^3 \end{align*} Since \(e^2 < 8 < 9\Rightarrow e < 2\sqrt2 < 3\) therefore: \begin{align*} && \frac{\ln 2 \sqrt2}{2 \sqrt 2} &> \frac{\ln 3}{3} \\ \Leftrightarrow && (2 \sqrt{2})^3 &> 3^{2 \sqrt{2}} \\ \Leftrightarrow && \sqrt{2}^9 &> 9^{\sqrt 2} \\ \end{align*}
7. \begin{align*} && 8^{\sqrt[3]{3}} & \stackrel{?}{>} \sqrt[3]{3}^8 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (\sqrt[3]{3}^4)^2 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (3\sqrt[3]{3})^2 \\ \end{align*} Since \(3\sqrt[3]{3} > 4\) we have \begin{align*} && \frac{\ln (3 \sqrt[3]3)}{3 \sqrt[3]3} &< \frac{\ln 4}{4} \\ &&&= \frac{\ln 2}{2}\\ \Rightarrow && (3 \sqrt[3]{3})^2 &< 2^{3 \sqrt[3]{3}} \\ \Rightarrow && \sqrt[3]3^8 &< 8^{\sqrt[3]3} \end{align*}

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Solution:

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   By symmetry we can observe that the parabola and circle will intersect \(0, 1\) (at the base), \(2, 4\) times. So setting up our system of equations we have: \begin{align*} &&& \begin{cases} x^2 + (y-a)^2 &= r^2 \\ 2ky &= x^2 \end{cases} \\ \Rightarrow && r^2 &= x^2 + \left (\frac{x^2}{2k} - a \right )^2 \\ \Rightarrow &&r^2 &= x^2 + a^2 - \frac{ax^2}{k} + \frac{x^4}{4k^2} \\ \Rightarrow &&0 &= \frac{1}{4k^2} x^4 + \left ( 1 - \frac{a}{k} \right) x^2 + a^2 - r^2 \\ \Rightarrow && \Delta &= \left ( 1 - \frac{a}{k} \right)^2-4 \cdot \frac{1}{4k^2} (a^2 - r^2) \\ &&&= 1 - \frac{2a}{k} + \frac{a^2}{k^2} - \frac{a^2}{k^2} + \frac{r^2}{k^2} \\ &&&= \frac{k^2-2ka+r^2}{k^2} \end{align*} Since there will be (at most) two solutions if \(\Delta = 0\) we must have if the circle and parabola are tangent \(r^2 - 2ka + k^2 = 0 \Rightarrow r^2 = k(2a-k)\). So long as there is a solution \(x^2 > 0\) there will be two tangent points, so if \(-\left(1 - \frac{a}{k}\right) > 0\) or \(a > k > 0\)
2. Since \(y = c \pm x\) are tangent to the circle with radius \(r\) and centre \((0,a)\) we have the following equations: \begin{align*} &&& \begin{cases} x^2 + (y-a)^2 &= r^2 \\ c \pm x &= y \end{cases} \\ \Rightarrow && r^2 &= x^2 + (c -a\pm x)^2 \\ &&&= 2x^2+(c-a)^2 \pm 2x(c-a) \\ \Rightarrow && \Delta &= 4(c-a)^2 -4 \cdot 2 \left ( (c-a)^2 -r^2 \right)\\ &&&= 8r^2-4(c-a)^2 \\ \Rightarrow && x &= \frac{\mp 2(c-a) \pm \sqrt{\Delta}}{4} \\ &&&= \mp \frac12 (c-a) \\ && y &= \pm \frac12 (c+a) \\ && (x,y) &= \left (\frac12 (c-a), \frac12 (c+a)\right), \left (-\frac12 (c-a), -\frac12 (c+a)\right) \end{align*}

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Solution:

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   \begin{align*} && y & = x^{1/x} = \exp \left ( \tfrac1x \ln x \right ) \\ \Rightarrow && y' &= \exp \left ( \tfrac1x \ln x \right ) \cdot \left ( \frac{1}{x^2} - \frac{\ln x}{x^2} \right ) \\ &&&= \frac{\exp \left ( \tfrac1x \ln x \right ) }{x^2}(1 - \ln x) \\ y' =0: && x &= e \end{align*} Therefore the largest integer values will be \(2\) or \(3\). Comparing \((2^{\frac12})^6 = 8 < 9 = (3^{1/3})^6\) we see the maximum value of \(n^{1/n}\) where \(n\) is an integer is \(\sqrt[3]{3}\)
2. The number of tests is \(r\) plus however many groups fail times \(k\). The probability of a group failing is \(g = 1-(1-p)^k\) and the number of failing groups is \(\sim B(r, g)\) so the expected number of additional groups is \(rg\) and the expected total number of tests is \[ \frac{N}{k} + N(1-(1-p)^k) \]
3. \(\,\) \begin{align*} && N &\geq E = \frac{N}{k} + N(1-(1-p)^k) \\ \Rightarrow && 1 &\geq \frac{1}{k} + 1-(1-p)^k \\ \Rightarrow && (1-p)^k &\geq \frac1k \\ \Rightarrow && k \ln(1-p) &\geq - \ln k \\ \Rightarrow && \ln(1 - p) &\geq -\frac{1}{k} \ln k \geq -\frac13 \ln 3 \\ \Rightarrow && 1-p &\geq \frac{1}{\sqrt[3]{3}} \\ \Rightarrow && p &\leq 1-\frac{1}{\sqrt[3]{3}} \end{align*} (taking \(k=3\)) Claim: \(1 - \frac{1}{\sqrt[3]{3}} > \frac14\) Proof: This is equivalent to \(\sqrt[3]{3} > \frac43\) or \(81 > 4^3 = 64\) which is clearly true.
4. If \(pk\) is small then \((1-p)^k \approx 1 - pk\) and so we obtain \(N \left ( \frac1k +pk \right)\) as required. If \(p = 0.01\) and \(k = 10\) then \(\frac{1}{10} + 0.01 \cdot 10 = 0.2\) so the expected number of tests is \(\sim 20\%\) of \(N\)

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Solution:

1. \(\,\) \begin{align*} && f_\alpha(x) &= \tan^{-1}\!\left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right) \\ && f'_\alpha(x) &= \frac{1}{1 + \left(\frac{x\tan\alpha + 1}{\tan\alpha - x}\right) ^2} \cdot \frac{\tan \alpha \cdot (\tan \alpha - x) - (x \tan \alpha + 1) \cdot (-1)}{(\tan \alpha - x)^2} \\ &&&= \frac{\tan^2 \alpha -1}{(\tan \alpha - x)^2 + (x \tan \alpha +1)^2} \\ &&&= \frac{\tan^2 \alpha +1}{\tan^2 \alpha - 2x \tan \alpha + x^2 + x^2 \tan^2 \alpha + 2 x \tan \alpha + 1} \\ &&&= \frac{1+\tan^2 \alpha}{(1+\tan^2 \alpha(x^2 + 1)} = \frac{1}{1+x^2} \end{align*}
   

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2. Let \(g(x) = \tanh^{-1}(\sin x) - \sinh^{-1}(\tan x)\) then \begin{align*} && g'(x) &= \frac{1}{1-\sin^2 x} \cdot \cos x - \frac{1}{\sqrt{\tan^2 +1}} \cdot \sec^2 x \\ &&&= \sec x - \frac{\sec^2 x}{|\sec x|} \\ &&& = \begin{cases} 0 &\text{if } \sec x \geq 0 \\ 2 \sec x &\text{ otherwise} \end{cases} \end{align*} Therefore \(g'(x) = 2\sec x\) if \(\tfrac12 \pi < x < \tfrac32\pi\) Therefore $\displaystyle g(x) = \begin{cases} 0 & \text{if } x \in [0, \frac{\pi}{2}] \cup [\frac{3\pi}{2}, 2\pi] \\ \ln( (\tan x + \sec x)^2) + C &\text{otherwise} \end{cases}$
   

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Solution: \(y = (-\tan \theta)(x-1)+k\) so when \(x = 0\), \(y = k + \tan \theta\), so \(Y = (0, k+\tan \theta)\). When \(y = 0\), \(x = 1 + \frac{k}{\tan \theta}\)

1. \(A = \frac12 (k+\tan \theta)\left ( 1 + \frac{k}{\tan \theta} \right) = k + \frac12 \left (\tan \theta + \frac{k^2}{\tan \theta} \right)\) Notice that \(x + \frac{k^2}{x} \geq 2 k\) by AM-GM, so the minimum is \(k + \frac12 \cdot 2k = 2k\)
2. \(\,\) \begin{align*} L &= k + \tan \theta + 1 + k \cot \theta + \sqrt{(k + \tan \theta)^2 + \left (1 + \frac{k}{\tan \theta} \right)^2} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{k^2 + 2 k \tan \theta +\tan^2 \theta + 1 + 2k \cot \theta + k^2\cot^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta + \sqrt{\sec^2 \theta+ 2k \sec\theta\cosec \theta + k^2\cosec^2 \theta} \\ &= k + \tan \theta + 1 + k \cot \theta +\sec \theta + k\cosec \theta\\ &= 1 + \tan \theta + \sec \theta + k (1 + \cot \theta + \cosec \theta) \end{align*} \begin{align*} && \frac{\d L}{\d \theta} &= \sec^2 \theta + \tan \theta \sec \theta + k(-\cosec^2 \theta - \cot \theta \cosec \theta ) \\ \Rightarrow && 0 &=\sec^2 \alpha+ \tan \theta \sec \alpha+ k(-\cosec^2 \alpha- \cot \alpha\cosec \alpha) \\ \Rightarrow && k &= \frac{\sec^2 \alpha+ \tan \alpha\sec \alpha}{\cosec^2 \alpha+ \cot \alpha\cosec \alpha} \\ &&&= \frac{\sin^2 \alpha(1 + \sin \alpha)}{\cos^2 \alpha (1+ \cos \alpha)} \\ &&&= \frac{(1-\cos^2 \alpha)(1 + \sin \alpha)}{(1-\sin^2 \alpha )(1+ \cos \alpha)} \\ &&&= \frac{1-\cos \alpha}{1-\sin \alpha} \\ \Rightarrow && L &= 1 + \tan \alpha + \sec \alpha + \frac{1-\cos \alpha}{1-\sin \alpha} \left (1 + \cot \alpha + \cosec \alpha \right) \\ &&&= \frac{1+\tan \alpha + \sec \alpha -\sin \alpha-\sin \alpha \tan \alpha-\tan \alpha}{1-\sin \alpha} + \\ &&&\quad \quad \frac{1+\cot \alpha + \cosec \alpha-\cos \alpha-\cos \alpha \cot \alpha -\cot \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\sec \alpha(1-\sin^2 \alpha)-\sin \alpha + \cosec \alpha(1-\cos^2 \alpha)-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2+\cos\alpha-\sin \alpha + \sin\alpha-\cos \alpha}{1-\sin \alpha} \\ &&&= \frac{2}{1-\sin \alpha} \end{align*}

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Solution:

1. \begin{align*}\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9}) &= \frac{\sin(\frac{2\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})}{2 \sin \frac{\pi}{9}} \\ &= \frac{\sin(\frac{4\pi}{9})\cos(\frac{4\pi}{9})}{4 \sin \frac{\pi}{9}} \\ &= \frac{\sin(\frac{8\pi}{9})}{8 \sin \frac{\pi}{9}} \\ &= \frac{1}{8} \end{align*}
2. Let \(\displaystyle P_n = \prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right)\). Claim: \(P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n} \r}\). Proof: This is true for \(n = 0\), assume true for \(n-1\) \begin{align*} \sin\l \frac{x}{2^{n}} \r P_n &= P_{n-1} \cos\l \frac{x}{2^{n}} \r \sin\l \frac{x}{2^{n}} \r \\ &= P_{n-1} \frac{1}{2} \sin\l \frac{x}{2^{n-1}} \r \\ &= \frac{\sin 2x}{2^{n} \sin \l \frac{x}{2^{n-1}}\r} \frac{1}{2} \sin\l \frac{x}{2^{n}} \r \\ &= \frac{\sin 2x}{2^{n+1}} \end{align*} Hence \(P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n}\r}\) Taking logs, we determine that: \begin{align*} && \sum_{k=0}^n \ln \cos \l \frac{x}{2^k} \r &= \ln \sin 2x - \ln \sin \l \frac{x}{2^n} \r - (n+1) \ln 2 \\ \Rightarrow && \sum_{k=0}^n \frac{1}{2^k} \tan \l \frac{x}{2^k} \r &= -2 \cot 2x + \frac{1}{2^n} \cot \l \frac{x}{2^n} \r - 0 \\ \end{align*} as required.
3. As \(n \to \infty\) \(\frac{x}{2^n} \to 0\), so \(\frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}} = \frac{2^n \sin \frac{x}{2^n}}{x} \to 1\) \begin{align*}\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) &= \lim_{n \to \infty} \frac{\sin x}{2^n \sin \l \frac{x}{2^n} \r} \\ &= \lim_{n \to \infty} \frac{\sin x}{x \frac{2^n \sin \l \frac{x}{2^n} \r}{x} } \\ &= \lim_{n \to \infty} \frac{\sin x}{x} \\ \end{align*} \begin{align*} \sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right) &= \sum_{j=0}^{\infty} \frac{1}{2^{j}} \tan\left(\frac{1}{2^j}\frac{\pi}{4}\right) \\ &= \lim_{n \to \infty} \l -2 \cot \frac{\pi}{2} + \frac{1}{2^n} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\ &= \frac{4}{\pi} \lim_{n \to \infty} \l \frac{1}{2^n} \frac{\pi}{4} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\ &\to \frac{\pi}{4} \end{align*}

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Solution: When \((\ln x)^2 = 1\) we have \(f(x) = \frac{1}{x\ln x}(1 - 1^2)^2 = 0\) \(f'(x) = \frac{2(1 - (\ln x)^2) \cdot (-2 \ln x ) \cdot \frac1x \cdot (x \ln x) - (\ln x +1)(1-(\ln x)^2)^2}{(x\ln x)^2} = \frac{2\cdot 0 \cdot (-2 \ln x ) \cdot \frac1x \cdot (x \ln x) - (\ln x +1) \cdot 0}{(x\ln x)^2} = 0\)

1. First consider \(0 < x < 1\), so \begin{align*} && F(x) &= \int_{1/e}^x f(t) \d t \\ &&&= \int_{1/e}^x \frac{1}{t\ln t} \left(1 - (\ln t)^2 \right)^2 \d t \\ u = \ln t, \d u = \frac1t \d t: &&&= \int_{u=-1}^{u=\ln x} \frac{1}{u}(1-u^2)^2 \d u \\ &&&= \int_{-1}^{\ln x} \left ( u^3 - 2u+\frac1u \right) \d u \\ &&&= \left [ \frac{u^4}{4} - u^2+ \ln |u| \right]_{-1}^{\ln x} \\ &&&= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln |\ln x| - \frac14+1 \end{align*} Now consider \(x > 1\) \begin{align*} && F(x) &= \int_{e}^x f(t) \d t \\ &&&= \int_{e}^x \frac{1}{t\ln t} \left(1 - (\ln t)^2 \right)^2 \d t \\ u = \ln t, \d u = \frac1t \d t: &&&= \int_{u=1}^{u=\ln x} \frac{1}{u}(1-u^2)^2 \d u \\ &&&= \int_{1}^{\ln x} \left ( u^3 - 2u+\frac1u \right) \d u \\ &&&= \left [ \frac{u^4}{4} - u^2+ \ln |u| \right]_{1}^{\ln x} \\ &&&= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln| \ln x| - \frac14+1 \end{align*} Notice that \begin{align*} F(x^{-1}) &= \frac{(\ln x^{-1})^4}{4} -(\ln x^{-1})^2 + \ln| \ln x^{-1}| - \frac14+1 \\ &= \frac{(-\ln x)^4}{4} -(-\ln x)^2 + \ln| -\ln x| - \frac14+1 \\ &= \frac{(\ln x)^4}{4} -(\ln x)^2 + \ln| \ln x| - \frac14+1 \\ &= F(x) \end{align*}
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Solution:

1. \(\,\) \begin{align*} && f(x) &= \frac1{1+\tan x} \\ && f'(x) &=-(1+\tan x)^{-2} \cdot \sec^2 x \\ &&&= - (\cos x+ \sin x)^{-2} \\ &&&= - (1 + 2 \sin x \cos x)^{-1} \\ &&&= - \frac{1}{1+\sin 2x} \end{align*} \(\sin 2x \in [0, 1]\) so \(1+\sin 2x \in [1,2]\) and \(f'(x) \in [-1, -\tfrac12]\)
   

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2. \(\displaystyle \int_{-1}^1 g(x) \d x = 2g(0) \)
3. Let \(g(x) = \frac{1}{1 + \tan^k x}\) then \(g(x)\) has rotational symmetry of order \(2\) about the point \((\frac{\pi}{4}, \frac12)\) which we can see since \begin{align*} g(x) + g(\tfrac12\pi - x) &= \frac{1}{1 + \tan^k x} + \frac{1}{1 + \tan^k(\tfrac12\pi - x)} \\ &= \frac{1}{1+\tan^k x} + \frac{1}{1+\cot^k x} \\ &= \frac{1}{1+\tan^k x} + \frac{\tan^k x}{\tan^k x + 1} \\ &= 1 = 2 \cdot \tfrac12 \end{align*} Therefore \[ \int_{\frac16\pi}^{\frac13\pi} \frac 1 {1+\tan^kx} \, \d x = \frac{\pi}{6} \cdot \frac12 = \frac{\pi}{12}\]

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Solution:

1. \begin{align*} \frac{\d y_n}{\d x} &= \frac{\d}{\d x} \l (-1)^n e^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r \r \\ &= (-1)^n 2xe^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r + (-1)^n e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - (-1)^{n+1} e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - y_{n+1} \end{align*}
2. \(y_0 = 1\), \(y_1 = (-1) e^{x^2} \cdot (-2x) \cdot e^{-x^2} = 2x\), \(y_2 = e^{x^2} \frac{\d^2}{\d x^2} \l e^{-x^2}\r = e^{x^2} \frac{\d }{\d x}\l -2xe^{-x^2} \r = e^{x^2} \l -2e^{-x^2}+4x^2e^{-x^2}\r = 4x^2-2\). Therefore \(2xy_1 - 2y_0 = 2x \cdot 2x - 2\cdot1 = 4x^2-2 = y_2\) so our statement is true for \(n=1\). Assume the statement is true for \(n=k\), then \begin{align*} && y_{k+1} &= 2xy_k - 2ky_{k-1} \\ \frac{\d }{\d x}: && \frac{\d y_{k+1}}{\d x} &= 2\frac{\d}{\d x}\l xy_k \r - 2k\frac{\d y_{k-1}}{\d x} \\ \Rightarrow && 2xy_{k+1}-y_{k+2} &= 2y_k+2x \l 2xy_k-y_{k+1}\r - 2k \l 2xy_{k-1}-y_k \r \\ \Rightarrow && y_{k+2} &=2y_k+ 4x \cdot y_{k+1}-(4x^2+2k)y_k+2x \cdot 2k y_{k-1} \\ &&&= 4x \cdot y_{k+1}-(4x^2+2(k+1))y_k+2x \l2xy_k - y_{k+1} \r \\ &&&= 2x \cdot y_{k+1} -2(k+1)y_k \end{align*} Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k=1\), therefore by the principle of mathematical induction it is true for all \(n \geq 1\). Since \(2x = \frac{y_{n+1}+2ny_{n-1}}{y_n}\) for all \(n\), we must have \begin{align*} && \frac{y_{n+1}+2ny_{n-1}}{y_n} &= \frac{y_{n+2}+2(n+1)y_{n}}{y_{n+1}} \\ \Leftrightarrow && y_{n+1}^2+2ny_{n-1}y_{n+1} &= y_ny_{n+2}+2ny_n^2+2y_n^2 \\ \Leftrightarrow && y_{n+1}^2-y_ny_{n+2} &= 2n(y_n^2-y_{n-1}y_{n+1})+2y_n^2 \end{align*}
3. Consider the functions \(f_n(x) = y_{n}^2-y_{n-1}y_{n+1}\) then clearly \(f_{n+1} = 2nf_{n} + 2y_n^2 \geq f_{n}\) so to prove \(f_n(x) > 0\) for \(n \geq 1\) it suffices to prove it for \(n = 1\). But \(f_1 = y_1^2 - y_0y_{2} = (2x)^2-(4x^2-2) = 2 > 0\) so we are done.

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Solution:

Clearly the distance \(PS\) is \(s - x\), so it remains to determine the heigh \(PQ\). Notice that \(\tan \beta = \frac{PQ}{OP}\) so the height is \(x \tan \beta\) and the area is \(x(s-x)\tan \beta \) Notice that \(R\) has a \(y\)-coordinate of \(x \tan \beta\), but is a distance \(a\) from the origin, so \(s^2 + x^2 \tan^2 \beta = a^2 \Rightarrow s = \sqrt{a^2-x^2 \tan^2 \beta}\) \begin{align*} && \frac{\d A}{\d x} &= (s-x)\tan \beta + x \left (\frac{\d s}{\d x} - 1 \right) \tan \beta \\ &&&= (s-x) \tan \beta + \left (\tfrac12(a^2-x^2\tan^2 \beta)^{-1/2} \cdot (-2x \tan^2 \beta) - 1\right) x \tan \beta \\ &&&= (s-x) \tan \beta + \left ( \frac{-x \tan^2 \beta}{s} -1\right)x \tan \beta \\ &&&= (s-2x) \tan \beta - \frac{x^2}{s}\tan^3\beta \\ \\ \frac{\d A}{\d x} = 0: && 0 &= s(s-2x)-x^2 \tan^2 \beta \\ &&&= s^2-(2x)s-x^2\tan^2 \beta \\ &&&= (s-x)^2-(1+\tan^2\beta)x^2 \\ \Rightarrow && s &= x + x \sec \beta \\ &&&= (1+\sec \beta)x \\ \\ && a^2 &= x^2(1+\sec\beta)^2 + x^2 \tan^2 \beta \\ &&&= x^2(2\sec \beta +2\sec^2 \beta ) \\ &&&= 2x^2 \sec \beta(1+\sec \beta) \\ \\ && A &= x^2\sec \beta \tan \beta \\ &&&= \frac12 a^2 \frac{\sec \beta \tan \beta}{\sec \beta(1+\sec \beta)} \\ &&&= \frac12 a^2 \frac{\tan \beta}{1+\sec \beta} = \frac12 a^2 \tan \frac{\beta}{2}\\ \end{align*} This occurs when \begin{align*} && \frac{RS}{SO} &= \frac{x \tan \beta}{s} \\ &&&= \frac{\tan \beta}{1+\sec \beta} = \tan \frac{\beta}2 \\ \Rightarrow&& \angle ROS &= \frac{\beta}2 \end{align*}