Let f() = ^2 + ^2 and g() = ^2 + ^2, where 0.

Question

Let $f(\theta) = \alpha\tan^2\theta + \beta\cot^2\theta$ and $g(\theta) = \alpha\sin^2\theta + \beta\cos^2\theta$, where $\alpha > \beta > 0$. If $$\min_{0<\theta<\pi/2} f(\theta) = \max_{0<\theta<\pi} g(\theta),$$ and the first term of a G.P. is $\dfrac{\alpha}{2\beta}$, its common ratio is $\dfrac{2\beta}{\alpha}$, and the sum of its first $10$ terms is $\dfrac{m}{n}$ (with $\gcd(m,n)=1$), then $m+n$ is equal to \_\_\_\_\_\_\_\_. Answer: 1279

Accepted Answer

Minimum of $f$: With $x=\tan^2\theta>0$, $f = \alpha x + \dfrac{\beta}{x} \ge 2\sqrt{\alpha\beta}$ by AM--GM, so $\min f = 2\sqrt{\alpha\beta}$.

Maximum of $g$: $g(\theta) = \beta + (\alpha-\beta)\sin^2\theta$. Since $\alpha>\beta$, $g$ increases with $\sin^2\theta$, so as $\theta \to \pi/2$, $\max g(\theta) = \alpha$.

Condition: $2\sqrt{\alpha\beta} = \alpha \Rightarrow 4\alpha\beta = \alpha^2 \Rightarrow \alpha = 4\beta$ (since $\alpha>0$).

G.P.: first term $a = \dfrac{\alpha}{2\beta} = \dfrac{4\beta}{2\beta} = 2$; common ratio $r = \dfrac{2\beta}{\alpha} = \dfrac{2\beta}{4\beta} = \dfrac12$.

Sum of first 10 terms:
$$S_{10} = \frac{a(1-r^{10})}{1-r} = \frac{2\left(1-\tfrac{1}{1024}\right)}{1/2} = 4\cdot\frac{1023}{1024} = \frac{1023}{256}.$$

Since $1023 = 3\cdot11\cdot31$ and $256=2^8$, $\gcd(1023,256)=1$, so $m=1023$, $n=256$.

$m+n = 1023+256 = 1279.$

Solutions · 1