每日一题[842]合理选参

已知边长为$1$的正三角形的中心为$O$，过$O$的直线与边$AB,AC$分别交于点$M,N$，求$\dfrac{1}{OM^2}+\dfrac{1}{ON^2}$的取值范围．

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正确答案是$[15,18]$．

分析与解　如图，注意到$\angle AMO+\angle ANO=\dfrac{2\pi}3$为定值，设$\angle AMO=\dfrac{\pi}3+x$，$\angle ANO=\dfrac{\pi}3-x$，其中$x$的取值范围是$\left[-\dfrac{\pi}6,\dfrac{\pi}6\right]$．

在$\triangle AMO$和$\triangle ANO$中分别应用正弦定理，可得

$$\begin{aligned} \dfrac{OA}{\sin\left(\dfrac{\pi}3+x\right)}=\dfrac{OM}{\sin\dfrac{\pi}6},\\\dfrac{OA}{\sin\left(\dfrac{\pi}3-x\right)}=\dfrac{ON}{\sin\dfrac{\pi}6},\end{aligned}$$ 这样就有 $$\begin{split}\dfrac{1}{OM^2}+\dfrac{1}{ON^2}&=\dfrac{\sin^2\left(\dfrac{\pi}3+x\right)+\sin^2\left(\dfrac{\pi}3-x\right)}{\sin^2\dfrac{\pi}6\cdot OA^2}\\&=12\left[\left(\dfrac{\sqrt 3}2\cos x+\dfrac 12\sin x\right)^2+\left(\dfrac{\sqrt 3}2\cos x-\dfrac 12\sin x\right)^2\right]\\&=12\left(\dfrac 32\cos^2x+\dfrac 12\sin^2x\right)\\&=12\left(\dfrac 12+\cos^2x\right), \end{split}$$
考虑到$\cos^2x$的取值范围是$\left[\dfrac 34,1\right]$，因此所求的取值范围是$[15,18]$．