As shown above, there is a regular triangle $ABC$ .

Let $M\text{, }N$ be midpoints of $\overline{AB}$ and $\overline{AC}$ .

$\overrightarrow{MN}$ and the circumcircle of $\triangle ABC$ intersect at point $P$ , which satisfies $\overline{NP}=1$ .

Define $x=\overline{MN}$ .

Find the value of $10\left(x^2+\dfrac{1}{x^2}\right)$ .

*
This problem is a part of
<Grade 10 CSAT Mock test> series
.
*

The answer is 30.

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Because of the property of a circle, $\overline{NA}\times\overline{NC}=\overline{NP}\times\overline{NQ}$ .

$x\times x=1\times(x+1)$

Simplify that and we get

$x^2-x-1=0 \\ x-\frac{1}{x}=1 \\ x^2+\frac{1}{x^2}=\left(x-\frac{1}{x}\right)^2+2=\text{ }3$

Therefore,

$10\left(x^2+\dfrac{1}{x^2}\right)=10\times3=\boxed{30}$