#28 of June 2015 Grade 10 CSAT(Korean SAT) Mock test

Geometry Level 4

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.

2 solutions

Boi (보이)
Jun 8, 2017

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,

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

I solved it using the assumption that the triangle is equilateral (and so did you) but you never mentioned in the problem statement that the triangle was equilateral.

H K - 4 years ago

Thanks.

I'm so first to Brilliant so I'm currently undergoing the process of incessantly making mistakes.

But again, thank you. I would've never noticed if you were not to tell me that.

Boi (보이) - 4 years ago

Niranjan Khanderia
Aug 11, 2018

#28 of June 2015 Grade 10 CSAT(Korean SAT) Mock test

The answer is 30.

2 solutions

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