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A point P lies inside a square ABCD

### Solution

To start, let’s add an angle \(\alpha\) like it is shown in the next figure

Applying the

**law of cosines**in the triangle \(\triangle BPA\), we getApplying the

**law of cosines**in the triangle \(\triangle BPC\), we getWe know that \(\cos^{2}\alpha+\sin^{2}\alpha=1\), thus

Let,

We can eliminate \(x =-\sqrt{17-4 \sqrt{14}}\) and \(x =-\sqrt{4 \sqrt{14}+17}\) because \(x>0\)

So the only possible solution is \[\large x=\sqrt{4 \sqrt{14}+17}
\]

Home -> Solved problems -> A point P lies inside a square ABCD

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