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

## 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 get
\begin{aligned} A P^{2} &=AB^{2}+BP^{2}-2AB\cdot BP \cos\alpha \\\\ 3^{2} &=x^{2}+4^{2}-8 x \cos \alpha \\\\ 9 &=x^{2}+16-8 x \cos \alpha \\\\ \cos \alpha &=\frac{x^{2}+7}{8 x} \end{aligned}
Applying the law of cosines in the triangle $$\triangle BPC$$, we get
\begin{aligned} PC^{2} &=BP^{2}+BC^{2}-2BP\cdot BC \cos \left(\frac{\pi}{2}-\alpha\right) \\\\ 5^{2} &=4^{2}+x^{2}-8 x \cos \left(\frac{\pi}{2}-\alpha\right) \\\\ & \cos \left(\frac{\pi}{2}-\alpha\right)=\sin \alpha \\\\ \Rightarrow 25 &=16+x^{2}-8 x \sin \alpha \\\\ \sin \alpha &=\frac{x^{2}-9}{8 x} \end{aligned}
We know that $$\cos^{2}\alpha+\sin^{2}\alpha=1$$, thus
$\Rightarrow\left(\frac{x^{2}+7}{8 x}\right)^{2}+\left(\frac{x^{2}-9}{8 x}\right)^{2}=1$ \begin{aligned} \frac{x^{4}+14 x^{2}+49}{64 x^{2}}+\frac{x^{4}-18 x^{2}+81}{64 x^{2}} &=1 \\\\ 2 x^{4}-4 x^{2}+130 &=64 x^{2} \\\\ 2 x^{4}-68 x^{2}+130 &=0 \\\\ x^{4}-34 x^{2}+65 &=0 \end{aligned}
Let,
\begin{aligned} y=x^{2} \\\\ \Rightarrow y^{2}-34 y+65=0\\\\ y \in\{17-4 \sqrt{14}, \;4 \sqrt{14}+17\} \end{aligned}
$\Rightarrow x \in\{-\sqrt{17-4 \sqrt{14}}, \sqrt{17-4 \sqrt{14}},-\sqrt{4 \sqrt{14}+17}, \sqrt{4 \sqrt{14}+17}\}$
We can eliminate $$x =-\sqrt{17-4 \sqrt{14}}$$ and $$x =-\sqrt{4 \sqrt{14}+17}$$ because $$x>0$$
We can also eliminate $$x =\sqrt{17-4 \sqrt{14}}$$ because $$x$$ must be larger than $$5$$ so $$P$$ is contained inside the square.
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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