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Question Number 72294 by mr W last updated on 27/Oct/19

Commented by mr W last updated on 27/Oct/19

$$thedistancesfromapointtothree$$$$verticesofasquareareknown.$$$$findthesidelengthofthesquare.$$

Commented by Prithwish sen last updated on 27/Oct/19

$${\mathbf{s}}_3^2+{\mathbf{s}}_2^2={\mathbf{a}}^2.....\left(\mathbf{i}\right)$$$${\mathbf{s}}_3^2+{\mathbf{s}}_1^2={\mathbf{b}}^2.....\left(\mathbf{ii}\right)$$$${\mathbf{s}}_1^2+{\mathbf{s}}_4^2={\mathbf{c}}^2......\left(\mathbf{iii}\right)$$$$\mathbf{and}{\mathbf{s}}_1+{\mathbf{s}}_2={\mathbf{s}}_3+{\mathbf{s}}_4=\mathbf{s}.....\left(\mathbf{iv}\right)$$$$\mathbf{solving}\mathbf{all}\mathbf{these}$$$${\mathbf{s}}_2=\frac{{\mathbf{s}}^2-{\mathbf{b}}^2+{\mathbf{a}}^2}{2\mathbf{s}}....\left(\mathbf{v}\right)$$$${\mathbf{s}}_3=\frac{{\mathbf{s}}^2+{\mathbf{b}}^2-{\mathbf{c}}^2}{2\mathbf{s}}$$$$\mathbf{putting}\mathbf{these}\mathbf{on}\left(\mathbf{i}\right)$$$${\left[\frac{{\mathbf{s}}^2-{\mathbf{b}}^2+{\mathbf{a}}^2}{2\mathbf{s}}\right]}^2+{\left[\frac{{\mathbf{s}}^2+{\mathbf{b}}^2-{\mathbf{c}}^2}{2\mathbf{s}}\right]}^2={\mathbf{a}}^2$$$$2{\mathbf{s}}^4-2{\mathbf{s}}^2({\mathbf{a}}^2+{\mathbf{c}}^2)+[{({\mathbf{a}}^2-{\mathbf{b}}^2)}^2+{({\mathbf{c}}^2-{\mathbf{b}}^2)}^2]=0$$$$\therefore {\mathbf{s}}^2=\frac{{\mathbf{a}}^2+{\mathbf{c}}^2}{2}\pm \sqrt{{\mathbf{b}}^2({\mathbf{a}}^2+{\mathbf{c}}^2-{\mathbf{b}}^2)-{\left[\frac{{\mathbf{a}}^2-{\mathbf{c}}^2}{2}\right]}^2}$$

Commented by Prithwish sen last updated on 27/Oct/19

Commented by mr W last updated on 27/Oct/19

$$thankssir!$$

Answered by mr W last updated on 27/Oct/19

$$a^2=s^2+b^2-2bs\cos \alpha$$$$\Rightarrow 2bs\cos \alpha =s^2+b^2-a^2...\left(i\right)$$$$c^2=s^2+b^2-2bs\cos (\frac{\pi }{2}-\alpha )$$$$\Rightarrow 2bs\sin \alpha =s^2+b^2-c^2...\left(ii\right)$$$${\left(i\right)}^2+{\left(ii\right)}^2:$$$$4b^2{s}^2={(s^2+b^2-a^2)}^2+{(s^2+b^2-c^2)}^2$$$$\Rightarrow s^4-(a^2+c^2)s^2-b^2(a^2+c^2-b^2)+\frac{a^4+c^4}{2}=0$$$$\Rightarrow s^2=\frac{1}{2}\{a^2+c^2\pm \sqrt{a^4+2a^2{c}^2+c^4+4b^2(a^2+c^2-b^2)-2a^4-2c^4}\}$$$$\Rightarrow s^2=\frac{a^2+c^2}{2}\pm \sqrt{b^2(a^2+c^2-b^2)-{\left(\frac{a^2-c^2}{2}\right)}^2}$$$$b^2(a^2+c^2-b^2)-{\left(\frac{a^2-c^2}{2}\right)}^2⩾0$$$$b^4-(a^2+c^2)b^2+{\left(\frac{a^2-c^2}{2}\right)}^2⩽0$$$$\Rightarrow b^2=\frac{1}{2}(a^2+c^2\pm 2ac)=\frac{{(a\pm c)}^2}{2}$$$$\Rightarrow \frac{\mid a-c\mid }{\sqrt{2}}⩽b⩽\frac{a+c}{\sqrt{2}}$$$$example:$$$$a=1,b=2,b=3$$$$s^2=\frac{1+9}{2}\pm \sqrt{4(1+9-4)-{\left(\frac{1-9}{2}\right)}^2}$$$$s^2=5\pm 2\sqrt{2}$$

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