# The-square-ABCD-has-side-equal-to-1-and-the-distance-AP-is-1-8-Calculate-the-side-of-the-equilateral-triangle-PMN-inscribed-in-the-square-

Question Number 68831 by Maclaurin Stickker last updated on 15/Sep/19
$$\mathrm{The}\:\mathrm{square}\:{ABCD}\:\mathrm{has}\:\mathrm{side}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1} \\$$$$\mathrm{and}\:\mathrm{the}\:\mathrm{distance}\:{AP}\:\:\mathrm{is}\:\:\frac{\mathrm{1}}{\mathrm{8}}. \\$$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral} \\$$$$\mathrm{triangle}\:{PMN}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{square}. \\$$
Commented by Maclaurin Stickker last updated on 15/Sep/19
Commented by ajfour last updated on 16/Sep/19
$${let}\:{AP}\:=\frac{\mathrm{1}}{\mathrm{8}}={a} \\$$$${triangle}\:{side}\:{s},\:{side}\:{of}\:{square}\:\mathrm{1}. \\$$$${s}\mathrm{cos}\:\alpha=\mathrm{1} \\$$$${DP}\:=\:{s}\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\mathrm{1}−{a} \\$$$$\Rightarrow\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\left(\frac{{s}}{\mathrm{2}}\right)\mathrm{sin}\:\alpha=\mathrm{1}−{a} \\$$$${s}\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{s}^{\mathrm{2}} }}=\frac{\mathrm{7}}{\mathrm{4}}−\sqrt{\mathrm{3}} \\$$$${s}^{\mathrm{2}} −\mathrm{1}=\left(\frac{\mathrm{7}}{\mathrm{4}}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\$$$$\:\:{s}^{\mathrm{2}} =\frac{\mathrm{49}}{\mathrm{16}}+\mathrm{3}−\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{2}}+\mathrm{1} \\$$$$\:{s}=\frac{\sqrt{\mathrm{113}−\mathrm{56}\sqrt{\mathrm{3}}}}{\mathrm{4}}\:. \\$$
Commented by Maclaurin Stickker last updated on 16/Sep/19
$${Wow}.\:{I}\:{loved}\:{how}\:{you}\:{used}\:{trigonometry}. \\$$
Answered by MJS last updated on 16/Sep/19
$${P}=\begin{pmatrix}{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{8}}}\end{pmatrix}\:\:{M}=\begin{pmatrix}{\mathrm{1}}\\{{y}}\end{pmatrix}\:\:{N}=\begin{pmatrix}{{x}}\\{\mathrm{1}}\end{pmatrix} \\$$$$\mid{PM}\mid^{\mathrm{2}} =\mid{PN}\mid^{\mathrm{2}} =\mid{MN}\mid^{\mathrm{2}} \\$$$${y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{y}+\frac{\mathrm{65}}{\mathrm{64}}={x}^{\mathrm{2}} +\frac{\mathrm{49}}{\mathrm{64}}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}{y}+\mathrm{2} \\$$$$\\$$$${y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{y}+\frac{\mathrm{65}}{\mathrm{64}}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{2}{y}+\mathrm{2} \\$$$$\Rightarrow\:{y}=\frac{\mathrm{4}}{\mathrm{7}}{x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{7}}{x}+\frac{\mathrm{9}}{\mathrm{16}} \\$$$$\\$$$${y}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}{y}+\frac{\mathrm{65}}{\mathrm{64}}={x}^{\mathrm{2}} +\frac{\mathrm{49}}{\mathrm{64}} \\$$$$\Rightarrow \\$$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\frac{\mathrm{79}}{\mathrm{32}}{x}^{\mathrm{2}} −\frac{\mathrm{49}}{\mathrm{16}}{x}+\frac{\mathrm{5341}}{\mathrm{4096}}=\mathrm{0} \\$$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{49}}{\mathrm{64}}\right)\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\frac{\mathrm{109}}{\mathrm{64}}\right)=\mathrm{0} \\$$$${x}\in\left[\mathrm{0};\:\mathrm{1}\right]\:\Rightarrow\:{x}=\mathrm{2}−\frac{\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\Rightarrow\:{y}=\frac{\mathrm{15}}{\mathrm{8}}−\sqrt{\mathrm{3}} \\$$$$\Rightarrow\:\mathrm{side}\:{s}=\frac{\sqrt{\mathrm{113}−\mathrm{56}\sqrt{\mathrm{3}}}}{\mathrm{4}} \\$$
Commented by Maclaurin Stickker last updated on 16/Sep/19
$${Your}\:{answer}\:{is}\:{correct}. \\$$
Commented by Maclaurin Stickker last updated on 16/Sep/19
$${How}\:{did}\:{you}\:{do}\:{that}\:{third}\:{equation}? \\$$
Commented by Maclaurin Stickker last updated on 16/Sep/19
$${Thank}\:{you}!\: \\$$
Commented by MJS last updated on 16/Sep/19
$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\frac{\mathrm{79}}{\mathrm{32}}{x}^{\mathrm{2}} −\frac{\mathrm{49}}{\mathrm{16}}{x}+\frac{\mathrm{5341}}{\mathrm{4096}}=\mathrm{0} \\$$$${x}={t}+\mathrm{1} \\$$$${t}^{\mathrm{4}} −\frac{\mathrm{113}}{\mathrm{32}}{t}^{\mathrm{2}} −\frac{\mathrm{49}}{\mathrm{8}}{t}−\frac{\mathrm{9379}}{\mathrm{4096}}=\mathrm{0} \\$$$$\left({t}^{\mathrm{2}} −\alpha{t}−\beta\right)\left({t}^{\mathrm{2}} +\alpha{t}−\gamma\right)=\mathrm{0} \\$$$$\Rightarrow \\$$$$\alpha^{\mathrm{2}} +\beta+\gamma−\frac{\mathrm{113}}{\mathrm{32}}=\mathrm{0}\wedge\alpha\beta−\alpha\gamma−\frac{\mathrm{49}}{\mathrm{8}}=\mathrm{0}\wedge\beta\gamma+\frac{\mathrm{9379}}{\mathrm{4096}}=\mathrm{0} \\$$$$\Rightarrow \\$$$$\alpha=\mathrm{2}\wedge\beta=\frac{\mathrm{83}}{\mathrm{64}}\wedge\gamma=−\frac{\mathrm{113}}{\mathrm{64}} \\$$$$\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\frac{\mathrm{83}}{\mathrm{64}}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\frac{\mathrm{113}}{\mathrm{64}}\right)=\mathrm{0} \\$$$${t}={x}−\mathrm{1} \\$$$$\left({x}^{\mathrm{2}} −\mathrm{4}{x}+\frac{\mathrm{109}}{\mathrm{64}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{49}}{\mathrm{64}}\right)=\mathrm{0} \\$$