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

The square A B C D has side equal to 1

and the distance A P is \frac{1}{8} .

Calculate the side of the equilateral

triangle P M N inscribed in the square .

Commented by Maclaurin Stickker last updated on 15/Sep/19

Commented by ajfour last updated on 16/Sep/19

l e t A P = \frac{1}{8} = a

t r i a n g l e s i d e s, s i d e o f s q u a r e 1 .

s \cos α = 1

D P = s \cos (\frac{π}{6} - α) = 1 - a

\Rightarrow \frac{\sqrt{3}}{2} + (\frac{s}{2}) \sin α = 1 - a

s \sqrt{1 - \frac{1}{s^{2}}} = \frac{7}{4} - \sqrt{3}

s^{2} - 1 = {(\frac{7}{4} - \sqrt{3})}^{2}

s^{2} = \frac{49}{16} + 3 - \frac{7 \sqrt{3}}{2} + 1

s = \frac{\sqrt{113 - 56 \sqrt{3}}}{4} .

Commented by Maclaurin Stickker last updated on 16/Sep/19

W o w . I l o v e d h o w y o u u s e d t r i g o n o m e t r y .

Answered by MJS last updated on 16/Sep/19

P = (\begin{matrix} 0 \\ \frac{1}{8} \end{matrix}) M = (\begin{matrix} 1 \\ y \end{matrix}) N = (\begin{matrix} x \\ 1 \end{matrix})

∣ P M ∣^{2} =∣ P N ∣^{2} =∣ M N ∣^{2}

y^{2} - \frac{1}{4} y + \frac{65}{64} = x^{2} + \frac{49}{64} = x^{2} + y^{2} - 2 x - 2 y + 2

y^{2} - \frac{1}{4} y + \frac{65}{64} = x^{2} + y^{2} - 2 x - 2 y + 2

\Rightarrow y = \frac{4}{7} x^{2} - \frac{8}{7} x + \frac{9}{16}

y^{2} - \frac{1}{4} y + \frac{65}{64} = x^{2} + \frac{49}{64}

\Rightarrow

x^{4} - 4 x^{3} + \frac{79}{32} x^{2} - \frac{49}{16} x + \frac{5341}{4096} = 0

(x^{2} + \frac{49}{64}) (x^{2} - 4 x + \frac{109}{64}) = 0

x \in [0; 1] \Rightarrow x = 2 - \frac{7 \sqrt{3}}{8} \Rightarrow y = \frac{15}{8} - \sqrt{3}

\Rightarrow side s = \frac{\sqrt{113 - 56 \sqrt{3}}}{4}

Commented by Maclaurin Stickker last updated on 16/Sep/19

Y o u r a n s w e r i s c o r r e c t .

Commented by Maclaurin Stickker last updated on 16/Sep/19

H o w d i d y o u d o t h a t t h i r d e q u a t i o n ?

Commented by Maclaurin Stickker last updated on 16/Sep/19

T h a n k y o u!

Commented by MJS last updated on 16/Sep/19

x^{4} - 4 x^{3} + \frac{79}{32} x^{2} - \frac{49}{16} x + \frac{5341}{4096} = 0

x = t + 1

t^{4} - \frac{113}{32} t^{2} - \frac{49}{8} t - \frac{9379}{4096} = 0

(t^{2} - α t - β) (t^{2} + α t - γ) = 0

\Rightarrow

α^{2} + β + γ - \frac{113}{32} = 0 \land α β - α γ - \frac{49}{8} = 0 \land β γ + \frac{9379}{4096} = 0

\Rightarrow

α = 2 \land β = \frac{83}{64} \land γ = - \frac{113}{64}

(t^{2} - 2 t - \frac{83}{64}) (t^{2} + 2 t + \frac{113}{64}) = 0

t = x - 1

(x^{2} - 4 x + \frac{109}{64}) (x^{2} + \frac{49}{64}) = 0

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