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# geometry problem

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Equilateral △ABC has a side length of 1 and the three inscribed squares have sides lengths a, 2a, and 3a as shown.  Compute a.

Sep 27, 2020

#1
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Equilateral △ABC has a side length of 1 and the three inscribed squares have sides lengths a, 2a, and 3a as shown.
Compute a.

Formula: $$\tan(60^\circ) = \sqrt{3}$$

$$\begin{array}{|rcll|} \hline \tan(60^\circ)&=& \dfrac{a}{x}\\ x &=& \dfrac{a}{\tan(60^\circ)} \quad | \quad \tan(60^\circ) = \sqrt{3} \\\\ \mathbf{x} &=& \mathbf{\dfrac{a}{\sqrt{3}}} \\\\ \hline \tan(60^\circ)&=& \dfrac{3a}{y}\\ y &=& \dfrac{3a}{\tan(60^\circ)} \quad | \quad \tan(60^\circ) = \sqrt{3} \\\\ \mathbf{y} &=& \mathbf{\dfrac{3a}{\sqrt{3}}} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline x+a+2a+3a+y &=& 1 \\\\ \dfrac{a}{\sqrt{3}} + 6a + \dfrac{3a}{\sqrt{3}}&=& 1 \\\\ a\left( \dfrac{1}{\sqrt{3}} + 6 + \dfrac{3}{\sqrt{3}} \right) &=& 1 \\\\ a\left( \dfrac{1}{\sqrt{3}} + 6\dfrac{\sqrt{3}}{\sqrt{3}} + \dfrac{3}{\sqrt{3}} \right) &=& 1 \\\\ \dfrac{a}{\sqrt{3}}\left( 4+ 6\sqrt{3} \right) &=& 1 \\\\ \mathbf{a} &=& \mathbf{ \dfrac{\sqrt{3}} {4+ 6\sqrt{3}} } \\\\ \mathbf{a} &=& \mathbf{0.12034561706} \\ \hline \end{array}$$

Sep 28, 2020