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 #4
avatar+26398 
+2

Which is larger, the blue area or the orange area?

\(\begin{array}{|rcll|} \hline {\color{orange}\text{orange}} +3\times {\color{blue}\text{blue}} &=& \pi r_{\text{circumcircle}}^2 \quad | \quad : {\color{orange}\text{orange}} \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { {\color{orange}\text{orange}} } \quad | \quad {\color{orange}\text{orange}} = \pi r_{\text{incircle}}^2 \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{ \pi r_{\text{circumcircle}}^2 } { \pi r_{\text{incircle}}^2 } \\ 1+3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 \\ 3\times \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ r_{\text{circumcircle}} } { r_{\text{incircle}} } \right)^2 -1 \right) \\ \\ && \boxed{\text{here, if triangle is equilateral: }\\ \mathbf{2\times r_{\text{incircle}} = r_{\text{circumcircle}}!!!} } \\\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\left( \left( \dfrac{ 2\times r_{\text{incircle}} } { r_{\text{incircle}} } \right)^2 -1 \right)\\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(2^2 -1) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& \dfrac{1}{3}\times(3) \\ \dfrac{ {\color{blue}\text{blue}} } { {\color{orange}\text{orange}} } &=& 1 \\ \mathbf{ {\color{blue}\text{blue}} } &=& \mathbf{ {\color{orange}\text{orange}} } \\ \hline \end{array}\)

 

laugh

15 нояб. 2019 г.
 #3
avatar+26398 
+2

What are the two smallest prime factors of \(2^{1024} - 1\)?

 

\(\begin{array}{|rcll|} \hline && 2^{1024} - 1 \\ &=& \left( 2^{512} - 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{256} - 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{128} - 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{64} - 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{32} - 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{16} - 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{8} - 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{4} - 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{2} - 1\right) \left( 2^{2} + 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& \left( 2^{1} - 1\right) \left( 2^{1} + 1\right) \left( 2^{2} + 1\right) \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ &=& 1\times 3 \times 5 \times \left( 2^{4} + 1\right) \left( 2^{8} + 1\right) \left( 2^{16} + 1\right) \left( 2^{32} + 1\right) \left( 2^{64} + 1\right) \left( 2^{128} + 1\right) \left( 2^{256} + 1\right) \left( 2^{512} + 1\right) \\ \hline \end{array} \)

 

The two smallest prime factors are 3 and 5, they are also the smallest possible odd prime numbers,

all factors are odd,

so that the 2 as the only even prime number cannot be included

 

laugh

15 нояб. 2019 г.
 #1
avatar+26398 
+2
13 нояб. 2019 г.
 #14
avatar+26398 
+1
13 нояб. 2019 г.
 #12
avatar+26398 
+3
12 нояб. 2019 г.
 #10
avatar+26398 
+3

Find \(\tan^2(20) + \tan^2(40) + \tan^2(80) \) .  All angles are in degrees.

 

Formula:
\(\begin{array}{rcll} \boxed{ \sum \limits_{l=1}^{n} \tan^2\left( \dfrac{180^\circ \cdot l}{2n+1} \right) = n(2n+1) } \\ \end{array}\)

Source (7): https://math.stackexchange.com/questions/173447/proving-sum-limits-l-1n-sum-limits-k-1n-1-tan-frac-lk-pi-2n1-t/173858#173858

 

\(\begin{array}{|lrcll|} \hline n=1: & \sum \limits_{l=1}^{1} \tan^2\left( \dfrac{180^\circ \cdot l}{2\cdot 1+1} \right) &=& 1\cdot(2\cdot 1 +1) \\ & \sum \limits_{l=1}^{1} \tan^2\left( \dfrac{180^\circ \cdot l}{3} \right) &=& 1\cdot(3) \\ & \mathbf{ \tan^2\left(60^\circ\right)} &=& \mathbf{ 3 } \\ \hline n=4: & \sum \limits_{l=1}^{4} \tan^2\left( \dfrac{180^\circ \cdot l}{2\cdot 4+1} \right) &=& 4\cdot(2\cdot 4 +1) \\ & \sum \limits_{l=1}^{4} \tan^2\left( \dfrac{180^\circ \cdot l}{9} \right) &=& 4\cdot(9) \\ & \mathbf{\tan^2\left(20^\circ\right)+\tan^2\left(40^\circ\right)+\tan^2\left(60^\circ\right)+\tan^2\left(80^\circ\right) } &=& \mathbf{ 36 } \\ \hline & \tan^2\left(20^\circ\right)+\tan^2\left(40^\circ\right)+\tan^2\left(80^\circ\right) &=& 36 - 3 \\ & \mathbf{\tan^2\left(20^\circ\right)+\tan^2\left(40^\circ\right)+\tan^2\left(80^\circ\right)} &=& \mathbf{33} \\ \hline \end{array}\)

 

laugh

12 нояб. 2019 г.