Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 2"

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When a diagonal with angle <math>22.5^\circ k</math> intersects with another diagonal with different angle <math>22.5^\circ m</math> where <math>m>k</math>, then the angle of the intersection is <math>22.5^\circ (m-k)</math>  Since all of the unique values of (m-k) will be <math>1,2,3,\cdots 7</math>, then let <math>n1,2,3,\cdots 7</math> and the sum of all distinct angle measures will be given by:
 
When a diagonal with angle <math>22.5^\circ k</math> intersects with another diagonal with different angle <math>22.5^\circ m</math> where <math>m>k</math>, then the angle of the intersection is <math>22.5^\circ (m-k)</math>  Since all of the unique values of (m-k) will be <math>1,2,3,\cdots 7</math>, then let <math>n1,2,3,\cdots 7</math> and the sum of all distinct angle measures will be given by:
  
<math>S=\sum_{n=1}^{7}22.5^\circ n=22.5^\circ\sum_{n=1}^{7} n=22.5^\circ \frac{7 \times 8}{2}=22.5^\circ \times 28 = /boxed{630^\circ}</math>
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<math>S=\sum_{n=1}^{7}22.5^\circ n=22.5^\circ\sum_{n=1}^{7} n=22.5^\circ \frac{7 \times 8}{2}=22.5^\circ \times 28 = \boxed{630^\circ}</math>
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 18:17, 26 November 2023

Problem

Draw in the diagonals of a regular octagon. What is the sum of all distinct angle measures, in degrees, formed by the intersections of the diagonals in the interior of the octagon?

Solution

Each diagonal will have an angle with the horizontal base of the octagon as $22.5^\circ k$ where $k=0,1,\cdots 7$

When a diagonal with angle $22.5^\circ k$ intersects with another diagonal with different angle $22.5^\circ m$ where $m>k$, then the angle of the intersection is $22.5^\circ (m-k)$ Since all of the unique values of (m-k) will be $1,2,3,\cdots 7$, then let $n1,2,3,\cdots 7$ and the sum of all distinct angle measures will be given by:

$S=\sum_{n=1}^{7}22.5^\circ n=22.5^\circ\sum_{n=1}^{7} n=22.5^\circ \frac{7 \times 8}{2}=22.5^\circ \times 28 = \boxed{630^\circ}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.