document.write( "Question 1136228: There is a regular decagon.Triangle are formed by joining the vertices of the polygon.What is the number of triangle that have no side common with any of the sides of the polygon? \n" ); document.write( "

Algebra.Com's Answer #754024 by greenestamps(13208) $\"\"$ $\"About$

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\n" ); document.write( "To help with the discussion, consider the consecutive vertices labeled 0 to 9.

\n" ); document.write( "(1) Pick one of the vertices as the first vertex of the triangle (10 choices).

\n" ); document.write( "(2a) The second vertex can be 2 away from the first in either direction (2 choices). If so, then there are 5 choices remaining for the third vertex. Total number of triangles for this case: 10*2*5 = 100.

\n" ); document.write( "(2b) The second vertex can be 3 away from the first in either direction (2 choices). If so, then there are 4 choices remaining for the third vertex. Total number of triangles for this case: 10*2*4 = 80.

\n" ); document.write( "(2c) The second vertex can be 4 away from the first in either direction (2 choices). If so, then there are 3 choices remaining for the third vertex. Total number of triangles for this case: 10*2*3 = 60.

\n" ); document.write( "(2d) The second vertex can be 5 away from the first in either direction (2 choices). If so, then there are 2 choices remaining for the third vertex. Total number of triangles for this case: 10*2*2 = 40.

\n" ); document.write( "(2e) The second vertex can be 6 away from the first in either direction (2 choices). If so, then there is 1 choice remaining for the third vertex. Total number of triangles for this case: 10*2*1 = 20.

\n" ); document.write( "Counting the triangles that way, there are a total of 100+80+60+40+20 = 300. But in counting them that way, each distinct triangle is counted 3*2*1=6 ways. So the number of distinct triangles is 300/6 = 50.

\n" ); document.write( "We can also arrive at this answer by listing all the triangles. It isn't as hard as it might seem, because there are nice patterns involved.

\n" ); document.write( "We can make an organized list of the triangles, using our 0 to 9 numbering of the vertices of the decagon. To ensure that we count each triangle only once, each entry in our list will be in strictly increasing order, with the restriction that no two adjacent vertices can be used.

\n" ); document.write( "The list....

\r\n" );
document.write( "024, 025, 026, 027, 028;\r\n" );
document.write( "035, 036, 037, 038;\r\n" );
document.write( "046, 047, 048;\r\n" );
document.write( "057, 058;\r\n" );
document.write( "068\r\n" );
document.write( "15 triangles there....\r\n" );
document.write( "135, 136, 137, 138, 139;\r\n" );
document.write( "146, 147, 148, 149;\r\n" );
document.write( "157, 158, 159;\r\n" );
document.write( "168, 169;\r\n" );
document.write( "179\r\n" );
document.write( "15 more there....\r\n" );
document.write( "246, 247, 248, 249;\r\n" );
document.write( "257, 258, 259;\r\n" );
document.write( "268, 269;\r\n" );
document.write( "279\r\n" );
document.write( "10 there....\r\n" );
document.write( "357, 358, 359;\r\n" );
document.write( "368, 369;\r\n" );
document.write( "379\r\n" );
document.write( "6 more there....\r\n" );
document.write( "468, 469;\r\n" );
document.write( "479\r\n" );
document.write( "3 more....\r\n" );
document.write( "579\r\n" );
document.write( "1 last one\r\n" );
document.write( "\r\n" );
document.write( "Total in our list: 15+15+10+6+3+1 = 50

\n" ); document.write( "ANSWER: There are 50 triangles that can be formed using the vertices of a decagon in which none of the sides of the decagon are used.

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\n" ); document.write( "I hope the reader enjoyed seeing the second solution from tutor @ikleyn as much as I did. The two methods are very different yet both completely logical.

\n" ); document.write( "I always enjoy learning new ways to solve problems that are simpler and easier to understand; for this problem I think hers is nicer. \n" ); document.write( "

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