SOLUTION: If every vertex of a regular pentagon is connected to every other vertex, how many triangles are formed?

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Question 621234: If every vertex of a regular pentagon is connected to every other vertex, how many triangles are formed?

Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!
It's too hard to count them when all the diagonals are drawn in there.



So we'll just draw each diagonal only until it just touches the next
diagonal or corner and does not cross another diagonal. We draw diagonal 
BE:



That introduces 1 triangle ABE.  We draw in diagonal BD:



That adds two triangles, BDE and BCD so we have 3 triangles so far. 
We draw just enough of it until it touches the first diagonal at G:




That adds 2 more triangles, ABG and AGE, so we now have 5 triangles so far.
Now we'll extend that diagonal until it touches the next diagonal at H:



That adds 2 more triangles, ABH and BGH so there are now 7 triangles.  Now we  complete drawing the diagonal AC:



That adds 4 more triangles, ABC, GBC, HBC, and HCD, so now we have 11 triangles. Now we begin the diagonal AD, drawn until it touches the 
other diagonal at F:



That adds 3 more triangles ABF, AGF, and AFE.  So now we have 14 triangles. Next we complete diagonal AD:



That adds 5 triangles, ABD, AHD, AED, FRD, and FED, so now we have
19 triangle. Now we start the last diagonal EC drawing just to the point 
where it touches the diagonal at J:



That adds 3 triangles, AEJ, FEJ, and EJD.  Now we have 22 triangles.
Now we extend the partial diagonal to where it just touches BD at I:


 
That adds 3 triangles BEI, EIJ, and JID.  So now we have 25 triangles.
Finally we complete the last diagonal.







That adds 6 triangles, BEC, BIC, HIC, DIC, JCD, and ECD.  So we 
end up with a grand total of 31 triangles. I don't think I've
missed any, or counted any twice.

Edwin

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