August 9th, 2010, 04:24 AM  #1 
Newbie Joined: Jun 2010 Posts: 22 Thanks: 0  Hard prove !
Hello ! Subjected to a square which has two isosceles triangles. One of them is: a triangle whose base is one of the sides of the square with base angles of 15. The other is a triangle whose base is the side opposite the base of the triangle it first touches the apex of the triangle at the end of the first node. Proved that the second triangle equilateral! 
August 9th, 2010, 04:45 AM  #2 
Senior Member Joined: Apr 2010 Posts: 105 Thanks: 0  Re: Hard prove !
Can you picture it?

August 9th, 2010, 06:01 AM  #3 
Senior Member Joined: Apr 2010 Posts: 105 Thanks: 0  Re: Hard prove !
Try to find tan(15) with the doubleangle formulae. So, 1/?3=tan(30)=... What has this to do with your problem.? By the way, there are 4 isosceles triangles? 
August 9th, 2010, 03:21 PM  #4  
Global Moderator Joined: Dec 2006 Posts: 18,048 Thanks: 1395  Quote:
 
August 9th, 2010, 05:41 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,593 Thanks: 937 Math Focus: Elementary mathematics and beyond  Re: Hard prove !
Trigonometry: Let the sides of the square be of length b units. Construct a segment from P perpendicular to AB meeting AB at point Q and passing through P to DC meeting DC at S. Then PQ = b * tan(15°)/2. tan(PDS) = ((2b  b * tan(15°))/2)/(b/2) = 2  tan(15°). By the tangent halfangle formula tan(15°) = csc(30°)  cot(30°) = 2  ?(3), so the tangent of angle PDS = ?(3), so angle PDS = 60°. By symmetry angle SCP is 60° so triangle PDC is equilateral. 
August 9th, 2010, 07:33 PM  #6  
Newbie Joined: Jun 2010 Posts: 22 Thanks: 0  Re: Hard prove ! Quote:
 
August 9th, 2010, 07:39 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,593 Thanks: 937 Math Focus: Elementary mathematics and beyond  Re: Hard prove ! 
August 9th, 2010, 08:08 PM  #8  
Newbie Joined: Jun 2010 Posts: 22 Thanks: 0  Re: Hard prove ! Quote:
 
August 9th, 2010, 08:37 PM  #9  
Newbie Joined: Jun 2010 Posts: 22 Thanks: 0  Re: Quote:
You are the king of geometric ! Thanks a lot friend !!!  

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