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August 9th, 2010, 04:24 AM   #1
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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!
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August 9th, 2010, 04:45 AM   #2
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Re: Hard prove !

Can you picture it?
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August 9th, 2010, 06:01 AM   #3
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Re: Hard prove !

Try to find tan(15) with the double-angle formulae.
So, 1/?3=tan(30)=...
What has this to do with your problem.?
By the way, there are 4 isosceles triangles?
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August 9th, 2010, 03:21 PM   #4
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Quote:
Originally Posted by yehoram
One of them is: a triangle whose base is one of the sides of the square with base angles of 15.
Construct a copy of this triangle, using as its base one of the two remaining sides of the square, then join the "apex" of this copy to the common apex of the previously constructed triangles. Prove that an equilateral triangle appears. It's easy to finish the problem from there. Each triangle's "apex" must lie inside the square. This problem appears in H. S. M. Coxeter's Introduction to Geometry.
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August 9th, 2010, 05:41 PM   #5
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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 half-angle 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.
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August 9th, 2010, 07:33 PM   #6
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Re: Hard prove !

Quote:
Originally Posted by greg1313
Trigonometry:


Without loss of generality let the sides of the square be of length 2 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 = tan(15). tan(PDS) = 2 - tan(15). By the tangent half-angle 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.
The prove must be with geometric way, no trigonometric !!!
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August 9th, 2010, 07:39 PM   #7
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Re: Hard prove !

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August 9th, 2010, 08:08 PM   #8
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Re: Hard prove !

Quote:
Originally Posted by greg1313
That it mean!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!
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August 9th, 2010, 08:37 PM   #9
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Re:

Quote:
Originally Posted by skipjack


Quote:
Originally Posted by yehoram
One of them is: a triangle whose base is one of the sides of the square with base angles of 15.
Construct a copy of this triangle, using as its base one of the two remaining sides of the square, then join the "apex" of this copy to the common apex of the previously constructed triangles. Prove that an equilateral triangle appears. It's easy to finish the problem from there. Each triangle's "apex" must lie inside the square. This problem appears in H. S. M. Coxeter's Introduction to Geometry.

You are the king of geometric ! Thanks a lot friend !!!
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