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Sir Col
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Triple Right Pentagons
« on: Dec 23rd, 2004, 3:00am » |
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I shall call a pentagon formed by drawing a square on the hypotenuse of a right triangle a triple right pentagon.
Problem...
If the perimeter of a triple right pentagon is 44 and the area is 124, find the exact length of the longest diagonal from any two vertices.
Extensions...
Let a and b be the lengths of the legs of the right triangle and c the length of the hypotenuse.
(1) Given the perimeter, P, and the area, A, of the triple right pentagon, find c in terms of P and A.
(2) Given a and b, find the length of the longest diagonal.
(3) Given that P and A are integer, determine if the longest diagonal can ever be integer.
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John_Gaughan
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Re: Triple Right Pentagons
« Reply #1 on: Dec 23rd, 2004, 6:15am » |
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Since the square has length of side c, I derived the following equations:
P = a + b + 3(a2 + b2)0.5
A = a2 + b2 + (1/2)ab
The triangle in this example is a 6/8/10 right triangle, which I found just by juggling the numbers in my head, comparing against known pythagorean triplets. So I guess this has nothing to do with the two equations. Anyway, from here I could see that the longest diagonal goes from the right angle of the triangle to the opposite corner of the square. This forms a triangle with sides 8 and 10, with a 2[pi]/3 angle between them. The Law of Cosines gives the third side, 15.62.
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Sir Col
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Re: Triple Right Pentagons
« Reply #2 on: Dec 23rd, 2004, 12:04pm » |
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Good work!
You're right about the lengths of the triangle, only I wish I'd picked less familiar numbers now, as solving that part by trial and improvement means you miss out on a key insight required for the extensions. Certainly if anyone wishes to solve the extensions, you will need to be able to solve c from the a general P and A.
Unfortunatley you've found the second longest diagonal; the other diagonal from the apex of the right triangle is longer. In addition I wanted the exact length, and by that I mean in surd form, not some inverse trigonometric value.
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John_Gaughan
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Re: Triple Right Pentagons
« Reply #3 on: Dec 23rd, 2004, 9:18pm » |
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Actually, that is from the apex of the triangle. It goes to the corner of the square that is further away. I am not able to find a longer diagonal.
I know there is an insight there, but I was in a rush this morning and it was not coming to me as quickly as I wanted it to. I answered this at work, and today was a half day, then I left early, so I only "worked" two hours. Solving it numerically was fast and easy, the general form is a little bit longer. I did not touch the extensions yet.
What you really wanted was the answer to (2), then for me to use it to solve the original problem. I'll see if I can figure it out.
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John_Gaughan
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Re: Triple Right Pentagons
« Reply #4 on: Dec 23rd, 2004, 9:51pm » |
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I must be exceptionally smart, because I just got a different answer: [sqrt]170.
Assume a is the longer side of the triangle, b is the shorter side, d is the length of the diagonal. This is what I get for (2):
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c = [sqrt](a2 + b2)
d1 = [sqrt](a2 + b2 - 2ac*cos ([pi]/2 + tan-1 (b/a)))
d2 = c[sqrt]2
d = max (d1, d2)
I know there is a relationship between the sides of the triangle and whether the longest diagonal attaches to the apex of the triangle or connects two corners of the square. I imagine it involves drawing a circle centered on one of the "far" corners of the square with its circumference through the diagonal corner, and seeing if the point at the apex of the triangle is outside the circle. Now that I mention it, there has to be an easy way to determine this.
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Sir Col
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When I said, "...the other diagonal from the apex of the right triangle is longer." I was referring to the fact that there are two diagonals from the apex to different corners of the square. Your original answer was the shorter of the two lengths.
I'm not sure where you got [sqrt]170 from? The diagonal of the square is [sqrt]200, but the longest diagonal is from the apex of the triangle. In fact, it is always in this position, but you will only be able to prove it by first deriving a formula for its length, in terms of a and b, and comparing with the diagonal of the square.
Although your formula for d1 will give the correct answer, it is dependent on the cos() part being rational. Otherwise you cannot obtain an exact answer (in surd form). In the case of the original numerical problem, however, you should now be able to use your formula to obtain the exact length of the longest diagonal; I think that the answer you gave originally for the other diagonal from the apex was [sqrt]232=2[sqrt]58.
The diagram below may help with the general case: similar triangles? 
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| « Last Edit: Dec 24th, 2004, 3:44am by Sir Col » |
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ThudnBlunder
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Re: Triple Right Pentagons
« Reply #6 on: Dec 24th, 2004, 8:20am » |
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(1) 12c2 - 6Pc + (P2 - 4A) = 0
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| « Last Edit: Dec 24th, 2004, 8:23am by ThudnBlunder » |
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Sir Col
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Re: Triple Right Pentagons
« Reply #7 on: Dec 24th, 2004, 8:32am » |
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Have you been peeking at my piece of paper, T&B? I knew I shouldn't have left it so close to the computer! 
Nice work, sir. But that equation has two solutions and only one applies to the problem. Which one, and why?
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ThudnBlunder
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Re: Triple Right Pentagons
« Reply #8 on: Dec 24th, 2004, 9:02am » |
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Quote:
c = (P/4) - {sqrt[3(16A - P2)]/12}
because obviously P > 4c
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| « Last Edit: Dec 24th, 2004, 10:49am by ThudnBlunder » |
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John_Gaughan
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Re: Triple Right Pentagons
« Reply #10 on: Dec 24th, 2004, 9:38pm » |
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I think maybe I misunderstand something
What is "surd" form?
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ThudnBlunder
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Re: Triple Right Pentagons
« Reply #11 on: Dec 24th, 2004, 9:38pm » |
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on Dec 24th, 2004, 2:03pm, Sir Col wrote:
Simply stunning! :clapsmilie: |
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Yes, it's really amazing what I am capable of when I have the forum to myself due to everybody else being busy attending Christmas Eve parties! (Getalife Thud!)
(2) Assume a [smiley=leqslant.gif] b; then x [smiley=leqslant.gif] y
Let length of longer diagonal = D
Let length of shorter diagonal = d
By similar triangles
y/a = b/c giving y = ab/c
a/c = x/a giving x = a2/c
D2 = (c + y)2 + (c - x)2
= (a + b)2 + b2
If we produce the side of length b downwards and to the right by a distance a, then the result becomes obvious.
Similarly d2 = (a + b)2 + a2
(3) Sir Col, you have described the figure as a triple right pentagon. If this means that a,b,c are Pythagorean triples then both the perimeter and the area are always integers (as a*b is always even).
on Dec 24th, 2004, 9:38pm, John_Gaughan wrote:I think maybe I misunderstand something
What is "surd" form? |
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It is a rather old-fashioned name for an irrational number, and usually refers to radicals.
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| « Last Edit: Dec 25th, 2004, 1:51am by ThudnBlunder » |
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Sir Col
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Re: Triple Right Pentagons
« Reply #12 on: Dec 25th, 2004, 9:59am » |
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Old fashioned! It is part of the staple diet of 14-16 year olds in England.
http://en.wikipedia.org/wiki/Radical_(mathematics)
Nice generalisation with the longest diagonal in terms of a and b. Perhaps you, or John, can now prove why the longest diagonal is always from the apex of the right triangle and not the diagonal of the square?
T&B, I don't quite follow your statement about (3)?! I never said that a,b,c were a Pythagorean triplet; certainly the first two generalisations, (1) and (2), were not dependent on this.
Although P and A being integer implies this, I don't see how this immediately addresses the challenge, "Given that P and A are integer, determine if the longest diagonal can ever be integer"?
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| « Last Edit: Dec 25th, 2004, 10:01am by Sir Col » |
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ThudnBlunder
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Re: Triple Right Pentagons
« Reply #13 on: Dec 26th, 2004, 5:31am » |
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Quote:| T&B, I don't quite follow your statement about (3)?! I never said that a,b,c were a Pythagorean triplet; |
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I was only asking.
Quote:| Perhaps you, or John, can now prove why the longest diagonal is always from the apex of the right triangle and not the diagonal of the square? |
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Good point, Sir Col.
Let length of longest diagonal = D
Let length of shortest diagonal = d
Let length of square diagonal = ds
Then ds2 = 2(a2 + b2)
D > ds
provided
(a + b)2 + b2 > 2(a2 + b2)
This reduces to
a(2b - a) > 0
Hence
0 < a < 2b
But wlog we had previously assumed that a [smiley=leqslant.gif] b
Hence D > ds
provided
a > 0
Similarly d > ds
provided
a < b < 2a
(3) A and P are both integers implies a,b,c are a Pythagorean triplet.
Assume D is an integer; then d is also an integer.
d2 = (a + b)2 + a2
D2 = (a + b)2 + b2
Adding
D2 + d2 = 3(a + b)2 - 2ab
D and d are both always odd; hence the LHS is even.
But a + b being always odd implies RHS is odd.
Hence neither D nor d can be integers.
Is this the proof you had in mind, Sir Col. Or was it more complicated?
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| « Last Edit: Dec 31st, 2004, 3:11am by ThudnBlunder » |
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Re: Triple Right Pentagons
« Reply #14 on: Dec 31st, 2004, 3:09am » |
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4) Given that both A and P are integers and that a + b is a perfect square, find the minimum possible value of A.
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Sir Col
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Re: Triple Right Pentagons
« Reply #15 on: Jan 1st, 2005, 2:38pm » |
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on Dec 26th, 2004, 5:31am, THUDandBLUNDER wrote:| Is this the proof you had in mind, Sir Col. Or was it more complicated? |
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Not quite; mine works!
If D is integer, why must d also be integer? (1+3)2+32=25, but (1+3)2+12=17.
Even if that were true, unfortunately you can't have d and D both odd. By assuming {a,b,c} is a primitive triplet (we can produce a primitive case by scaling down if need be), then if one of a or b is odd, the other must be even. So (a+b)2 will be odd and adding a2 or b2 will produce one even sum and one odd sum.
My proof is not very nice, and I was rather hoping that someone would find a better one...
Working back to D2=(c+y)2+(c-x)2, we can see that, as c is integer, x and y must be integer. If x=a2/c is integer then c|a2 and if y=ab/c is integer then c|ab. As {a,b,c} is a primitive Pythagorean triplet, c can only divide ab is c=1. This is impossible, hence D can never be integer.
With regards to your question...
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If (a+b) is a perfect square, we're looking for a Pythagorean triplet, {a,b,c}, such k(a+b)=j2. It can be seen that {3,4,5} provides a solution 7(3+4)=72, so {21,28,35} is the least such triplet and A=c2+ab/2=1519.
The next triplet is {9,40,41}, for which A=1821.
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ThudnBlunder
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Re: Triple Right Pentagons
« Reply #16 on: Jan 1st, 2005, 10:48pm » |
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As for my claim that D is an integer => d is an integer, after doing 1) and 2) I must have had too much Christmas spirit!
Quote:| Even if that were true, unfortunately you can't have d and D both odd. By assuming {a,b,c} is a primitive triplet (we can produce a primitive case by scaling down if need be), then if one of a or b is odd, the other must be even. So (a+b)2 will be odd and adding a2 or b2 will produce one even sum and one odd sum. |
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Yes, I got my wires crossed there. I was assuming for no reason whatsoever that {D, a+b, b} and {d, a+b, a} were Pythagorean triplets and hence D and d must both be odd.
Sorry, for 4) I meant "Find a primitive Pythagorean triplet a,b,c such that a+b is a perfect square."
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| « Last Edit: Jan 1st, 2005, 10:52pm by ThudnBlunder » |
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