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Thread: Could there be more than three?

20080419, 22:59 #1
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Could there be more than three?
I've got three right triangles in mind, whereby:
 <LI>the lengths of the three sides are integers, and
<LI>the area of the triangle is exactly twice the perimeter.
 Ricky

20080420, 00:03 #2
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Re: Could there be more than three?
<P ID="edit" class=small>(Edited by Jezza on 19Apr08 22:03. Changed my mind :))</P>Hi there
I am assuming a right triangle <img src=/S/flags/USA.gif border=0 alt=USA width=30 height=18> is a rightangle triangle <img src=/S/flags/UK.gif border=0 alt=UK width=30 height=18> <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
how about:<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold"> 9, 40, 41</span hide>Jerry

20080420, 01:12 #3
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Re: Could there be more than three?
I'm fairly certain you misunderstood me correctly <img src=/S/sarcasm.gif border=0 alt=sarcasm width=15 height=15> <img src=/S/scratch.gif border=0 alt=scratch width=25 height=29>
One would have to ask to be sure  I could've been referring to a triangle with two right angles <img src=/S/nuts.gif border=0 alt=nuts width=15 height=15>
BTW  You have correctly picked one of my three solutions. <img src=/S/trophy.gif border=0 alt=trophy width=15 height=15> Ricky

20080420, 01:19 #4
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Re: Could there be more than three?
OK, on a run now <img src=/S/grin.gif border=0 alt=grin width=15 height=15>
Howzabout
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">
10,24,26
12,16,20
</span hide>
<img src=/S/drop.gif border=0 alt=drop width=23 height=23>Jerry

20080420, 01:44 #5
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Re: Could there be more than three?
There are only three solutions.
<span style="backgroundcolor: #FFFF00; color: #FFFF00; fontweight: bold">Let's say the legs have length a and b.
Area = a*b/2
Perimeter = a+b+SQRT(a^2+b^2)
From area = 2*perimeter we get a*b/2 = 2*(a+b+SQRT(a^2+b^2))
a*b = 4*(a+b+SQRT(a^2+b^2))
a*b4*a4*b = 4*SQRT(a^2+b^2)
(a*b4*a4*^2 = 16*(a^2+b^2)
a^2*b^28*a^2*b8*a*b^2+16*a^2+32*a*b+16*b^2 = 16*a^2+16*b^2
a^2*b^28*a^2*b8*a*b^2+32*a*b = 0
Since we don't want a = 0 or b = 0, we can divide by a*b:
a*b8*a8*b+32 = 0
(a8)*(b8) = 32
Since a and b are whole numbers, a8 and b8 must be integer dividers of 32. The only solutions are:
1) a8 = 1 and b8 = 32, i.e. a=9 and b=40.
2) a8 = 2 and b8 = 16, i.e. a=10 and b=24.
3) a8 = 4 and b8 = 8, i.e. a=12 and b=16.
4) a8 = 8 and b8 = 4: this is the same as 3) with a and b reversed.
5) a8 = 16 and b8 = 2; this is the same as 2) with a and b reversed.
6) a8 = 32 and b8 = 1; this is the same as 1) with a and b reversed.</span hide>

20080420, 20:47 #6
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Re: Could there be more than three?
Looks like JB (being on the roll he was) has supplied the three correct solutions,
while Hans has proved beyond a doubt that there are only three.
On the farm, we refer to that as "shucking the corn right down to the cob".
Choccy bars for the bear and for Hans. <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16> Ricky