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Nir Dahan Inner circle Munich, Germany 1390 Posts |
How many RIGHT triangles exist whose edges sizes (expressed in cm) are integers and whose circumference (expressed in cm) equals their area (expressed in cm^2)
n. |
landmark Inner circle within a triangle 5194 Posts |
Circumference of a triangle? Do you mean perimeter of the triangle? Or circumference of the circumscribed (or inscribed) circle?
can you clarify this please? Jack Shalom
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
stanalger Special user St. Louis, MO 998 Posts |
According to my dictionary, "perimeter" and
"circumference" are synonyms. So I'll go with "two." Stan Alger |
Nir Dahan Inner circle Munich, Germany 1390 Posts |
Quote:
On 2006-05-22 19:38, landmark wrote: sum of all edges... sorry for my lack of language skills |
Nir Dahan Inner circle Munich, Germany 1390 Posts |
Quote:
On 2006-05-23 00:46, stanalger wrote: very nice stan. I assume you combined symmetrical solutions... can you elaborate on the way to the answer? I think it is pretty neat. nir |
stanalger Special user St. Louis, MO 998 Posts |
I'll bet there's a more elegant way to arrive at the solution.
But I got it this way: Let A, B be the legs and let C be the hypotenuse. C=sqrt(A^2 + B^2). We want (1/2)A*B = A + B + sqrt(A^2 + B^2). So -sqrt(A^2 + B^2) = (A + B) - (1/2)A*B. Squaring both sides gives: A^2 + B^2 = A^2 + 2A*B + B^2 -A*B*(A + B) + (1/4)A^2*B^2. Subtract (A^2 + B^2) from both sides: 0 = 2A*B - A*B*(A+B) + (1/4)A^2*B^2. Factor: 0=A*B*(2-A-B+(1/4)A*B) Since neither A nor B can equal zero, 0= 2-A-B+(1/4)A*B. Since A, B are integers, (1/4)A*B must also be an integer. This means 4 divides A*B. So either (i) 4 divides A or (ii) 4 divides B or (iii) 2 divides both A and B. Let's examine case (iii) in detail. If 2 divides both A and B, then A + B is even... so (1/4)A*B must also be even. This implies 8 divides A*B. So either 4 divides A or 4 divides B. Thus all three cases allow us to conclude that (at least) one of the legs must be divisible by 4. Without loss of generality, assume A is divisible by 4. So there is a positive integer D for which 4*D = A. Substituting 4*D for A in the last equation gives: 0 = 2 - 4*D - B + (1/4)4*D*B which, when solved for B, becomes B = (2 - 4*D)/(1 - D). Since B is an integer, (1 - D) divides (2 - 4*D). But (2-4*D)/(1 - D) = 4 - 2/(1 - D) (by long division) so 2/(1 - D) must be an integer. Since 2 only has 4 divisors (namely 1, 2, -1, -2), there are only four possibilities for D: -1, 0, 3, and 2. Only D = 2 and D = 3 lead to positive values for A. When D = 2, A = 4*D = 4*2 = 8 ........and B = (2 - 4*D)/ (1 - D) = (2 - 8) / ( 1 - 2) = -6/(-1) = 6. When D = 3, A = 4*D = 4*3 = 12 ........and B = (2 - 4*D) / (1 - D) = (2-4*3)/(1-3) = -10/(-2) = 5. So we either have a 6-8-10 right triangle or a 5-12-13 right triangle. I now await a more elegant solution from Nir...or some other Café member. Stan |
stanalger Special user St. Louis, MO 998 Posts |
Quote:
On 2006-05-23 11:29, stanalger wrote: Clarification was requested in a PM. 0 = 2 - A - B + (1/4)*A*B so (1/4)A*B = A + B - 2. If A and B are both even, so is A + B - 2. I.e. (1/4)A*B must be even. Since 2 divides (1/4)A*B, 4*2 divides A*B. ------------------------------------------------- Also, there's no need to introduce D. Solving (1/4)*A*B = A + B - 2 for A gives: A = (4*B -8) / (B -4). Using long division, this quotient is equal to 4 + (8/(B - 4)). For this to be an integer, (B - 4) must divide 8. 8 only has 8 divisors: -8, -4, -2, -1, 1, 2, 4, 8. Etc. |
landmark Inner circle within a triangle 5194 Posts |
Legs and hyp of a RT with integer sides can always be expressed as m^2 - n^2, 2mn, and m^2 +n^2, where m and n are pos integers m>n.
So (m^2 -n^2) (mn) = 2m^2 + 2mn; Factoring: (m+n)(m-n)(mn) =2m (m+n); Dividing:( m-n)(n) =2 One factor on the left must equal 2, the other must equal 1. Therefore either m=3, n=1 in which case we have a 6-8-10 triangle or m=3, n=2 in which case we have a 5-12-13 triangle. Jack Shalom
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
stanalger Special user St. Louis, MO 998 Posts |
Jack,
Thanks. Much nicer! |
Wolflock Inner circle South Africa 2257 Posts |
Ok. Some of you guys are just too clever for me and also have way to much time on your hands. :o)
Regards Wolflock
Wolflock
Pro Magician & Escapologist Member of JMC (Johannesburg Magic Circle) South Africa |
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