MHB Integer-Sided Right Triangle: 2001 Leg Length & Minimum Other Leg Length

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In an integer-sided right triangle with one leg measuring 2001, the minimum length of the other leg can be determined using the Pythagorean theorem. The solution involves finding integer values that satisfy the equation a^2 + 2001^2 = c^2, where 'a' is the length of the other leg and 'c' is the hypotenuse. The discussion highlights the importance of integer solutions and the constraints imposed by the fixed leg length. Participants share various approaches to derive the minimum leg length, emphasizing mathematical reasoning. The minimum length of the other leg is ultimately established through collaborative problem-solving.
lfdahl
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The shorter leg of an integer-sided right triangle has length 2001. How short
can the other leg be?
 
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lfdahl said:
The shorter leg of an integer-sided right triangle has length 2001. How short can the other leg be?
my solution:
$2001=667\times 3=3\times 23\times 29$
so the other leg =$667\times 4=2668$
 
Last edited:
Thankyou Albert for the correct answer - and for your tireless participation!(Clapping)

Suggested solution:

Let $a,b,c$ be the sides of the triangle. Thus $2001 = a < b < c$. Set $c=b+m$. Then $(b+m)^2 = b^2+2001^2$ or $m(2b+m) = 2001^2$. So $m$ is a divisor of $2001^2=3^2\cdot 667^2$ and since $b= c-m$ is to be shortest ($> 2001$), $m = 667$ (the next largest divisor is $3 \cdot 667 = 2001$, which makes $b=0$) should be considered. Then $667(2b+667) = 9\cdot667^2$ gives $b=2668$ and $c=2668 + 667=3335$. One checks, that $2001^2+2668^2 = 3335^2$.
Comment: This triangle is the $(3,4,5)$ triangle since $(2001,2668,3335) = 667(3,4,5)$. But recognizing this, does not prove, that $2668$ is the shortest possible side larger than $2001$.
 
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