Let 5a + 12b and 12a + 5b be the side lengths of a right-angled triangle and 13a + kb be the hypotenuse, where a, b and k are positive integers. Find the smallest possible value of k and the smallest values of a and b for that k
Using Pythagoras, we get
(5a+12b)2+(12a+5b)2 = (13a+kb)2
Solving for k:
k=(sqrt(169*b^2+240*a*b+169*a^2)-13*a)/b
or
k=-(sqrt(169*b^2+240*a*b+169*a^2)+13*a)/b
The second solution is negative and does not fit the requirements that k is a positive integer.
We are required to find pairs of positive integers a and b such that k is also a positive integer.
I have not been able to find an analytical solution of the integer problem, although the minimum value of k appears to be ten, with the minimum values of a and b being 69 and 20.
Thus the triplet (a,b,k) of (69,20,10) is the minimal solution.
The solution for k=11 is (24,23,11), and for k=12, (25,72,12). Other solutions exist for the same k where a and b are multiples of the basic case.
the answer is 3==D
maths... keewwllllll
To find the smallest possible value of k and the smallest values of a and b for that k, we need to determine the values of a, b, and k that satisfy the conditions of a right-angled triangle.
In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Using this information, we can set up an equation:
(5a + 12b)^2 + (12a + 5b)^2 = (13a + kb)^2
Expanding and simplifying this equation, we get:
25a^2 + 120ab + 144b^2 + 144a^2 + 60ab + 25b^2 = 169a^2 + 26akb + k^2b^2
Combining like terms, we have:
25a^2 + 144a^2 + 169a^2 + 120ab + 60ab + 26akb + 144b^2 + 25b^2 - k^2b^2 = 0
Collecting the like terms, we get:
338a^2 + 180ab + (169 - k^2)b^2 - 26akb = 0
Since a, b, and k are all positive integers, the equation above implies that each individual term must be zero. We will solve them separately:
1) 338a^2 + 180ab - 26akb = 0
Dividing both sides by a (since a must be nonzero), we get:
338a + 180b - 26kb = 0
Rearranging the terms, we have:
338a = 26kb - 180b
From this equation, we notice that 338a must be divisible by b, and since a and b are positive integers, the smallest possible value of a is 169 (since 338 is divisible by 2, 169 is divisible by 2, and it's the smallest possible positive value of a). Substitute a = 169 into the equation:
338(169) = 26kb - 180b
57122 = (26k - 180)b
To find the smallest possible value of b, we try dividing 57122 by all the possible values of (26k - 180).
2) (169 - k^2)b^2 = 0
Since b is a positive integer, the only possible solution for this equation is b = 0. However, this is not an acceptable solution since b cannot be zero in the context of a right-angled triangle.
Considering all the equations, the smallest possible value of k is the smallest value that makes the equation in step 1) solvable, while the smallest values of a and b will be the ones that satisfy all the given conditions.