To find a new Pythagorean triple that contains 183 using the original triple (11, 60, 61), we can use a scaling factor.
First, notice that the original triple represents the relationship \(11^2 + 60^2 = 61^2\). To create a new Pythagorean triple, we can scale the values in the original triple by a factor \(k\).
To find a suitable scaling factor \(k\) such that one of the values becomes 183, we can set one of the components equal to 183:
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Set \(11k = 183\): \[ k = \frac{183}{11} = 16.6364 \quad (\text{not an integer, discard}) \]
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Set \(60k = 183\): \[ k = \frac{183}{60} = 3.05 \quad (\text{not an integer, discard}) \]
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Set \(61k = 183\): \[ k = \frac{183}{61} = 3 \quad (\text{integer, keep}) \]
Now we scale the entire triple (11, 60, 61) by the integer \(k = 3\):
\[ (11 \times 3, 60 \times 3, 61 \times 3) = (33, 180, 183) \]
Therefore, the new Pythagorean triple that contains 183 is:
\[ (33, 183) \]
We can check this as follows:
\[ 33^2 + 180^2 = 1089 + 32400 = 33489 \] \[ 183^2 = 33489 \]
Since both sides are equal, the new Pythagorean triple is confirmed.
So, the final answer is:
The new Pythagorean triple is \( (33, 183) \).