Question

Given that (11,60 62) is a pythagorean triple, use (11, 60, 61) to find a new pythagorean triple that contains 183.

Answer 1

To find a new Pythagorean triple containing 183 using the Pythagorean triple (11, 60, 61), we can utilize the method of scaling the original triple.

A Pythagorean triple is generated by the formula:

\[
(a, b, c) = (m^2 - n^2, 2mn, m^2 + n^2)
\]

However, in this case, we will simply multiply the entire triple (11, 60, 61) by a factor \( k \) so that one of the numbers in the new triple equals 183.

### Step 1: Find the scaling factor \( k \)

We first determine which number will be set to 183. The simplest option is to set \( 11k = 183 \):

\[
k = \frac{183}{11} = 16.63636
\]

However, since \( k \) should ideally be an integer and can be used to scale the triple efficiently, we check which of the other two numbers would work better.

Next, let's set \( 60k = 183 \):

\[
k = \frac{183}{60} = 3.05
\]

Likewise, setting \( 61k = 183 \):

\[
k = \frac{183}{61} = 3
\]

### Step 2: Using integer scaling

The integer value \( k = 3 \) works because it translates into the original triple as:

- \( 11 \cdot 3 = 33 \)
- \( 60 \cdot 3 = 180 \)
- \( 61 \cdot 3 = 183 \)

### Step 3: Confirm the new Pythagorean triple

Now, let's see if \( (33, 180, 183) \) forms a Pythagorean triple:

\[
33^2 + 180^2 = 1089 + 32400 = 33489
\]

Now, we check \( 183^2 \):

\[
183^2 = 33489
\]

Since both calculations are equal, we confirm that:

\[
33^2 + 180^2 = 183^2
\]

### Conclusion

Thus, a new Pythagorean triple containing 183 is:

\[
\boxed{(33, 180, 183)}
\]