Alice was having a conversation with her friend Trina, who had a discovery to share: Pick any two integers. Look at the sum of their squares, the difference of their squares,
and twice the product of the two integers you chose. Those three numbers are the
sides of a right triangle.
a. Write an equation that models this conjecture using the variables x and y.
b. Investigate this conjecture for at least three pairs of integers. Does her trick
appear to work in all cases, or only in some cases? Explain.
c. Use Trina’s trick to find an example of a right triangle in which all of the sides
have integer length, all three sides are longer than 100 units, and the three side
lengths do not have common factors.
BONUS: If Trina’s conjecture is true, use the equation found in part a to prove the
conjecture. If it is not true, modify it so it is a true statement, and prove the new
statement

a. (x² + y²)² = (x² - y²)² + (2xy)²

b. The conjecture works in all cases.

c. Sides of 119, 120, and 169

Step-by-step explanation:

a. Equation that models this conjecture

x² + y² = sum of squares

x² - y² = difference of square

  2xy = twice the product of the integers

If these are the sides of a right triangle then

(x² + y²)² = (x² - y²)² + (2xy)²

b. Test the conjecture

(i) Try x = 2, y = 1

(2² + 1²)² = (2² - 1²)² + (2×2×1)²

       5² = 3² + 4²

      25 = 9 + 16

(ii) Try x = 3, y = 1

(3² + 1²)² = (3² - 1²)² + (2×3×1)²

      10² = 8² + 6²

    100 = 64 + 36

(iii) Try x = 3, y = 2

(3² + 2²)² = (3² - 2²)² + (2×3×2)²

       13² = 5² + 12²

     169 = 25 + 144

The conjecture appears to work in all cases.

c. A possible triangle

We must have one side greater than 100. That means,  

x² > 100 or x >1 0.

                                      Let x = 12

                               One side = 12² + y²

                  The second side = 12² - y²

The third side must have 2xy > 100

                                        24y > 100

                                            y > 4.2

                                      Try y = 5

                              (12² + 5²)² = (12² - 5²)² + (2 × 12 × 5)²

                                       169² = 119² + 120²

So, one right triangle could have sides of 119, 120, and 169.

Furthermore, these sides have no common factors.

Check:

 169² = 119² + 120²

28561 = 14161 + 14400

28561 = 28561

452,489 to the nearest thousand is 452,000

location: point 0

step-by-step explanation:

well, -8 to 10 on a number line is 18 spaces.

1/9 of 18 is 2.

so, 4/9 of 18 must be 8.

8 spaces from -8 (a) is 0.

so, 4/9 of the way from -8 to 10 is 0!