Let a=2 mn, n=m^2-n^2, and c = m^2 + n^2 be the sides of a pythagorean triangle. suppose that b=a+1, show that (m-n)^2 -2n^2 = 1. and determine all such triangles

Part A

When a=2mn, b=m^2-n^2, c=m^2+n^2 and b=a+1 we can substitute for a and b to get

... b = a+1

... m^2 -n^2 = 2mn +1

Subtracting 2mn, we have

... m^2 - 2mn - n^2 = 1

Adding 2n^2-2n^2 = 0 on the left, we can rearrange to get

... (m^2 -2mn -n^2 +2n^2) -2n^2 = 1

... (m^2 -2mn +n^2) -2n^2 = 1

... (m-n)^2 -2n^2 = 1 . . . . . . . the desired conclusion

Part B

This is a form of the Pell equation, which has solutions

... (m-n)/n are alternate convergents of the continued fraction for Sqrt[2].

The continued fraction for a square root is quite regular, so we can write recursive relations for m and n as follows.

... m[k] = 6m[k-1] -m[k-2] . . . . . . . m[1] = 5, m[2] = 29

... n[k] = 6n[k-1] -n[k-2] . . . . . . . . . n[1] = 2, n[2] = 12

The first ten values of m and n are

... m = {5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965}

... n = {2, 12, 70, 408, 2378, 13860, 80782, 470832, 2744210, 15994428}

These give rise to the Pythagorean triples

... {a, b, c} =

{20, 21, 29},

{696, 697, 985},

{23660, 23661, 33461},

{803760, 803761, 1136689},

{27304196, 27304197, 38613965},

{927538920, 927538921, 1311738121},

{31509019100, 31509019101, 44560482149},

{1070379110496, 1070379110497, 1513744654945},

{36361380737780, 36361380737781, 51422757785981},

{1235216565974040, 1235216565974041, 1746860020068409}, ...

The values grow almost exponentially, and there are an infinite number of such triples.

The {3, 4, 5} triangle arises when m=2, n=1. In terms of the above, that triple is {4, 3, 5}, where b = a-1, not a+1.

95.03 is the answer. hope this !

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.