Re: Geometry with right triangle..



On May 1, 10:33 am, bill <b92...@xxxxxxxxx> wrote:
On Apr 30, 10:34 pm, "mina_world" <mina_wo...@xxxxxxxxxxx> wrote:





Hello teacher~

The perimeter of a right triangel ABC is 2a.

Find the scope of hypotenuse x.

Answer : 2{sqrt(2) - 1}a <= x < a.

---------------------------------------------------
Ok, let's go...

hypotenuse : x
base : c
altitude : d

c + d + x = 2a
c^2 + d^2 = x^2
c + d > x

(i) If x >= a,   then c + d = 2a - x <= a <= x.
contradiction.

(ii) By Cauchy-Schwarz inequality,
(c + d)^2 <= (1^2 + 1^2)(c^2 + d^2)
so, (c + d)^2 <= 2.(x^2)
so, (2a - x)^2 <= 2.(x^2)
so, 2.(x^2) - (2a - x)^2 >= 0
so, 2.(x^2) - [4a^2 - 4ax + x^2] >= 0
so, x^2 + 4ax - 4a^2 >= 0
so, x <= 2a{-1 - sqrt(2)} , x >= 2a(-1 + sqrt(2)}

so, 2{sqrt(2) - 1}a <= x < a.

x is minimun when c+d is maximum.
Guessing that c+d is maximum when
c = d; we surmise that

     2*c^2 = x^2  Then
     Sqrt(2)*c = x = sqrt(2)*d
      c = x/sqrt(2) = d
    2*a = x/sqrt(2) + x/sqrt(2) +x
          = sqrt(2)*x  + x
     2*a = x*(1 +sqrt*2)
     2*a/[1 +sqrt(2)]  < = x

Bill J- Hide quoted text -

- Show quoted text -

Trig 6 relationships sides to angles
The pythagorean theorem.
.



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