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Science Forum Index » Mathematics Forum » Geometry with right triangle..
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| mina_world |
 Posted: Thu May 01, 2008 12:34 am |
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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. |
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| Julio Di Egidio |
| Posted: Thu May 01, 2008 12:34 am |
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Quote: 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.
Unless: a = 0.
In that case: a <= x <= a, that is: a = x.
That assuming: a = a.
No guarantee...
Julio
Quote:
---------------------------------------------------
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.
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| bill |
| Posted: Thu May 01, 2008 8:33 am |
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On Apr 30, 10:34 pm, "mina_world" <mina_wo...@hanmail.net> wrote:
Quote: 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 |
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| Guest |
| Posted: Thu May 01, 2008 11:44 am |
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On May 1, 10:33 am, bill <b92...@yahoo.com> wrote:
Quote: On Apr 30, 10:34 pm, "mina_world" <mina_wo...@hanmail.net> 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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