Geometry with right triangle..
- From: "mina_world" <mina_world@xxxxxxxxxxx>
- Date: Thu, 1 May 2008 14:34:31 +0900
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.
.
- Follow-Ups:
- Re: Geometry with right triangle..
- From: Philippe 92
- Re: Geometry with right triangle..
- From: bill
- Re: Geometry with right triangle..
- From: Julio Di Egidio
- Re: Geometry with right triangle..
- Prev by Date: Re: Seminars on New Math
- Next by Date: Re: Geometry with right triangle..
- Previous by thread: How do I proof a Subset relation
- Next by thread: Re: Geometry with right triangle..
- Index(es):
Relevant Pages
|
