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Subject: "Missing information"     Previous Topic | Next Topic
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Conferences The CTK Exchange High school Topic #219
Reading Topic #219
roboknight
Member since Dec-17-02
Dec-17-02, 11:51 AM (EST)
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"Missing information"
 
   The problem is the following:

Given that in a right triangle with sides of length x,y, and z respectively where x < y < z (therefore z is the hypotenuse)

x^2 + y^2 = z^2 (by Pythagoras)
and given that the area of the triangle is 666666, so
1/2*x*y = 666666

compute x,y, and z such that the triple (x,y,z) is a pythagorean primative (this is a triple that follows x^2 + y^2 = z^2 AND gcd(x,y,z)=1)

I'm not looking for a direct answer. I am looking for some paths to FIND the answer. I am missing something about this problem and I don't know what I am missing (obviously otherwise I would have solved it). So my question is what relationships would I need, or information might I need in order to solve this problem.


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alexb
Charter Member
2237 posts
Dec-17-02, 11:58 AM (EST)
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1. "RE: Missing information"
In response to message #0
 
   >I'm not looking for a direct answer. I am looking for some
>paths to FIND the answer. I am missing something about this
>problem and I don't know what I am missing (obviously
>otherwise I would have solved it). So my question is what
>relationships would I need, or information might I need in
>order to solve this problem.

Just factor 666666. There are not many of them:

xy/2 = 666666

xy = 2·666·1001.

666 = 2·3·3·37
1001 = 7·11·13

All you have to do now is split the factors between x and y and see what their squares add up to.


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daves
guest
Jun-22-08, 07:32 AM (EST)
 
2. "RE: Missing information"
In response to message #1
 
   Remember the Pythagorean triples?

A simpler way to start with is to let
z = m^2 + n^2
x = m^2 - n^2
y = 2mn

Now the area of the right triangle is
(1/2)*(m^2 - n^2)*2mn
= mn(m + n)(m - n)

Set up the following equation
mn(m + n)(m - n)= 666666
mn(m + n)(m - n)= 2·3·3·7·11·13.37


https://www.idealmath.com


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