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anonymous

  • one year ago

Geometry :D Suppose f and g are two isometries such that f(A) = g(A), f(B) = g(B), and f(C) = g(C) for some nondegenerate triangle ΔABC. Show that f = g

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  1. anonymous
    • one year ago
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    I feel like this should be easy, but still getting used to all the geometry theorems and what not.

  2. anonymous
    • one year ago
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    @freckles know anything about this stuff?

  3. anonymous
    • one year ago
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    @ganeshie8 Know any of this stuff?

  4. ganeshie8
    • one year ago
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    idk much but i found this #3 http://www.ms.uky.edu/~droyster/courses/spring04/homework/restricted/Homework2%20Solns.pdf @ikram002p did non eucledean geometries a couple of years ago

  5. anonymous
    • one year ago
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    Ah, nice find! I found something else online, but it used a theorem that I couldnt use based on the sequence of my text. But alright, if ikram knows some of this stuff, Ill see what he can do when I have questions. I think I can just use what that link gives. Thanks :)

  6. ganeshie8
    • one year ago
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    good luck! :)

  7. ikram002p
    • one year ago
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    hey first u need to know what isometries functions means, it means a rigid transforms function which maps a shape to some where else but without changing size in other way both origin and transforms are congruent.

  8. ikram002p
    • one year ago
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    lets do it in geometry style |dw:1433400064930:dw|

  9. ikram002p
    • one year ago
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    so since both are isometries we have 3 congruent triangles, now here is a thing why it called rigid transformation it since u do not change it size and shape , like triangle do not become a circle for example or dont become another triangle with same size but different sides that cannot happen.

  10. ikram002p
    • one year ago
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    now step two, it says if f(A) = g(A), f(B) = g(B), and f(C) = g(C), then show Show that f = g in euclidean geometry its ok if u wanna only graph it that works like this it would be a proof by itself |dw:1433400511306:dw|

  11. ikram002p
    • one year ago
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    now since you wanna it theoretically f(A)=g(A) g^-1o f(A)=g^_1 o g(A) =A so if u apply the function g^-1 o f on the other two points u got g^-1 o f(B)=B g^-1 o f(C)=C according to a theorem (idk if ts given in ur book or u need to prove it let me know in both cases ) g^-1o f must be the identity which means :- g^-1 o f(P)=P , for any P on the triangle g o g^-1 o f(P)= g(P) f(P)=g(P) which is done :D let me know if something is not clear enough, good luck !!

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