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Thread: polygons and triangles

  1. #1
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    polygons and triangles

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  2. #2
    MHF Contributor red_dog's Avatar
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    Question 1:
    $\displaystyle AC=\sqrt{AB^2+AC^2}=\sqrt{2}$
    $\displaystyle AD=\sqrt{AC^2+CD^2}=\sqrt{2+1}=\sqrt{3}$
    $\displaystyle AE=\sqrt{AD^2+DE^2}=\sqrt{3+1}=2$
    $\displaystyle AF=\sqrt{AE^2+EF^2}=\sqrt{4+1}=\sqrt{5}$
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  3. #3
    MHF Contributor red_dog's Avatar
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    Question 2:
    $\displaystyle \widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}\Ri ghtarrow 2(x+y+z)=180^{\circ}\Rightarrow x+y+z=90^{\circ}$
    $\displaystyle \widehat{ADB}=180^{\circ}-2z=180^{\circ}-2(90^{\circ}-(x+y))=$
    $\displaystyle =180^{\circ}-180^{\circ}+2(x+y)=2\widehat{ACB}$
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  4. #4
    MHF Contributor red_dog's Avatar
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    Question 3:
    If $\displaystyle \displaystyle \frac{AC}{AB}=k$ then $\displaystyle \displaystyle \overrightarrow{OM}=\frac{1}{1+k}\cdot\overrightar row{OC}+\frac{k}{1+k}\cdot\overrightarrow{OB}$
    In this case $\displaystyle \displaystyle k=\frac{7}{3}$
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  5. #5
    MHF Contributor red_dog's Avatar
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    Question 4:
    $\displaystyle \widehat{FBG}$ is exterior to triangle $\displaystyle BDG$. So $\displaystyle \widehat{FBG}>\widehat{BDG}$ (1).
    $\displaystyle \widehat{BDG}$ is exterior to triangle $\displaystyle ADE$. So $\displaystyle \widehat{BDG}>\widehat{AED}$ (2).
    From (1) and (2) via tranzitivity, yields $\displaystyle \widehat{FBC}>\widehat{AED}$
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  6. #6
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    Hello,

    to question 2:

    I've modified your drawing (see attachment):

    The angle in red is an exterior angle of the isosceles triangle ACD;
    the angle in blue is an exterior angle of the isosceles triangle ABD.

    The angle at the centre is 2x + 2y = 2(x+y) that means it is twice as large as the angle at C.
    Attached Thumbnails Attached Thumbnails polygons and triangles-puz_beweis.gif  
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  7. #7
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    Hello, Raiden_11!

    Basically the same as red_dog . . .


    Click image for larger version. 

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    In $\displaystyle \Delta ABC\!:\;2x + 2y + 2z \;=\;180^o\quad\Rightarrow\quad x + y + z \;=\;90^o$

    . . $\displaystyle \angle BCA \;=\;x + y \;=\;90^o - z$


    In $\displaystyle \Delta ADB\!:\;\angle ADB + 2z \;=\;180^o$

    Therefore: .$\displaystyle \angle ADB \;=\;180 - 2z \;=\;2(90^o-z) \;=\;2\cdot\angle BCA$

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  8. #8
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    Can u help me with these?

    2. Observe the relationship between the elements of Diagonal 1 and Diagonal 2 in Pascalís Triangle, shown on the next page. The sum of the first two elements in Diagonal 1 is equal to the second element in Diagonal 2. Similarly, the sum of the first 5 elements of Diagonal 1 is equal to the 5th element of Diagonal 2. Suppose we wish to use Pascalís Triangle to find the sum of the first n natural numbers. Use combinatorial notation to express the sum of the n elements of Diagonal 1 in terms of the corresponding value in Diagonal 2.



    3. Using the method of mathematical induction, prove that the following statements are false.


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  9. #9
    MHF Contributor red_dog's Avatar
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    2.
    $\displaystyle C_1^1+C_2^1=C_3^2$
    $\displaystyle C_1^1+C_2^1+C_3^1+C_4^1+C_5^1=C_6^2$
    $\displaystyle C_1^1+C_2^1+\ldots C_n^1=C_{n+1}^2$

    3.
    For the first identity:
    For $\displaystyle n=1\Rightarrow 2=2$
    For $\displaystyle n=2\Rightarrow 6=6$
    For $\displaystyle n=3\Rightarrow 12=14$, false. So the identity is not true for all $\displaystyle n\in\mathbf{N^*}$.

    For the second identity:
    For $\displaystyle n=1\Rightarrow 2=4$, false. So the identity is not true for all $\displaystyle n\in\mathbf{N^*}$.
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  10. #10
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    Last thing i'll ever need from you!?

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