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Thread: Circles

  1. #1
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    Circles

    Let $\displaystyle \bold{r} $ be a point on a circle with diameter from $\displaystyle - \bold{a} $ to $\displaystyle \bold{a} $ (so that $\displaystyle |\bold{r}| = |\bold{a}| $). Draw the chord through $\displaystyle \bold{r} $ perpendicular to $\displaystyle \bold{a} $ (the length and direction of this chord is $\displaystyle 2 \bold{r}_{a \perp} $). Let $\displaystyle \bold{t} = \bold{r}_{a} $ be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from $\displaystyle -\bold{a} $ to $\displaystyle \bold{t} $ times the distance from $\displaystyle \bold{t} $ to $\displaystyle \bold{a} $: $\displaystyle |\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t}) $.
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  2. #2
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    So I got $\displaystyle \hat{\bold{a}} = \frac{\bold{a}}{|\bold{a}|} $

    and $\displaystyle \bold{t} = \bold{r} \cdot \hat{\bold{a}} $

    and $\displaystyle \bold{r}_{a \perp} = \bold{t} - \bold{r} $.

    Then $\displaystyle \bold{a} + \bold{r} $ gives one hypotenuse. And so $\displaystyle |\bold{a} + \bold{r}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}+\bold{t}|^{2} $.

    So $\displaystyle |\bold{r} - \bold{a}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}-\bold{t}|^{2} $.
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  3. #3
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    Quote Originally Posted by heathrowjohnny View Post
    Let $\displaystyle \bold{r} $ be a point on a circle with diameter from $\displaystyle - \bold{a} $ to $\displaystyle \bold{a} $ (so that $\displaystyle |\bold{r}| = |\bold{a}| $). Draw the chord through $\displaystyle \bold{r} $ perpendicular to $\displaystyle \bold{a} $ (the length and direction of this chord is $\displaystyle 2 \bold{r}_{a \perp} $). Let $\displaystyle \bold{t} = \bold{r}_{a} $ be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from $\displaystyle -\bold{a} $ to $\displaystyle \bold{t} $ times the distance from $\displaystyle \bold{t} $ to $\displaystyle \bold{a} $: $\displaystyle |\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t}) $.
    Hello,

    you are dealing with a right triangle (see attachment)

    If the legs of a right triangle are a and b and the hypotenuse is c then you know:

    $\displaystyle a^2+b^2 = c^2$ ....... and $\displaystyle c = p+q$. That means $\displaystyle c^2 = p^2+2pq+q^2$

    Therefore:

    $\displaystyle a^2+b^2 = p^2+2pq+q^2$

    According to the drawing you have:

    $\displaystyle a^2 = h^2 + p^2$ ....... and $\displaystyle b^2 = h^2 +q^2$

    So finally you get:

    $\displaystyle h^2 + p^2 + h^2 +q^2 = p^2+2pq+q^2~\iff~ 2h^2 = 2pq~\iff~\boxed{h^2 = pq}$
    Attached Thumbnails Attached Thumbnails Circles-euklid_theorem.gif  
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