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Thread: Circle problem

  1. #1
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    Circle problem

    Have a circle with a radius of 1. Split the circle into 4 equal quadrants. Inside 1, draw a circle that touches both radius and the circle arc. How do I calculate the radius of smaller circle ?
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  2. #2
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    Re: Circle problem

    Quote Originally Posted by Archiecrab View Post
    Have a circle with a radius of 1. Split the circle into 4 equal quadrants. Inside 1, draw a circle that touches both radius and the circle arc. How do I calculate the radius of smaller circle ?
    Cheers
    1. Draw a sketch! (see attachment)

    2. You've got 2 right triangles. The larger one has the hypotenuse

    $\displaystyle 2r+x=1$ .................... [A]

    3. Use proportions:

    $\displaystyle \frac{r}{\frac12 \sqrt{2}}=\frac{r+x}1$ .................... [B]

    4. From [A] you get: $\displaystyle x=1-2r$

    Thus your proportion becomes:

    $\displaystyle \frac{r}{\frac12 \sqrt{2}}=\frac{r+1-2r}1$

    $\displaystyle \frac{r}{\frac12 \sqrt{2}}=1-r$

    $\displaystyle r+r \cdot \frac12 \sqrt{2}=\frac12 \sqrt{2}~\implies~\boxed{r=\frac{\frac12 \sqrt{2}}{1+\frac12 \sqrt{2}}}$
    Attached Thumbnails Attached Thumbnails Circle problem-radiusberuehrkreis.png  
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Circle problem

    I used an identical drawing and determined the point where the two circles touch must have equal coordinates, and so we may determine:

    $\displaystyle x=y=\frac{1}{\sqrt{2}}$

    And thus we may state:

    $\displaystyle 2(x-r)^2=r^2$

    After simplification we find:

    $\displaystyle r=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$
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  4. #4
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    Re: Circle problem

    And, remarkably, those two answers are the same!
    Starting with earboth's result, $\displaystyle \frac{\frac{1}{2}\sqrt{2}}{1+ \frac{1}{2}\sqrt{2}}$, multiply both numerator and denominator by 2 to get $\displaystyle \frac{\sqrt{2}}{2+ \sqrt{2}}$. Now, "rationalize" the denominator by multiplying both numerator and denominator by $\displaystyle 2- \sqrt{2}$:
    $\displaystyle \frac{\sqrt{2}}{2+ \sqrt{2}}\frac{2- \sqrt{2}}{2- \sqrt{2}}= \frac{2\sqrt{2}- 2}{4- 2}= \frac{2(\sqrt{2}- 1)}{2}= \sqrt{2}- 1$
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