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Math Help - Tangent circle questions

  1. #1
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    Tangent circle questions

    1.Two semicircles of radius 6 units are inscribed in a semicircle of radius 12 units. A circle is inscribed so that it is tangent to all three semicircles. Find the circumference of the circle

    2. In triangle ABC, AB=260, AC=400, and BC=520. Point D is chosen on BC so that the circle inscribed in triangle ABD and ADC are tangent to AD at the same point. What is the length of BD?
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  2. #2
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    Quote Originally Posted by chengbin View Post
    1.Two semicircles of radius 6 units are inscribed in a semicircle of radius 12 units. A circle is inscribed so that it is tangent to all three semicircles. Find the circumference of the circle

    ...
    1. Draw a rough sketch (see attachment)

    2. Calculate the length of r using Pythagorean theorem:

    (12-r)^2+6^2=(6+r)^2~\implies~r=4

    3. Therefore the circumference is c = 2\cdot \pi \cdot r = 8 \pi
    Attached Thumbnails Attached Thumbnails Tangent circle questions-tankrs_an2krse.png  
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  3. #3
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    Quote Originally Posted by chengbin View Post
    2. In triangle ABC, AB=260, AC=400, and BC=520. Point D is chosen on BC so that the circle inscribed in triangle ABD and ADC are tangent to AD at the same point. What is the length of BD?

    We have MD=ND=RD=x, \ BM=BQ=y, \ AQ=AR=AP=z, \ CP=CN=t

    Then 2(x+y+z+t)=AB+AC+BC=1180\Rightarrow x+y+z+t=590

    But x+y=BD, \ z+t=AC=400\Rightarrow BD+400=590\Rightarrow BD=190
    Attached Thumbnails Attached Thumbnails Tangent circle questions-circle.jpg  
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  4. #4
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    Quote Originally Posted by earboth View Post
    1. Draw a rough sketch (see attachment)

    2. Calculate the length of r using Pythagorean theorem:

    (12-r)^2+6^2=(6+r)^2~\implies~r=4

    3. Therefore the circumference is c = 2\cdot \pi \cdot r = 8 \pi
    I don't get it. Do you mind showing me your logic? Thanks.
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  5. #5
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    Quote Originally Posted by chengbin View Post
    I don't get it. Do you mind showing me your logic? Thanks.
    Not sure what exactly you don't get ....

    I've marked the right triangle I use for my calculations (see attachment)

    <br />
(12-r)^2+6^2=(6+r)^2~\implies~ 144 - 24r +r^2 + 36 = 36+12r + r^2

    Collect like terms and separate the variable and the constants on different sides of the equation:

    r^2-r^2-24r-12r = -144 - 36 + 36~\implies -36r = -144~\implies~r=4<br />

    The original question asked to calculate the circumference of the tangent circle. Since you know the radius you can use the formula:

    c = 2 \cdot \pi \cdot r

    Plug in the value of r and you'll get:

    c = 2 \cdot \pi \cdot 4 = 8 \pi
    Attached Thumbnails Attached Thumbnails Tangent circle questions-tangkrs_inhalbkrs.png  
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  6. #6
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    I got it. You made a right triangle to find the length. I didn't see that before.

    Thanks.
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