Three mutually-tangent circles have centers at given distances from each other; find each radius, and find...











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Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?










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  • What did you tried so far ?
    – servabat
    Jan 24 '15 at 13:01








  • 9




    The first time I read the title I thought it said "Solid menstruation (cycles)".
    – Pp..
    Jan 25 '15 at 14:16










  • $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    – Akiva Weinberger
    Nov 3 '15 at 17:05















up vote
6
down vote

favorite
1












Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?










share|cite|improve this question
























  • What did you tried so far ?
    – servabat
    Jan 24 '15 at 13:01








  • 9




    The first time I read the title I thought it said "Solid menstruation (cycles)".
    – Pp..
    Jan 25 '15 at 14:16










  • $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    – Akiva Weinberger
    Nov 3 '15 at 17:05













up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?










share|cite|improve this question















Three circles of different radii are tangent to each other externally. The distance between their centers are $9 cm$, $8 cm$, and $11 cm$.




  1. Find the radius of each circle.


  2. Find the area in between the three circles.




The distance between the centers of two tangent circles should be the sum of their radii. If we call the radii $r_1,r_2,r_3$ then we get a system of equations $$begin{align}r_1+r_2&=9\r_2+r_3&=8\r_3+r_1&=11end{align}$$ From this we can find $r_1,r_2,r_3$.



Let us call $O_1,O_2,O_3$ the centers of the circles. For the area between the centers we could compute it as the difference of the area of the triangle $O_1O_2O_3$ minus the three circular sectors that the circles cover. We have the area of the triangle because we have its sides $9,8,11$ and we could use Heron's formula with them. For the circular sectors we know their radii but we also need their amplitude. Ah, but the amplitude is the same as the angles of the triangle. I guess we can compute the angles, because we have the sides of the triangle and we can use The law of cosines to solve for the angles.




Is this correct?



Would I have to use the $arccos(x)$ function to get the final answer or there is a simpler way?







euclidean-geometry






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edited Oct 2 at 20:44









Blue

46.9k870147




46.9k870147










asked Jan 24 '15 at 12:59









richmond

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312












  • What did you tried so far ?
    – servabat
    Jan 24 '15 at 13:01








  • 9




    The first time I read the title I thought it said "Solid menstruation (cycles)".
    – Pp..
    Jan 25 '15 at 14:16










  • $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    – Akiva Weinberger
    Nov 3 '15 at 17:05


















  • What did you tried so far ?
    – servabat
    Jan 24 '15 at 13:01








  • 9




    The first time I read the title I thought it said "Solid menstruation (cycles)".
    – Pp..
    Jan 25 '15 at 14:16










  • $(r_1,r_2,r_3)=(6,3,5)$, by the way.
    – Akiva Weinberger
    Nov 3 '15 at 17:05
















What did you tried so far ?
– servabat
Jan 24 '15 at 13:01






What did you tried so far ?
– servabat
Jan 24 '15 at 13:01






9




9




The first time I read the title I thought it said "Solid menstruation (cycles)".
– Pp..
Jan 25 '15 at 14:16




The first time I read the title I thought it said "Solid menstruation (cycles)".
– Pp..
Jan 25 '15 at 14:16












$(r_1,r_2,r_3)=(6,3,5)$, by the way.
– Akiva Weinberger
Nov 3 '15 at 17:05




$(r_1,r_2,r_3)=(6,3,5)$, by the way.
– Akiva Weinberger
Nov 3 '15 at 17:05










3 Answers
3






active

oldest

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up vote
0
down vote













here is one way to do this:



(a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



(b) find the incenter $I$ of $ABC$



(c) the common value $AI = BI = CI$ is the radius you want.



the same can be done algebraically by finding



(a) two angles using the cosine rule



(b) use herons formula to find the area of $ABC$



(c) in-radius = $dfrac{area}{semi perimeter}$



(d) $AI = dfrac{r}{tan A/2}$






share|cite|improve this answer




























    up vote
    0
    down vote













    Mathematica solving for 3 variables.



    Clear[a, b, c];
    Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]

    (* {{a -> 6, b -> 3, c -> 5}} *)





    share|cite|improve this answer




























      up vote
      0
      down vote













      Solving for the radii, we get ${3,5,6}$.



      For the area we get
      $$
      begin{align}
      &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
      &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
      &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
      &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
      end{align}
      $$
      which is $3.05537320587455$.



      enter image description here






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        0
        down vote













        here is one way to do this:



        (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



        (b) find the incenter $I$ of $ABC$



        (c) the common value $AI = BI = CI$ is the radius you want.



        the same can be done algebraically by finding



        (a) two angles using the cosine rule



        (b) use herons formula to find the area of $ABC$



        (c) in-radius = $dfrac{area}{semi perimeter}$



        (d) $AI = dfrac{r}{tan A/2}$






        share|cite|improve this answer

























          up vote
          0
          down vote













          here is one way to do this:



          (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



          (b) find the incenter $I$ of $ABC$



          (c) the common value $AI = BI = CI$ is the radius you want.



          the same can be done algebraically by finding



          (a) two angles using the cosine rule



          (b) use herons formula to find the area of $ABC$



          (c) in-radius = $dfrac{area}{semi perimeter}$



          (d) $AI = dfrac{r}{tan A/2}$






          share|cite|improve this answer























            up vote
            0
            down vote










            up vote
            0
            down vote









            here is one way to do this:



            (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



            (b) find the incenter $I$ of $ABC$



            (c) the common value $AI = BI = CI$ is the radius you want.



            the same can be done algebraically by finding



            (a) two angles using the cosine rule



            (b) use herons formula to find the area of $ABC$



            (c) in-radius = $dfrac{area}{semi perimeter}$



            (d) $AI = dfrac{r}{tan A/2}$






            share|cite|improve this answer












            here is one way to do this:



            (a) construct a triangle $ABC$ with sides $8, 9$ and $11.$



            (b) find the incenter $I$ of $ABC$



            (c) the common value $AI = BI = CI$ is the radius you want.



            the same can be done algebraically by finding



            (a) two angles using the cosine rule



            (b) use herons formula to find the area of $ABC$



            (c) in-radius = $dfrac{area}{semi perimeter}$



            (d) $AI = dfrac{r}{tan A/2}$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 25 '15 at 16:21









            abel

            26.5k11948




            26.5k11948






















                up vote
                0
                down vote













                Mathematica solving for 3 variables.



                Clear[a, b, c];
                Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]

                (* {{a -> 6, b -> 3, c -> 5}} *)





                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  Mathematica solving for 3 variables.



                  Clear[a, b, c];
                  Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]

                  (* {{a -> 6, b -> 3, c -> 5}} *)





                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Mathematica solving for 3 variables.



                    Clear[a, b, c];
                    Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]

                    (* {{a -> 6, b -> 3, c -> 5}} *)





                    share|cite|improve this answer












                    Mathematica solving for 3 variables.



                    Clear[a, b, c];
                    Solve[9 - a == b && 8 - b == c && 11 - c == a, {a, b, c}]

                    (* {{a -> 6, b -> 3, c -> 5}} *)






                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 9 '16 at 10:57









                    Fred Kline

                    54721038




                    54721038






















                        up vote
                        0
                        down vote













                        Solving for the radii, we get ${3,5,6}$.



                        For the area we get
                        $$
                        begin{align}
                        &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
                        &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
                        &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
                        &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
                        end{align}
                        $$
                        which is $3.05537320587455$.



                        enter image description here






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          Solving for the radii, we get ${3,5,6}$.



                          For the area we get
                          $$
                          begin{align}
                          &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
                          &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
                          &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
                          &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
                          end{align}
                          $$
                          which is $3.05537320587455$.



                          enter image description here






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Solving for the radii, we get ${3,5,6}$.



                            For the area we get
                            $$
                            begin{align}
                            &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
                            &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
                            &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
                            &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
                            end{align}
                            $$
                            which is $3.05537320587455$.



                            enter image description here






                            share|cite|improve this answer












                            Solving for the radii, we get ${3,5,6}$.



                            For the area we get
                            $$
                            begin{align}
                            &sqrt{14(14-8)(14-9)(14-11)}\[9pt]
                            &-frac{(8+9-11)^2}8arccosleft(frac{8^2+9^2-11^2}{2cdot8cdot9}right)\
                            &-frac{(11+8-9)^2}8arccosleft(frac{11^2+8^2-9^2}{2cdot11cdot8}right)\
                            &-frac{(9+11-8)^2}8arccosleft(frac{9^2+11^2-8^2}{2cdot9cdot11}right)
                            end{align}
                            $$
                            which is $3.05537320587455$.



                            enter image description here







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 21 '16 at 6:11









                            robjohn

                            263k27301622




                            263k27301622






























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