Prove that radius of circle $r$ exceeds $3/2$












2












$begingroup$


Let $T$ be a circle with diameter $AB$. Let $P$ be a point inside the circle such that P lies on the line $AB$. Consider the circles wit diameters $PA=6$ and $PB=4$. A fourth circle $r$ is drawn such that it is tangent to the previous three circles. Prove that the radius of $r$ exceeds $3/2$.



I assumed radius of $r$ to be $k$. Then as the line joining the centers of two circles touching each other at one point passes through the point of contact, we obtain a triangle with sides $3+a, 5$ and $2+a$ (if I havent gone wrong anywhere). But after that I dont know how to proceed. Please help.






geometry







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    $begingroup$


    Let $T$ be a circle with diameter $AB$. Let $P$ be a point inside the circle such that P lies on the line $AB$. Consider the circles wit diameters $PA=6$ and $PB=4$. A fourth circle $r$ is drawn such that it is tangent to the previous three circles. Prove that the radius of $r$ exceeds $3/2$.



    I assumed radius of $r$ to be $k$. Then as the line joining the centers of two circles touching each other at one point passes through the point of contact, we obtain a triangle with sides $3+a, 5$ and $2+a$ (if I havent gone wrong anywhere). But after that I dont know how to proceed. Please help.






    geometry







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    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Let $T$ be a circle with diameter $AB$. Let $P$ be a point inside the circle such that P lies on the line $AB$. Consider the circles wit diameters $PA=6$ and $PB=4$. A fourth circle $r$ is drawn such that it is tangent to the previous three circles. Prove that the radius of $r$ exceeds $3/2$.



      I assumed radius of $r$ to be $k$. Then as the line joining the centers of two circles touching each other at one point passes through the point of contact, we obtain a triangle with sides $3+a, 5$ and $2+a$ (if I havent gone wrong anywhere). But after that I dont know how to proceed. Please help.






      geometry







      share|cite|improve this question









      $endgroup$




      Let $T$ be a circle with diameter $AB$. Let $P$ be a point inside the circle such that P lies on the line $AB$. Consider the circles wit diameters $PA=6$ and $PB=4$. A fourth circle $r$ is drawn such that it is tangent to the previous three circles. Prove that the radius of $r$ exceeds $3/2$.



      I assumed radius of $r$ to be $k$. Then as the line joining the centers of two circles touching each other at one point passes through the point of contact, we obtain a triangle with sides $3+a, 5$ and $2+a$ (if I havent gone wrong anywhere). But after that I dont know how to proceed. Please help.






      geometry




      geometry






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      asked Jan 10 at 7:36









      YellowYellow

      16011




      16011






















          2 Answers
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          3












          $begingroup$

          Both the Soddy-Gosset Theorem and Descartes' Theorem say
          $$
          left(frac12+frac13-frac15+frac1rright)^2=2left(frac1{2^2}+frac1{3^2}+frac1{5^2}+frac1{r^2}right)
          $$
          which means that $r=frac{30}{19}gtfrac32$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Hint. Let $A=(-5,0)$ and let $B=(5,0)$. Then $P=(1,0)$ and if $(x,y)$ is the center of the fourth circle and $k$ is its radius, it follows that
            $$begin{cases}
            (x+2)^2+y^2=(k+3)^2\(x-3)^2+y^2=(k+2)^2\x^2+y^2=(5-k)^2
            end{cases}.$$
            Now you should be able to find the precise value of $k$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Ah, coordinate bashing (if I am not wrong)
              $endgroup$
              – Yellow
              Jan 10 at 8:08












            • $begingroup$
              See my edit. Note that by solving the system we get $k=30/19$.
              $endgroup$
              – Robert Z
              Jan 10 at 8:24










            • $begingroup$
              Oh wait, what do you mean by $k$ is its center?
              $endgroup$
              – Yellow
              Jan 10 at 8:37










            • $begingroup$
              Sorry $k$ is the radius. The same notation you used in your question.
              $endgroup$
              – Robert Z
              Jan 10 at 8:40













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            2 Answers
            2






            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Both the Soddy-Gosset Theorem and Descartes' Theorem say
            $$
            left(frac12+frac13-frac15+frac1rright)^2=2left(frac1{2^2}+frac1{3^2}+frac1{5^2}+frac1{r^2}right)
            $$
            which means that $r=frac{30}{19}gtfrac32$.






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Both the Soddy-Gosset Theorem and Descartes' Theorem say
              $$
              left(frac12+frac13-frac15+frac1rright)^2=2left(frac1{2^2}+frac1{3^2}+frac1{5^2}+frac1{r^2}right)
              $$
              which means that $r=frac{30}{19}gtfrac32$.






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Both the Soddy-Gosset Theorem and Descartes' Theorem say
                $$
                left(frac12+frac13-frac15+frac1rright)^2=2left(frac1{2^2}+frac1{3^2}+frac1{5^2}+frac1{r^2}right)
                $$
                which means that $r=frac{30}{19}gtfrac32$.






                share|cite|improve this answer











                $endgroup$



                Both the Soddy-Gosset Theorem and Descartes' Theorem say
                $$
                left(frac12+frac13-frac15+frac1rright)^2=2left(frac1{2^2}+frac1{3^2}+frac1{5^2}+frac1{r^2}right)
                $$
                which means that $r=frac{30}{19}gtfrac32$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 10 at 9:37

























                answered Jan 10 at 9:30









                robjohn♦robjohn

                266k27305626




                266k27305626























                    1












                    $begingroup$

                    Hint. Let $A=(-5,0)$ and let $B=(5,0)$. Then $P=(1,0)$ and if $(x,y)$ is the center of the fourth circle and $k$ is its radius, it follows that
                    $$begin{cases}
                    (x+2)^2+y^2=(k+3)^2\(x-3)^2+y^2=(k+2)^2\x^2+y^2=(5-k)^2
                    end{cases}.$$
                    Now you should be able to find the precise value of $k$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Ah, coordinate bashing (if I am not wrong)
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:08












                    • $begingroup$
                      See my edit. Note that by solving the system we get $k=30/19$.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:24










                    • $begingroup$
                      Oh wait, what do you mean by $k$ is its center?
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:37










                    • $begingroup$
                      Sorry $k$ is the radius. The same notation you used in your question.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:40


















                    1












                    $begingroup$

                    Hint. Let $A=(-5,0)$ and let $B=(5,0)$. Then $P=(1,0)$ and if $(x,y)$ is the center of the fourth circle and $k$ is its radius, it follows that
                    $$begin{cases}
                    (x+2)^2+y^2=(k+3)^2\(x-3)^2+y^2=(k+2)^2\x^2+y^2=(5-k)^2
                    end{cases}.$$
                    Now you should be able to find the precise value of $k$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Ah, coordinate bashing (if I am not wrong)
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:08












                    • $begingroup$
                      See my edit. Note that by solving the system we get $k=30/19$.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:24










                    • $begingroup$
                      Oh wait, what do you mean by $k$ is its center?
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:37










                    • $begingroup$
                      Sorry $k$ is the radius. The same notation you used in your question.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:40
















                    1












                    1








                    1





                    $begingroup$

                    Hint. Let $A=(-5,0)$ and let $B=(5,0)$. Then $P=(1,0)$ and if $(x,y)$ is the center of the fourth circle and $k$ is its radius, it follows that
                    $$begin{cases}
                    (x+2)^2+y^2=(k+3)^2\(x-3)^2+y^2=(k+2)^2\x^2+y^2=(5-k)^2
                    end{cases}.$$
                    Now you should be able to find the precise value of $k$.






                    share|cite|improve this answer











                    $endgroup$



                    Hint. Let $A=(-5,0)$ and let $B=(5,0)$. Then $P=(1,0)$ and if $(x,y)$ is the center of the fourth circle and $k$ is its radius, it follows that
                    $$begin{cases}
                    (x+2)^2+y^2=(k+3)^2\(x-3)^2+y^2=(k+2)^2\x^2+y^2=(5-k)^2
                    end{cases}.$$
                    Now you should be able to find the precise value of $k$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 10 at 8:56

























                    answered Jan 10 at 8:07









                    Robert ZRobert Z

                    94.9k1063134




                    94.9k1063134












                    • $begingroup$
                      Ah, coordinate bashing (if I am not wrong)
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:08












                    • $begingroup$
                      See my edit. Note that by solving the system we get $k=30/19$.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:24










                    • $begingroup$
                      Oh wait, what do you mean by $k$ is its center?
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:37










                    • $begingroup$
                      Sorry $k$ is the radius. The same notation you used in your question.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:40




















                    • $begingroup$
                      Ah, coordinate bashing (if I am not wrong)
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:08












                    • $begingroup$
                      See my edit. Note that by solving the system we get $k=30/19$.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:24










                    • $begingroup$
                      Oh wait, what do you mean by $k$ is its center?
                      $endgroup$
                      – Yellow
                      Jan 10 at 8:37










                    • $begingroup$
                      Sorry $k$ is the radius. The same notation you used in your question.
                      $endgroup$
                      – Robert Z
                      Jan 10 at 8:40


















                    $begingroup$
                    Ah, coordinate bashing (if I am not wrong)
                    $endgroup$
                    – Yellow
                    Jan 10 at 8:08






                    $begingroup$
                    Ah, coordinate bashing (if I am not wrong)
                    $endgroup$
                    – Yellow
                    Jan 10 at 8:08














                    $begingroup$
                    See my edit. Note that by solving the system we get $k=30/19$.
                    $endgroup$
                    – Robert Z
                    Jan 10 at 8:24




                    $begingroup$
                    See my edit. Note that by solving the system we get $k=30/19$.
                    $endgroup$
                    – Robert Z
                    Jan 10 at 8:24












                    $begingroup$
                    Oh wait, what do you mean by $k$ is its center?
                    $endgroup$
                    – Yellow
                    Jan 10 at 8:37




                    $begingroup$
                    Oh wait, what do you mean by $k$ is its center?
                    $endgroup$
                    – Yellow
                    Jan 10 at 8:37












                    $begingroup$
                    Sorry $k$ is the radius. The same notation you used in your question.
                    $endgroup$
                    – Robert Z
                    Jan 10 at 8:40






                    $begingroup$
                    Sorry $k$ is the radius. The same notation you used in your question.
                    $endgroup$
                    – Robert Z
                    Jan 10 at 8:40




















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