Equilateral triangle on a concentric circle












2












$begingroup$


Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of tbe equilateral triangle?



My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ i will manipulate the formula afterwards,,,






geometry euclidean-geometry circle







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$endgroup$












  • $begingroup$
    What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle?
    $endgroup$
    – астон вілла олоф мэллбэрг
    4 hours ago


















2












$begingroup$


Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of tbe equilateral triangle?



My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ i will manipulate the formula afterwards,,,






geometry euclidean-geometry circle







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle?
    $endgroup$
    – астон вілла олоф мэллбэрг
    4 hours ago
















2












2








2


1



$begingroup$


Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of tbe equilateral triangle?



My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ i will manipulate the formula afterwards,,,






geometry euclidean-geometry circle







share|cite|improve this question











$endgroup$




Is my idea correct? 3 concentric circles of radius 1, 2 and 3 are given. An equilateral triangle is formed having its vertices lie on the side of the three concentric circles. What is the length of tbe equilateral triangle?



My idea is to set a point at the middle of the triangle, then use the distance of it to the vertices given that the three concentric circles are set as $$x^2 + y^2 = 1$$$$x^2 + y^2 = 4$$ and $$x^2 + y^2 = 9$$ i will manipulate the formula afterwards,,,






geometry euclidean-geometry circle




geometry euclidean-geometry circle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 4 hours ago









Michael Rozenberg

108k1895200




108k1895200










asked 4 hours ago









rosarosa

587516




587516












  • $begingroup$
    What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle?
    $endgroup$
    – астон вілла олоф мэллбэрг
    4 hours ago




















  • $begingroup$
    What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle?
    $endgroup$
    – астон вілла олоф мэллбэрг
    4 hours ago


















$begingroup$
What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle?
$endgroup$
– астон вілла олоф мэллбэрг
4 hours ago






$begingroup$
What does "the side of the concentric circles" mean? Do you mean that the three vertices lie one on each circle?
$endgroup$
– астон вілла олоф мэллбэрг
4 hours ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

Using the construction that @Michael Rozenberg suggested



enter image description here



I will leave the following exercise for you (which isn't that hard)




Prove that the quadrilateral $ABCD$ is cyclic.




Thus $angle BDC=180°-angle CAB=120°$. In virtue of the law of Cosines $$begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·cos(angle BDC)\ &=1+4-2·1·2·(-0.5)\ &=5+2=7 end{array}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice! I thought about algebraic solution only.
    $endgroup$
    – Michael Rozenberg
    2 hours ago



















2












$begingroup$

The hint.



Take $A$ on the biggest circle and rotate the smallest circle by $60^{circ}$ around $A$.



Now, take an intersection point $B$ with the middle circle.



Thus, $AB$ is a side of the needed triangle.



I took $A(-3,0)$ and got $AB=sqrt7.$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Here's the image



    I'll use the process of Dr. Mathva in a different way.



    We'll first prove $ABCD$ is cyclic.



    Let $AB=AC=BC=s$



    $DC=1, DB=2, DA=3$



    We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$



    By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.



    After this you can find out a through pure trigonometric means. I got $s = sqrt{7}$






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Using the construction that @Michael Rozenberg suggested



      enter image description here



      I will leave the following exercise for you (which isn't that hard)




      Prove that the quadrilateral $ABCD$ is cyclic.




      Thus $angle BDC=180°-angle CAB=120°$. In virtue of the law of Cosines $$begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·cos(angle BDC)\ &=1+4-2·1·2·(-0.5)\ &=5+2=7 end{array}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Nice! I thought about algebraic solution only.
        $endgroup$
        – Michael Rozenberg
        2 hours ago
















      3












      $begingroup$

      Using the construction that @Michael Rozenberg suggested



      enter image description here



      I will leave the following exercise for you (which isn't that hard)




      Prove that the quadrilateral $ABCD$ is cyclic.




      Thus $angle BDC=180°-angle CAB=120°$. In virtue of the law of Cosines $$begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·cos(angle BDC)\ &=1+4-2·1·2·(-0.5)\ &=5+2=7 end{array}$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Nice! I thought about algebraic solution only.
        $endgroup$
        – Michael Rozenberg
        2 hours ago














      3












      3








      3





      $begingroup$

      Using the construction that @Michael Rozenberg suggested



      enter image description here



      I will leave the following exercise for you (which isn't that hard)




      Prove that the quadrilateral $ABCD$ is cyclic.




      Thus $angle BDC=180°-angle CAB=120°$. In virtue of the law of Cosines $$begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·cos(angle BDC)\ &=1+4-2·1·2·(-0.5)\ &=5+2=7 end{array}$$






      share|cite|improve this answer









      $endgroup$



      Using the construction that @Michael Rozenberg suggested



      enter image description here



      I will leave the following exercise for you (which isn't that hard)




      Prove that the quadrilateral $ABCD$ is cyclic.




      Thus $angle BDC=180°-angle CAB=120°$. In virtue of the law of Cosines $$begin{array}a [CB]^2&=[CD]^2+[DB]^2-2·[CD]·[DB]·cos(angle BDC)\ &=1+4-2·1·2·(-0.5)\ &=5+2=7 end{array}$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 4 hours ago









      Dr. MathvaDr. Mathva

      2,461526




      2,461526












      • $begingroup$
        Nice! I thought about algebraic solution only.
        $endgroup$
        – Michael Rozenberg
        2 hours ago


















      • $begingroup$
        Nice! I thought about algebraic solution only.
        $endgroup$
        – Michael Rozenberg
        2 hours ago
















      $begingroup$
      Nice! I thought about algebraic solution only.
      $endgroup$
      – Michael Rozenberg
      2 hours ago




      $begingroup$
      Nice! I thought about algebraic solution only.
      $endgroup$
      – Michael Rozenberg
      2 hours ago











      2












      $begingroup$

      The hint.



      Take $A$ on the biggest circle and rotate the smallest circle by $60^{circ}$ around $A$.



      Now, take an intersection point $B$ with the middle circle.



      Thus, $AB$ is a side of the needed triangle.



      I took $A(-3,0)$ and got $AB=sqrt7.$






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        The hint.



        Take $A$ on the biggest circle and rotate the smallest circle by $60^{circ}$ around $A$.



        Now, take an intersection point $B$ with the middle circle.



        Thus, $AB$ is a side of the needed triangle.



        I took $A(-3,0)$ and got $AB=sqrt7.$






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          The hint.



          Take $A$ on the biggest circle and rotate the smallest circle by $60^{circ}$ around $A$.



          Now, take an intersection point $B$ with the middle circle.



          Thus, $AB$ is a side of the needed triangle.



          I took $A(-3,0)$ and got $AB=sqrt7.$






          share|cite|improve this answer











          $endgroup$



          The hint.



          Take $A$ on the biggest circle and rotate the smallest circle by $60^{circ}$ around $A$.



          Now, take an intersection point $B$ with the middle circle.



          Thus, $AB$ is a side of the needed triangle.



          I took $A(-3,0)$ and got $AB=sqrt7.$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          Michael RozenbergMichael Rozenberg

          108k1895200




          108k1895200























              1












              $begingroup$

              Here's the image



              I'll use the process of Dr. Mathva in a different way.



              We'll first prove $ABCD$ is cyclic.



              Let $AB=AC=BC=s$



              $DC=1, DB=2, DA=3$



              We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$



              By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.



              After this you can find out a through pure trigonometric means. I got $s = sqrt{7}$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Here's the image



                I'll use the process of Dr. Mathva in a different way.



                We'll first prove $ABCD$ is cyclic.



                Let $AB=AC=BC=s$



                $DC=1, DB=2, DA=3$



                We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$



                By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.



                After this you can find out a through pure trigonometric means. I got $s = sqrt{7}$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Here's the image



                  I'll use the process of Dr. Mathva in a different way.



                  We'll first prove $ABCD$ is cyclic.



                  Let $AB=AC=BC=s$



                  $DC=1, DB=2, DA=3$



                  We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$



                  By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.



                  After this you can find out a through pure trigonometric means. I got $s = sqrt{7}$






                  share|cite|improve this answer









                  $endgroup$



                  Here's the image



                  I'll use the process of Dr. Mathva in a different way.



                  We'll first prove $ABCD$ is cyclic.



                  Let $AB=AC=BC=s$



                  $DC=1, DB=2, DA=3$



                  We see that $AB×DC+BD×AC=1s+2s=3s=AD×BC$



                  By Converse of Ptolemy's theorem, we conclude that $ABCD$ is cyclic.



                  After this you can find out a through pure trigonometric means. I got $s = sqrt{7}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Shashwat AsthanaShashwat Asthana

                  527




                  527






























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