For $A_i$ on $y=sqrt{x}$ and $B_i$ on $x$-axis, with $triangle B_{i-1}B_iA_i$ equilateral of side $ell_i$....












2












$begingroup$



Let $O$ be the origin, $A_1,A_2,A_3,ldots$ be distinct points on the curve $y=sqrt{x}$ and $B_1,B_2,B_3,cdots$ be points on the positive $X$-axis such that the triangles $OB_1A_1,B_1B_2A_2,B_2B_3A_3,ldots$ are all equilateral triangles with side lengths $l_1,l_2,l_3,cdots$ respectively. Find the value of $l_1+l_2+ldots+l_{300}$.




My Attempt:
We have $OA_1:y=tan (60^{circ})x=sqrt{3}x$ so
$$
A_1=left(frac{1}{3},frac{sqrt{3}}{3}right)
quad text{and} quad
B_1=left(frac{2}{3},0right),
l_1=frac{2}{3}.
$$



Let $B_i=(x_i,0)$ then
begin{align*}
B_iA_{i+1}:& y=sqrt{3}(x-x_i) \
iff & sqrt{3}(x-x_i)=sqrt{x} \
iff & 3x^2-x(6x_i+1)+3x_i^2=0 \
iff & x=frac{6x_i+1+sqrt{12x_i+1}}{6}.
end{align*}
and so $A_{i+1}:left(frac{6x_i+1+sqrt{12x_i+1}}{6},sqrt{frac{6x_i+1+sqrt{12x_i+1}}{6}}=frac{sqrt{3}(1+sqrt{12x_i+1})}{6}right)$ and
begin{align*}
x_{i+1}
& =2left(frac{6x_i+1+sqrt{12x_i+1}}{6}right)-x_i \
& =frac{3x_i+1+sqrt{12x_i+1}}{3}
end{align*}
begin{align*}
l_{i+1}
& = x_{i+1}-x_i \
& = frac{1+sqrt{12x_i+1}}{3} \
& = frac{1+sqrt{12(ell_1+ell_2+ldots+ell_i)+1}}{3}.
end{align*}



We will prove, that $ell_i=frac{2i}{3}$



For $i=1$ it is true.



For $i=n+1$:
begin{align*}
3 ell_{n+1}
& = 1+sqrt{12(ell_1+ell_2+ldots+ell_n)+1} \
& = 1+sqrt{8(1+2+ldots+n)+1} \
& = 1+sqrt{(4n^2+4n+1)} \
& = 2n+2.
end{align*}
so
$$
ell_{n+1}=
frac{2(n+1)}{3}.
$$
Hence
$$
ell_1+ldots+ell_{300}
=frac{2}{3} (1+ldots+300)
=30100.
$$



Am I true for this?






geometry







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    Let $O$ be the origin, $A_1,A_2,A_3,ldots$ be distinct points on the curve $y=sqrt{x}$ and $B_1,B_2,B_3,cdots$ be points on the positive $X$-axis such that the triangles $OB_1A_1,B_1B_2A_2,B_2B_3A_3,ldots$ are all equilateral triangles with side lengths $l_1,l_2,l_3,cdots$ respectively. Find the value of $l_1+l_2+ldots+l_{300}$.




    My Attempt:
    We have $OA_1:y=tan (60^{circ})x=sqrt{3}x$ so
    $$
    A_1=left(frac{1}{3},frac{sqrt{3}}{3}right)
    quad text{and} quad
    B_1=left(frac{2}{3},0right),
    l_1=frac{2}{3}.
    $$



    Let $B_i=(x_i,0)$ then
    begin{align*}
    B_iA_{i+1}:& y=sqrt{3}(x-x_i) \
    iff & sqrt{3}(x-x_i)=sqrt{x} \
    iff & 3x^2-x(6x_i+1)+3x_i^2=0 \
    iff & x=frac{6x_i+1+sqrt{12x_i+1}}{6}.
    end{align*}
    and so $A_{i+1}:left(frac{6x_i+1+sqrt{12x_i+1}}{6},sqrt{frac{6x_i+1+sqrt{12x_i+1}}{6}}=frac{sqrt{3}(1+sqrt{12x_i+1})}{6}right)$ and
    begin{align*}
    x_{i+1}
    & =2left(frac{6x_i+1+sqrt{12x_i+1}}{6}right)-x_i \
    & =frac{3x_i+1+sqrt{12x_i+1}}{3}
    end{align*}
    begin{align*}
    l_{i+1}
    & = x_{i+1}-x_i \
    & = frac{1+sqrt{12x_i+1}}{3} \
    & = frac{1+sqrt{12(ell_1+ell_2+ldots+ell_i)+1}}{3}.
    end{align*}



    We will prove, that $ell_i=frac{2i}{3}$



    For $i=1$ it is true.



    For $i=n+1$:
    begin{align*}
    3 ell_{n+1}
    & = 1+sqrt{12(ell_1+ell_2+ldots+ell_n)+1} \
    & = 1+sqrt{8(1+2+ldots+n)+1} \
    & = 1+sqrt{(4n^2+4n+1)} \
    & = 2n+2.
    end{align*}
    so
    $$
    ell_{n+1}=
    frac{2(n+1)}{3}.
    $$
    Hence
    $$
    ell_1+ldots+ell_{300}
    =frac{2}{3} (1+ldots+300)
    =30100.
    $$



    Am I true for this?






    geometry







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      Let $O$ be the origin, $A_1,A_2,A_3,ldots$ be distinct points on the curve $y=sqrt{x}$ and $B_1,B_2,B_3,cdots$ be points on the positive $X$-axis such that the triangles $OB_1A_1,B_1B_2A_2,B_2B_3A_3,ldots$ are all equilateral triangles with side lengths $l_1,l_2,l_3,cdots$ respectively. Find the value of $l_1+l_2+ldots+l_{300}$.




      My Attempt:
      We have $OA_1:y=tan (60^{circ})x=sqrt{3}x$ so
      $$
      A_1=left(frac{1}{3},frac{sqrt{3}}{3}right)
      quad text{and} quad
      B_1=left(frac{2}{3},0right),
      l_1=frac{2}{3}.
      $$



      Let $B_i=(x_i,0)$ then
      begin{align*}
      B_iA_{i+1}:& y=sqrt{3}(x-x_i) \
      iff & sqrt{3}(x-x_i)=sqrt{x} \
      iff & 3x^2-x(6x_i+1)+3x_i^2=0 \
      iff & x=frac{6x_i+1+sqrt{12x_i+1}}{6}.
      end{align*}
      and so $A_{i+1}:left(frac{6x_i+1+sqrt{12x_i+1}}{6},sqrt{frac{6x_i+1+sqrt{12x_i+1}}{6}}=frac{sqrt{3}(1+sqrt{12x_i+1})}{6}right)$ and
      begin{align*}
      x_{i+1}
      & =2left(frac{6x_i+1+sqrt{12x_i+1}}{6}right)-x_i \
      & =frac{3x_i+1+sqrt{12x_i+1}}{3}
      end{align*}
      begin{align*}
      l_{i+1}
      & = x_{i+1}-x_i \
      & = frac{1+sqrt{12x_i+1}}{3} \
      & = frac{1+sqrt{12(ell_1+ell_2+ldots+ell_i)+1}}{3}.
      end{align*}



      We will prove, that $ell_i=frac{2i}{3}$



      For $i=1$ it is true.



      For $i=n+1$:
      begin{align*}
      3 ell_{n+1}
      & = 1+sqrt{12(ell_1+ell_2+ldots+ell_n)+1} \
      & = 1+sqrt{8(1+2+ldots+n)+1} \
      & = 1+sqrt{(4n^2+4n+1)} \
      & = 2n+2.
      end{align*}
      so
      $$
      ell_{n+1}=
      frac{2(n+1)}{3}.
      $$
      Hence
      $$
      ell_1+ldots+ell_{300}
      =frac{2}{3} (1+ldots+300)
      =30100.
      $$



      Am I true for this?






      geometry







      share|cite|improve this question











      $endgroup$





      Let $O$ be the origin, $A_1,A_2,A_3,ldots$ be distinct points on the curve $y=sqrt{x}$ and $B_1,B_2,B_3,cdots$ be points on the positive $X$-axis such that the triangles $OB_1A_1,B_1B_2A_2,B_2B_3A_3,ldots$ are all equilateral triangles with side lengths $l_1,l_2,l_3,cdots$ respectively. Find the value of $l_1+l_2+ldots+l_{300}$.




      My Attempt:
      We have $OA_1:y=tan (60^{circ})x=sqrt{3}x$ so
      $$
      A_1=left(frac{1}{3},frac{sqrt{3}}{3}right)
      quad text{and} quad
      B_1=left(frac{2}{3},0right),
      l_1=frac{2}{3}.
      $$



      Let $B_i=(x_i,0)$ then
      begin{align*}
      B_iA_{i+1}:& y=sqrt{3}(x-x_i) \
      iff & sqrt{3}(x-x_i)=sqrt{x} \
      iff & 3x^2-x(6x_i+1)+3x_i^2=0 \
      iff & x=frac{6x_i+1+sqrt{12x_i+1}}{6}.
      end{align*}
      and so $A_{i+1}:left(frac{6x_i+1+sqrt{12x_i+1}}{6},sqrt{frac{6x_i+1+sqrt{12x_i+1}}{6}}=frac{sqrt{3}(1+sqrt{12x_i+1})}{6}right)$ and
      begin{align*}
      x_{i+1}
      & =2left(frac{6x_i+1+sqrt{12x_i+1}}{6}right)-x_i \
      & =frac{3x_i+1+sqrt{12x_i+1}}{3}
      end{align*}
      begin{align*}
      l_{i+1}
      & = x_{i+1}-x_i \
      & = frac{1+sqrt{12x_i+1}}{3} \
      & = frac{1+sqrt{12(ell_1+ell_2+ldots+ell_i)+1}}{3}.
      end{align*}



      We will prove, that $ell_i=frac{2i}{3}$



      For $i=1$ it is true.



      For $i=n+1$:
      begin{align*}
      3 ell_{n+1}
      & = 1+sqrt{12(ell_1+ell_2+ldots+ell_n)+1} \
      & = 1+sqrt{8(1+2+ldots+n)+1} \
      & = 1+sqrt{(4n^2+4n+1)} \
      & = 2n+2.
      end{align*}
      so
      $$
      ell_{n+1}=
      frac{2(n+1)}{3}.
      $$
      Hence
      $$
      ell_1+ldots+ell_{300}
      =frac{2}{3} (1+ldots+300)
      =30100.
      $$



      Am I true for this?






      geometry




      geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Jan 19 at 21:47









      Viktor Glombik

      1,0411527




      1,0411527










      asked Jan 19 at 9:47









      Shane Dizzy SukardyShane Dizzy Sukardy

      60819




      60819






















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          $begingroup$

          Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
          $$
          frac{y_i}{x_i-z_i}=frac{y_i}{z_{i+1}-x_i}=tanfrac{pi}{3}=sqrt{3}
          $$
          then
          $$
          z_i=y_i^2-frac{sqrt {3}y_i}{3}\
          z_{i+1}=y_i^2+frac{sqrt {3}y_i}{3}\
          l_i=frac{2sqrt{3}}{3}y_i
          $$



          Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
          $$
          y_{i+1}^2-frac{sqrt {3}y_{i+1}}{3}=y_i^2+frac{sqrt {3}y_i}{3}Rightarrow u_i^2+2u_iy_i-frac{sqrt {3}u_i}{3}=frac{2sqrt{3}}{3}y_i\
          Rightarrow u_i=-y_i+frac{sqrt 3}{6}+sqrt{y_i^2+frac{1}{12}-frac{sqrt 3}{3}y_i+frac{2sqrt{3}}{3}y_i}=-y_i+frac{sqrt 3}{6}+y_i+frac{sqrt 3}{6}=frac{sqrt 3}{3}
          $$
          then set $x_0=y_0=0$ we have
          $$
          y_{i+1}=y_i+frac{sqrt 3}{3}=frac{sqrt 3}{3}(i+1)Rightarrow l_i=frac{2sqrt{3}}{3}frac{sqrt 3}{3}i=frac{2}{3}i
          $$
          and this is an arithmetic succession and you can easily find its sum.






          share|cite|improve this answer









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            $begingroup$

            Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
            $$
            frac{y_i}{x_i-z_i}=frac{y_i}{z_{i+1}-x_i}=tanfrac{pi}{3}=sqrt{3}
            $$
            then
            $$
            z_i=y_i^2-frac{sqrt {3}y_i}{3}\
            z_{i+1}=y_i^2+frac{sqrt {3}y_i}{3}\
            l_i=frac{2sqrt{3}}{3}y_i
            $$



            Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
            $$
            y_{i+1}^2-frac{sqrt {3}y_{i+1}}{3}=y_i^2+frac{sqrt {3}y_i}{3}Rightarrow u_i^2+2u_iy_i-frac{sqrt {3}u_i}{3}=frac{2sqrt{3}}{3}y_i\
            Rightarrow u_i=-y_i+frac{sqrt 3}{6}+sqrt{y_i^2+frac{1}{12}-frac{sqrt 3}{3}y_i+frac{2sqrt{3}}{3}y_i}=-y_i+frac{sqrt 3}{6}+y_i+frac{sqrt 3}{6}=frac{sqrt 3}{3}
            $$
            then set $x_0=y_0=0$ we have
            $$
            y_{i+1}=y_i+frac{sqrt 3}{3}=frac{sqrt 3}{3}(i+1)Rightarrow l_i=frac{2sqrt{3}}{3}frac{sqrt 3}{3}i=frac{2}{3}i
            $$
            and this is an arithmetic succession and you can easily find its sum.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
              $$
              frac{y_i}{x_i-z_i}=frac{y_i}{z_{i+1}-x_i}=tanfrac{pi}{3}=sqrt{3}
              $$
              then
              $$
              z_i=y_i^2-frac{sqrt {3}y_i}{3}\
              z_{i+1}=y_i^2+frac{sqrt {3}y_i}{3}\
              l_i=frac{2sqrt{3}}{3}y_i
              $$



              Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
              $$
              y_{i+1}^2-frac{sqrt {3}y_{i+1}}{3}=y_i^2+frac{sqrt {3}y_i}{3}Rightarrow u_i^2+2u_iy_i-frac{sqrt {3}u_i}{3}=frac{2sqrt{3}}{3}y_i\
              Rightarrow u_i=-y_i+frac{sqrt 3}{6}+sqrt{y_i^2+frac{1}{12}-frac{sqrt 3}{3}y_i+frac{2sqrt{3}}{3}y_i}=-y_i+frac{sqrt 3}{6}+y_i+frac{sqrt 3}{6}=frac{sqrt 3}{3}
              $$
              then set $x_0=y_0=0$ we have
              $$
              y_{i+1}=y_i+frac{sqrt 3}{3}=frac{sqrt 3}{3}(i+1)Rightarrow l_i=frac{2sqrt{3}}{3}frac{sqrt 3}{3}i=frac{2}{3}i
              $$
              and this is an arithmetic succession and you can easily find its sum.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
                $$
                frac{y_i}{x_i-z_i}=frac{y_i}{z_{i+1}-x_i}=tanfrac{pi}{3}=sqrt{3}
                $$
                then
                $$
                z_i=y_i^2-frac{sqrt {3}y_i}{3}\
                z_{i+1}=y_i^2+frac{sqrt {3}y_i}{3}\
                l_i=frac{2sqrt{3}}{3}y_i
                $$



                Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
                $$
                y_{i+1}^2-frac{sqrt {3}y_{i+1}}{3}=y_i^2+frac{sqrt {3}y_i}{3}Rightarrow u_i^2+2u_iy_i-frac{sqrt {3}u_i}{3}=frac{2sqrt{3}}{3}y_i\
                Rightarrow u_i=-y_i+frac{sqrt 3}{6}+sqrt{y_i^2+frac{1}{12}-frac{sqrt 3}{3}y_i+frac{2sqrt{3}}{3}y_i}=-y_i+frac{sqrt 3}{6}+y_i+frac{sqrt 3}{6}=frac{sqrt 3}{3}
                $$
                then set $x_0=y_0=0$ we have
                $$
                y_{i+1}=y_i+frac{sqrt 3}{3}=frac{sqrt 3}{3}(i+1)Rightarrow l_i=frac{2sqrt{3}}{3}frac{sqrt 3}{3}i=frac{2}{3}i
                $$
                and this is an arithmetic succession and you can easily find its sum.






                share|cite|improve this answer









                $endgroup$



                Let $A_i=(x_i, y_i)$ and $B_i=(z_i, 0)$ then because $y_i>0$, $x_i-z_i>0$ and $z_{i+1}-x_i>0$ we must have
                $$
                frac{y_i}{x_i-z_i}=frac{y_i}{z_{i+1}-x_i}=tanfrac{pi}{3}=sqrt{3}
                $$
                then
                $$
                z_i=y_i^2-frac{sqrt {3}y_i}{3}\
                z_{i+1}=y_i^2+frac{sqrt {3}y_i}{3}\
                l_i=frac{2sqrt{3}}{3}y_i
                $$



                Confronting them let $y_{i+1}=y_i+u_i$ with $u_i>0$ it follows
                $$
                y_{i+1}^2-frac{sqrt {3}y_{i+1}}{3}=y_i^2+frac{sqrt {3}y_i}{3}Rightarrow u_i^2+2u_iy_i-frac{sqrt {3}u_i}{3}=frac{2sqrt{3}}{3}y_i\
                Rightarrow u_i=-y_i+frac{sqrt 3}{6}+sqrt{y_i^2+frac{1}{12}-frac{sqrt 3}{3}y_i+frac{2sqrt{3}}{3}y_i}=-y_i+frac{sqrt 3}{6}+y_i+frac{sqrt 3}{6}=frac{sqrt 3}{3}
                $$
                then set $x_0=y_0=0$ we have
                $$
                y_{i+1}=y_i+frac{sqrt 3}{3}=frac{sqrt 3}{3}(i+1)Rightarrow l_i=frac{2sqrt{3}}{3}frac{sqrt 3}{3}i=frac{2}{3}i
                $$
                and this is an arithmetic succession and you can easily find its sum.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 at 10:50









                P De DonatoP De Donato

                4897




                4897






























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                    Npm cannot find a required file even through it is in the searched directory

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                    up vote 0 down vote favorite I got that error: users-Air:pdfuploader user$ npm run start client@0.1.0 start /Users/user/Documents/pdfuploader react-scripts start Could not find a required file. Name: index.html Searched in: /Users/user/Documents/pdfuploader/public npm ERR! code ELIFECYCLE npm ERR! errno 1 npm ERR! client@0.1.0 start: `react-scripts start` npm ERR! Exit status 1 npm ERR! npm ERR! Failed at the client@0.1.0 start script. npm ERR! This is probably not a problem with npm. There is likely additional logging output above. npm ERR! A complete log of this run can be found in: npm ERR! /Users/user/.npm/_logs/2018-11-19T10_19_14_555Z-debug.log users-Air:pdfuploader user$ And the file is in /Users/user/Documents/pdfuploader/public as you can see in my github repository: https://github.c...
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