Series of Squares Under Triangle

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A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.



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    A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.



    enter image description here






    summation geometric-series







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      up vote
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      down vote

      favorite









      up vote
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      down vote

      favorite











      A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.



      enter image description here






      summation geometric-series







      share|cite|improve this question















      A line crosses the $x$ and $y$ axes at $(a,0)$ and $(0,1)$ respectively, where $a>0$. Square are placed successively inside the right angled triangle thus formed. What is the area enclosed by all squares when their number goes to infinity.



      enter image description here






      summation geometric-series




      summation geometric-series






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      mwt

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          Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
          We conclude that the total area occupied by the squares is
          $$ frac2a2a+1cdot frac a2=fraca^22a+1.$$






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            1 Answer
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            Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
            We conclude that the total area occupied by the squares is
            $$ frac2a2a+1cdot frac a2=fraca^22a+1.$$






            share|cite|improve this answer
























              up vote
              5
              down vote



              accepted










              Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
              We conclude that the total area occupied by the squares is
              $$ frac2a2a+1cdot frac a2=fraca^22a+1.$$






              share|cite|improve this answer






















                up vote
                5
                down vote



                accepted







                up vote
                5
                down vote



                accepted






                Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
                We conclude that the total area occupied by the squares is
                $$ frac2a2a+1cdot frac a2=fraca^22a+1.$$






                share|cite|improve this answer












                Each square of side length $s$ has as "hat" a right triangle with horizontal leg $s$ and vertical leg $frac sa$. Within a single instance of "square + hat", the square occupies a share of $fracs^2s^2+frac12cdot scdot frac sa=frac11+frac12a=frac2a2a+1$. This proportion remains constant, no matter how many squares with hat we consider, hence also in the limit.
                We conclude that the total area occupied by the squares is
                $$ frac2a2a+1cdot frac a2=fraca^22a+1.$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 3 hours ago









                Hagen von Eitzen

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