Finding all possible proofs












11














I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).



I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...



The problem is the following




Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).



Proof that the following relation holds $$a=frac{8r}{5}$$




enter image description here



My proofs so far are the following




Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.



It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.




$$$$




An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).




As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.



Thanks in advance






proof-explanation euclidean-geometry circle ratio







share|cite|improve this question




















  • 1




    see math.stackexchange.com/questions/1188845/… (possible duplicate?)
    – Carmeister
    Dec 2 '18 at 22:42


















11














I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).



I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...



The problem is the following




Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).



Proof that the following relation holds $$a=frac{8r}{5}$$




enter image description here



My proofs so far are the following




Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.



It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.




$$$$




An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).




As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.



Thanks in advance






proof-explanation euclidean-geometry circle ratio







share|cite|improve this question




















  • 1




    see math.stackexchange.com/questions/1188845/… (possible duplicate?)
    – Carmeister
    Dec 2 '18 at 22:42
















11












11








11


3





I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).



I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...



The problem is the following




Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).



Proof that the following relation holds $$a=frac{8r}{5}$$




enter image description here



My proofs so far are the following




Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.



It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.




$$$$




An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).




As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.



Thanks in advance






proof-explanation euclidean-geometry circle ratio







share|cite|improve this question















I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).



I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...



The problem is the following




Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).



Proof that the following relation holds $$a=frac{8r}{5}$$




enter image description here



My proofs so far are the following




Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.



It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.




$$$$




An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).




As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.



Thanks in advance






proof-explanation euclidean-geometry circle ratio




proof-explanation euclidean-geometry circle ratio






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 2 '18 at 17:44

























asked Dec 2 '18 at 17:36









Dr. Mathva

919316




919316








  • 1




    see math.stackexchange.com/questions/1188845/… (possible duplicate?)
    – Carmeister
    Dec 2 '18 at 22:42
















  • 1




    see math.stackexchange.com/questions/1188845/… (possible duplicate?)
    – Carmeister
    Dec 2 '18 at 22:42










1




1




see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42






see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42












3 Answers
3






active

oldest

votes


















11














I find the following quite instructive:



Obviously
begin{align}
a&=r+y \
&=r+sqrt{r^2 - left(frac{a}{2}right)^2}
end{align}
and solving gives $$frac{r}{a} = frac{5}{8} , .$$



circle in square and vice versa






share|cite|improve this answer





























    8














    Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
    to deduce $a=8r/5$.






    share|cite|improve this answer





















    • It might be nice to clarify where you got $a-r$ from.
      – Servaes
      Dec 2 '18 at 20:31






    • 2




      @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
      – egreg
      Dec 2 '18 at 23:46





















    0














    Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      11














      I find the following quite instructive:



      Obviously
      begin{align}
      a&=r+y \
      &=r+sqrt{r^2 - left(frac{a}{2}right)^2}
      end{align}
      and solving gives $$frac{r}{a} = frac{5}{8} , .$$



      circle in square and vice versa






      share|cite|improve this answer


























        11














        I find the following quite instructive:



        Obviously
        begin{align}
        a&=r+y \
        &=r+sqrt{r^2 - left(frac{a}{2}right)^2}
        end{align}
        and solving gives $$frac{r}{a} = frac{5}{8} , .$$



        circle in square and vice versa






        share|cite|improve this answer
























          11












          11








          11






          I find the following quite instructive:



          Obviously
          begin{align}
          a&=r+y \
          &=r+sqrt{r^2 - left(frac{a}{2}right)^2}
          end{align}
          and solving gives $$frac{r}{a} = frac{5}{8} , .$$



          circle in square and vice versa






          share|cite|improve this answer












          I find the following quite instructive:



          Obviously
          begin{align}
          a&=r+y \
          &=r+sqrt{r^2 - left(frac{a}{2}right)^2}
          end{align}
          and solving gives $$frac{r}{a} = frac{5}{8} , .$$



          circle in square and vice versa







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 2 '18 at 18:09









          Diger

          1,5921413




          1,5921413























              8














              Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
              to deduce $a=8r/5$.






              share|cite|improve this answer





















              • It might be nice to clarify where you got $a-r$ from.
                – Servaes
                Dec 2 '18 at 20:31






              • 2




                @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
                – egreg
                Dec 2 '18 at 23:46


















              8














              Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
              to deduce $a=8r/5$.






              share|cite|improve this answer





















              • It might be nice to clarify where you got $a-r$ from.
                – Servaes
                Dec 2 '18 at 20:31






              • 2




                @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
                – egreg
                Dec 2 '18 at 23:46
















              8












              8








              8






              Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
              to deduce $a=8r/5$.






              share|cite|improve this answer












              Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
              to deduce $a=8r/5$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 2 '18 at 18:06









              BPP

              2,169927




              2,169927












              • It might be nice to clarify where you got $a-r$ from.
                – Servaes
                Dec 2 '18 at 20:31






              • 2




                @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
                – egreg
                Dec 2 '18 at 23:46




















              • It might be nice to clarify where you got $a-r$ from.
                – Servaes
                Dec 2 '18 at 20:31






              • 2




                @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
                – egreg
                Dec 2 '18 at 23:46


















              It might be nice to clarify where you got $a-r$ from.
              – Servaes
              Dec 2 '18 at 20:31




              It might be nice to clarify where you got $a-r$ from.
              – Servaes
              Dec 2 '18 at 20:31




              2




              2




              @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
              – egreg
              Dec 2 '18 at 23:46






              @Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
              – egreg
              Dec 2 '18 at 23:46













              0














              Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.






              share|cite|improve this answer


























                0














                Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.






                  share|cite|improve this answer












                  Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 5:37









                  akhmeteli

                  34616




                  34616






























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