Perimeter of an equilateral triangle drawn with respect to a square.

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Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:




If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?






What I did:



Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.





By Pythagoras' Theorem,
$$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$



So perimeter of the triangle is:
$$beginalign
P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
&= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
&= frac52+sqrt5
endalign$$



However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?



Thanks for the attention.










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    up vote
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    Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:




    If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?






    What I did:



    Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.





    By Pythagoras' Theorem,
    $$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$



    So perimeter of the triangle is:
    $$beginalign
    P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
    &= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
    &= frac52+sqrt5
    endalign$$



    However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?



    Thanks for the attention.










    share|cite|improve this question























      up vote
      4
      down vote

      favorite
      3









      up vote
      4
      down vote

      favorite
      3






      3





      Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:




      If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?






      What I did:



      Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.





      By Pythagoras' Theorem,
      $$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$



      So perimeter of the triangle is:
      $$beginalign
      P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
      &= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
      &= frac52+sqrt5
      endalign$$



      However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?



      Thanks for the attention.










      share|cite|improve this question













      Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:




      If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?






      What I did:



      Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.





      By Pythagoras' Theorem,
      $$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$



      So perimeter of the triangle is:
      $$beginalign
      P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
      &= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
      &= frac52+sqrt5
      endalign$$



      However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?



      Thanks for the attention.







      geometry analytic-geometry






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      asked 4 hours ago









      Faiq Irfan

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          5 Answers
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          $M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
          $$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
          Moreover $|ED|=|MB|$ (why?).
          Can you take it from here?






          share|cite|improve this answer






















          • Wish I'd refreshed before posting my answer, I may have given too much away!
            – Tartaglia's Stutter
            4 hours ago

















          up vote
          1
          down vote













          Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.






          share|cite|improve this answer








          New contributor




          Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.

















          • Nice answer. BTW I like your username.
            – Robert Z
            4 hours ago

















          up vote
          1
          down vote













          Given a horizontal segment with length $L$ and two lines



          $$
          begincases
          y_1 = x tan(fracpi3)\
          y_2 = L-xtan(fracpi2)
          endcases
          $$



          their intersection is at



          $$
          y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
          $$



          hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$






          share|cite



























            up vote
            0
            down vote













            Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:



            $hspace3cm$enter image description here



            For the equilateral triangle:
            $$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$



            From the similarity of triangles:
            $$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$



            Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.






            share|cite|improve this answer





























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              We only need to know that the $angle ABC=30°$, the rest is just straight forward.
              enter image description here






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                5 Answers
                5






                active

                oldest

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                5 Answers
                5






                active

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                active

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                up vote
                2
                down vote













                $M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
                $$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
                Moreover $|ED|=|MB|$ (why?).
                Can you take it from here?






                share|cite|improve this answer






















                • Wish I'd refreshed before posting my answer, I may have given too much away!
                  – Tartaglia's Stutter
                  4 hours ago














                up vote
                2
                down vote













                $M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
                $$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
                Moreover $|ED|=|MB|$ (why?).
                Can you take it from here?






                share|cite|improve this answer






















                • Wish I'd refreshed before posting my answer, I may have given too much away!
                  – Tartaglia's Stutter
                  4 hours ago












                up vote
                2
                down vote










                up vote
                2
                down vote









                $M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
                $$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
                Moreover $|ED|=|MB|$ (why?).
                Can you take it from here?






                share|cite|improve this answer














                $M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
                $$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
                Moreover $|ED|=|MB|$ (why?).
                Can you take it from here?







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 4 hours ago

























                answered 4 hours ago









                Robert Z

                87.5k1056127




                87.5k1056127











                • Wish I'd refreshed before posting my answer, I may have given too much away!
                  – Tartaglia's Stutter
                  4 hours ago
















                • Wish I'd refreshed before posting my answer, I may have given too much away!
                  – Tartaglia's Stutter
                  4 hours ago















                Wish I'd refreshed before posting my answer, I may have given too much away!
                – Tartaglia's Stutter
                4 hours ago




                Wish I'd refreshed before posting my answer, I may have given too much away!
                – Tartaglia's Stutter
                4 hours ago










                up vote
                1
                down vote













                Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.






                share|cite|improve this answer








                New contributor




                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.

















                • Nice answer. BTW I like your username.
                  – Robert Z
                  4 hours ago














                up vote
                1
                down vote













                Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.






                share|cite|improve this answer








                New contributor




                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.

















                • Nice answer. BTW I like your username.
                  – Robert Z
                  4 hours ago












                up vote
                1
                down vote










                up vote
                1
                down vote









                Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.






                share|cite|improve this answer








                New contributor




                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.







                share|cite|improve this answer








                New contributor




                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer






                New contributor




                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                answered 4 hours ago









                Tartaglia's Stutter

                944




                944




                New contributor




                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                New contributor





                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.











                • Nice answer. BTW I like your username.
                  – Robert Z
                  4 hours ago
















                • Nice answer. BTW I like your username.
                  – Robert Z
                  4 hours ago















                Nice answer. BTW I like your username.
                – Robert Z
                4 hours ago




                Nice answer. BTW I like your username.
                – Robert Z
                4 hours ago










                up vote
                1
                down vote













                Given a horizontal segment with length $L$ and two lines



                $$
                begincases
                y_1 = x tan(fracpi3)\
                y_2 = L-xtan(fracpi2)
                endcases
                $$



                their intersection is at



                $$
                y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
                $$



                hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$






                share|cite
























                  up vote
                  1
                  down vote













                  Given a horizontal segment with length $L$ and two lines



                  $$
                  begincases
                  y_1 = x tan(fracpi3)\
                  y_2 = L-xtan(fracpi2)
                  endcases
                  $$



                  their intersection is at



                  $$
                  y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
                  $$



                  hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$






                  share|cite






















                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Given a horizontal segment with length $L$ and two lines



                    $$
                    begincases
                    y_1 = x tan(fracpi3)\
                    y_2 = L-xtan(fracpi2)
                    endcases
                    $$



                    their intersection is at



                    $$
                    y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
                    $$



                    hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$






                    share|cite












                    Given a horizontal segment with length $L$ and two lines



                    $$
                    begincases
                    y_1 = x tan(fracpi3)\
                    y_2 = L-xtan(fracpi2)
                    endcases
                    $$



                    their intersection is at



                    $$
                    y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
                    $$



                    hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$







                    share|cite












                    share|cite



                    share|cite










                    answered 5 mins ago









                    Cesareo

                    7,1072415




                    7,1072415




















                        up vote
                        0
                        down vote













                        Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:



                        $hspace3cm$enter image description here



                        For the equilateral triangle:
                        $$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$



                        From the similarity of triangles:
                        $$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$



                        Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:



                          $hspace3cm$enter image description here



                          For the equilateral triangle:
                          $$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$



                          From the similarity of triangles:
                          $$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$



                          Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:



                            $hspace3cm$enter image description here



                            For the equilateral triangle:
                            $$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$



                            From the similarity of triangles:
                            $$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$



                            Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.






                            share|cite|improve this answer














                            Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:



                            $hspace3cm$enter image description here



                            For the equilateral triangle:
                            $$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$



                            From the similarity of triangles:
                            $$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$



                            Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 4 hours ago

























                            answered 4 hours ago









                            farruhota

                            16.8k2735




                            16.8k2735




















                                up vote
                                0
                                down vote













                                We only need to know that the $angle ABC=30°$, the rest is just straight forward.
                                enter image description here






                                share|cite|improve this answer
























                                  up vote
                                  0
                                  down vote













                                  We only need to know that the $angle ABC=30°$, the rest is just straight forward.
                                  enter image description here






                                  share|cite|improve this answer






















                                    up vote
                                    0
                                    down vote










                                    up vote
                                    0
                                    down vote









                                    We only need to know that the $angle ABC=30°$, the rest is just straight forward.
                                    enter image description here






                                    share|cite|improve this answer












                                    We only need to know that the $angle ABC=30°$, the rest is just straight forward.
                                    enter image description here







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 1 hour ago









                                    Seyed

                                    6,11531222




                                    6,11531222



























                                         

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