What is the radius of this circle?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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Polygon $ABC$ is an equilateral triangle. Three congruent semicircular arcs are drawn on the three sides and towards the interior of the triangle. The terminal points of the semicircular arcs are the three vertices of the triangle $ABC$.
Semicircle $C_1$ of diameter $AB$
Semicircle $C_2$ of diameter $BC$
Semicircle $C_3$ of diameter $CA$



A small circle $F$ is drawn such that it is tangent internally to $C_1$ and $C_3$ and externally to $C_2$



What is the radius of circle $ F$?



Here is the picture










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  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 15 at 11:20










  • how can the arcs be congruent if they have different diameters? Also, you can insert a link to a picture.
    – Vasya
    Aug 15 at 11:48











  • @FareedAF How can $F$ be internally tangent to both $C_1$ to $C_3$? I used GeoGebra to draw out what you mean.
    – Toby Mak
    Aug 15 at 11:52







  • 1




    @Vasya ABC is equilateral, so the diameters are all congruent.
    – Toby Mak
    Aug 15 at 11:53










  • @Vasya thank you for the note, I added a link
    – Fareed AF
    Aug 15 at 12:35














up vote
9
down vote

favorite
1












Polygon $ABC$ is an equilateral triangle. Three congruent semicircular arcs are drawn on the three sides and towards the interior of the triangle. The terminal points of the semicircular arcs are the three vertices of the triangle $ABC$.
Semicircle $C_1$ of diameter $AB$
Semicircle $C_2$ of diameter $BC$
Semicircle $C_3$ of diameter $CA$



A small circle $F$ is drawn such that it is tangent internally to $C_1$ and $C_3$ and externally to $C_2$



What is the radius of circle $ F$?



Here is the picture










share|cite|improve this question























  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 15 at 11:20










  • how can the arcs be congruent if they have different diameters? Also, you can insert a link to a picture.
    – Vasya
    Aug 15 at 11:48











  • @FareedAF How can $F$ be internally tangent to both $C_1$ to $C_3$? I used GeoGebra to draw out what you mean.
    – Toby Mak
    Aug 15 at 11:52







  • 1




    @Vasya ABC is equilateral, so the diameters are all congruent.
    – Toby Mak
    Aug 15 at 11:53










  • @Vasya thank you for the note, I added a link
    – Fareed AF
    Aug 15 at 12:35












up vote
9
down vote

favorite
1









up vote
9
down vote

favorite
1






1





Polygon $ABC$ is an equilateral triangle. Three congruent semicircular arcs are drawn on the three sides and towards the interior of the triangle. The terminal points of the semicircular arcs are the three vertices of the triangle $ABC$.
Semicircle $C_1$ of diameter $AB$
Semicircle $C_2$ of diameter $BC$
Semicircle $C_3$ of diameter $CA$



A small circle $F$ is drawn such that it is tangent internally to $C_1$ and $C_3$ and externally to $C_2$



What is the radius of circle $ F$?



Here is the picture










share|cite|improve this question















Polygon $ABC$ is an equilateral triangle. Three congruent semicircular arcs are drawn on the three sides and towards the interior of the triangle. The terminal points of the semicircular arcs are the three vertices of the triangle $ABC$.
Semicircle $C_1$ of diameter $AB$
Semicircle $C_2$ of diameter $BC$
Semicircle $C_3$ of diameter $CA$



A small circle $F$ is drawn such that it is tangent internally to $C_1$ and $C_3$ and externally to $C_2$



What is the radius of circle $ F$?



Here is the picture







geometry circle






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edited Aug 15 at 12:42









Deepesh Meena

3,8912825




3,8912825










asked Aug 15 at 11:16









Fareed AF

916




916











  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 15 at 11:20










  • how can the arcs be congruent if they have different diameters? Also, you can insert a link to a picture.
    – Vasya
    Aug 15 at 11:48











  • @FareedAF How can $F$ be internally tangent to both $C_1$ to $C_3$? I used GeoGebra to draw out what you mean.
    – Toby Mak
    Aug 15 at 11:52







  • 1




    @Vasya ABC is equilateral, so the diameters are all congruent.
    – Toby Mak
    Aug 15 at 11:53










  • @Vasya thank you for the note, I added a link
    – Fareed AF
    Aug 15 at 12:35
















  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Aug 15 at 11:20










  • how can the arcs be congruent if they have different diameters? Also, you can insert a link to a picture.
    – Vasya
    Aug 15 at 11:48











  • @FareedAF How can $F$ be internally tangent to both $C_1$ to $C_3$? I used GeoGebra to draw out what you mean.
    – Toby Mak
    Aug 15 at 11:52







  • 1




    @Vasya ABC is equilateral, so the diameters are all congruent.
    – Toby Mak
    Aug 15 at 11:53










  • @Vasya thank you for the note, I added a link
    – Fareed AF
    Aug 15 at 12:35















Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 15 at 11:20




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Aug 15 at 11:20












how can the arcs be congruent if they have different diameters? Also, you can insert a link to a picture.
– Vasya
Aug 15 at 11:48





how can the arcs be congruent if they have different diameters? Also, you can insert a link to a picture.
– Vasya
Aug 15 at 11:48













@FareedAF How can $F$ be internally tangent to both $C_1$ to $C_3$? I used GeoGebra to draw out what you mean.
– Toby Mak
Aug 15 at 11:52





@FareedAF How can $F$ be internally tangent to both $C_1$ to $C_3$? I used GeoGebra to draw out what you mean.
– Toby Mak
Aug 15 at 11:52





1




1




@Vasya ABC is equilateral, so the diameters are all congruent.
– Toby Mak
Aug 15 at 11:53




@Vasya ABC is equilateral, so the diameters are all congruent.
– Toby Mak
Aug 15 at 11:53












@Vasya thank you for the note, I added a link
– Fareed AF
Aug 15 at 12:35




@Vasya thank you for the note, I added a link
– Fareed AF
Aug 15 at 12:35










3 Answers
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3
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From the red triangle in the picture we infer
$$(1-r)^2=left(1over2right)^2+left(r+left(1-sqrt3over2right)right)^2 ,$$
and this leads to
$$r=3sqrt3-1over13 .$$



enter image description here






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    up vote
    3
    down vote













    enter image description here



    Using coordinates . . .



    For convenience of notation, let $h=sqrt3$.



    Let $B = (-1,0),;C=(1,0),;A=(0,h)$.



    Let $P$ be the center of the required circle.



    Then $P=(0,1+r)$, where $r$ is the unknown radius.



    Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.



    Let $M$ be the midpoint of segment $CA$.



    Then $M=bigl(largefrac12,largefrach2bigr)$.



    Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.



    Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.



    Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.



    enter image description here



    Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula
    beginalign*
    &MP^2=(1-r)^2\[4pt]
    implies;&left(smallfrac12-0right)^!2+left(smallfrach2-(1+r)right)^!!2=(1-r)^2\[4pt]
    implies;&r=frac4h-1-h^24(4-h)\[4pt]
    &phantomr=frac4sqrt3-44(4-sqrt3)\[4pt]
    &phantomr=frac3sqrt3-113approx .3227809558\[4pt]
    endalign*






    share|cite|improve this answer





























      up vote
      0
      down vote













      Although I think using coordinates is the best way to solve these types of problems, it's fun to find other solution.

      The figure below is pretty apparent and gives us the construction for the circle. Let the side of the triangle be $2R$


      Using the Law of Cosines we get $MH=Rsqrt2-sqrt3, MD=Rsqrt5-2sqrt3 $ (note that we are using $cos 105^circ=dfracsqrt2+sqrt64$). We also have $DEcdot MD=ADcdot DH$, which gives us $$fracR^2(sqrt3-1)MD^2=fracDEMDRightarrow fracsqrt3-15-2sqrt3=fracIER-IE$$
      That last equation simplified to $$IE=fracsqrt3-14-sqrt3R=frac3sqrt3-113R$$






      share|cite|improve this answer




















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

        votes









        active

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        active

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        up vote
        3
        down vote













        From the red triangle in the picture we infer
        $$(1-r)^2=left(1over2right)^2+left(r+left(1-sqrt3over2right)right)^2 ,$$
        and this leads to
        $$r=3sqrt3-1over13 .$$



        enter image description here






        share|cite|improve this answer
























          up vote
          3
          down vote













          From the red triangle in the picture we infer
          $$(1-r)^2=left(1over2right)^2+left(r+left(1-sqrt3over2right)right)^2 ,$$
          and this leads to
          $$r=3sqrt3-1over13 .$$



          enter image description here






          share|cite|improve this answer






















            up vote
            3
            down vote










            up vote
            3
            down vote









            From the red triangle in the picture we infer
            $$(1-r)^2=left(1over2right)^2+left(r+left(1-sqrt3over2right)right)^2 ,$$
            and this leads to
            $$r=3sqrt3-1over13 .$$



            enter image description here






            share|cite|improve this answer












            From the red triangle in the picture we infer
            $$(1-r)^2=left(1over2right)^2+left(r+left(1-sqrt3over2right)right)^2 ,$$
            and this leads to
            $$r=3sqrt3-1over13 .$$



            enter image description here







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Aug 15 at 15:41









            Christian Blatter

            166k7110312




            166k7110312




















                up vote
                3
                down vote













                enter image description here



                Using coordinates . . .



                For convenience of notation, let $h=sqrt3$.



                Let $B = (-1,0),;C=(1,0),;A=(0,h)$.



                Let $P$ be the center of the required circle.



                Then $P=(0,1+r)$, where $r$ is the unknown radius.



                Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.



                Let $M$ be the midpoint of segment $CA$.



                Then $M=bigl(largefrac12,largefrach2bigr)$.



                Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.



                Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.



                Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.



                enter image description here



                Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula
                beginalign*
                &MP^2=(1-r)^2\[4pt]
                implies;&left(smallfrac12-0right)^!2+left(smallfrach2-(1+r)right)^!!2=(1-r)^2\[4pt]
                implies;&r=frac4h-1-h^24(4-h)\[4pt]
                &phantomr=frac4sqrt3-44(4-sqrt3)\[4pt]
                &phantomr=frac3sqrt3-113approx .3227809558\[4pt]
                endalign*






                share|cite|improve this answer


























                  up vote
                  3
                  down vote













                  enter image description here



                  Using coordinates . . .



                  For convenience of notation, let $h=sqrt3$.



                  Let $B = (-1,0),;C=(1,0),;A=(0,h)$.



                  Let $P$ be the center of the required circle.



                  Then $P=(0,1+r)$, where $r$ is the unknown radius.



                  Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.



                  Let $M$ be the midpoint of segment $CA$.



                  Then $M=bigl(largefrac12,largefrach2bigr)$.



                  Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.



                  Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.



                  Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.



                  enter image description here



                  Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula
                  beginalign*
                  &MP^2=(1-r)^2\[4pt]
                  implies;&left(smallfrac12-0right)^!2+left(smallfrach2-(1+r)right)^!!2=(1-r)^2\[4pt]
                  implies;&r=frac4h-1-h^24(4-h)\[4pt]
                  &phantomr=frac4sqrt3-44(4-sqrt3)\[4pt]
                  &phantomr=frac3sqrt3-113approx .3227809558\[4pt]
                  endalign*






                  share|cite|improve this answer
























                    up vote
                    3
                    down vote










                    up vote
                    3
                    down vote









                    enter image description here



                    Using coordinates . . .



                    For convenience of notation, let $h=sqrt3$.



                    Let $B = (-1,0),;C=(1,0),;A=(0,h)$.



                    Let $P$ be the center of the required circle.



                    Then $P=(0,1+r)$, where $r$ is the unknown radius.



                    Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.



                    Let $M$ be the midpoint of segment $CA$.



                    Then $M=bigl(largefrac12,largefrach2bigr)$.



                    Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.



                    Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.



                    Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.



                    enter image description here



                    Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula
                    beginalign*
                    &MP^2=(1-r)^2\[4pt]
                    implies;&left(smallfrac12-0right)^!2+left(smallfrach2-(1+r)right)^!!2=(1-r)^2\[4pt]
                    implies;&r=frac4h-1-h^24(4-h)\[4pt]
                    &phantomr=frac4sqrt3-44(4-sqrt3)\[4pt]
                    &phantomr=frac3sqrt3-113approx .3227809558\[4pt]
                    endalign*






                    share|cite|improve this answer














                    enter image description here



                    Using coordinates . . .



                    For convenience of notation, let $h=sqrt3$.



                    Let $B = (-1,0),;C=(1,0),;A=(0,h)$.



                    Let $P$ be the center of the required circle.



                    Then $P=(0,1+r)$, where $r$ is the unknown radius.



                    Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.



                    Let $M$ be the midpoint of segment $CA$.



                    Then $M=bigl(largefrac12,largefrach2bigr)$.



                    Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.



                    Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.



                    Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.



                    enter image description here



                    Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula
                    beginalign*
                    &MP^2=(1-r)^2\[4pt]
                    implies;&left(smallfrac12-0right)^!2+left(smallfrach2-(1+r)right)^!!2=(1-r)^2\[4pt]
                    implies;&r=frac4h-1-h^24(4-h)\[4pt]
                    &phantomr=frac4sqrt3-44(4-sqrt3)\[4pt]
                    &phantomr=frac3sqrt3-113approx .3227809558\[4pt]
                    endalign*







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Aug 15 at 19:18

























                    answered Aug 15 at 14:21









                    quasi

                    33.9k22461




                    33.9k22461




















                        up vote
                        0
                        down vote













                        Although I think using coordinates is the best way to solve these types of problems, it's fun to find other solution.

                        The figure below is pretty apparent and gives us the construction for the circle. Let the side of the triangle be $2R$


                        Using the Law of Cosines we get $MH=Rsqrt2-sqrt3, MD=Rsqrt5-2sqrt3 $ (note that we are using $cos 105^circ=dfracsqrt2+sqrt64$). We also have $DEcdot MD=ADcdot DH$, which gives us $$fracR^2(sqrt3-1)MD^2=fracDEMDRightarrow fracsqrt3-15-2sqrt3=fracIER-IE$$
                        That last equation simplified to $$IE=fracsqrt3-14-sqrt3R=frac3sqrt3-113R$$






                        share|cite|improve this answer
























                          up vote
                          0
                          down vote













                          Although I think using coordinates is the best way to solve these types of problems, it's fun to find other solution.

                          The figure below is pretty apparent and gives us the construction for the circle. Let the side of the triangle be $2R$


                          Using the Law of Cosines we get $MH=Rsqrt2-sqrt3, MD=Rsqrt5-2sqrt3 $ (note that we are using $cos 105^circ=dfracsqrt2+sqrt64$). We also have $DEcdot MD=ADcdot DH$, which gives us $$fracR^2(sqrt3-1)MD^2=fracDEMDRightarrow fracsqrt3-15-2sqrt3=fracIER-IE$$
                          That last equation simplified to $$IE=fracsqrt3-14-sqrt3R=frac3sqrt3-113R$$






                          share|cite|improve this answer






















                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            Although I think using coordinates is the best way to solve these types of problems, it's fun to find other solution.

                            The figure below is pretty apparent and gives us the construction for the circle. Let the side of the triangle be $2R$


                            Using the Law of Cosines we get $MH=Rsqrt2-sqrt3, MD=Rsqrt5-2sqrt3 $ (note that we are using $cos 105^circ=dfracsqrt2+sqrt64$). We also have $DEcdot MD=ADcdot DH$, which gives us $$fracR^2(sqrt3-1)MD^2=fracDEMDRightarrow fracsqrt3-15-2sqrt3=fracIER-IE$$
                            That last equation simplified to $$IE=fracsqrt3-14-sqrt3R=frac3sqrt3-113R$$






                            share|cite|improve this answer












                            Although I think using coordinates is the best way to solve these types of problems, it's fun to find other solution.

                            The figure below is pretty apparent and gives us the construction for the circle. Let the side of the triangle be $2R$


                            Using the Law of Cosines we get $MH=Rsqrt2-sqrt3, MD=Rsqrt5-2sqrt3 $ (note that we are using $cos 105^circ=dfracsqrt2+sqrt64$). We also have $DEcdot MD=ADcdot DH$, which gives us $$fracR^2(sqrt3-1)MD^2=fracDEMDRightarrow fracsqrt3-15-2sqrt3=fracIER-IE$$
                            That last equation simplified to $$IE=fracsqrt3-14-sqrt3R=frac3sqrt3-113R$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Aug 15 at 15:21









                            cortek

                            7564




                            7564



























                                 

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