When do three disks in the plane intersect?Help reducing the following expression involving Heron's formula for the area of a triangle.Radius ratio for four packed circlesTriangle Area problemGeometric inequality $fracR_a2a+b+fracR_b2b+c+fracR_c2c+ageqfrac1sqrt3$Inequality involving inradius, exradii, sides, area, semiperimeterIf $r$ is the inradius of $triangle ABC$,then prove that $frac2r=frac1r_a+frac1r_b+frac1r_c$Three tangent circlesDetermine if 3 circles intersect at a common pointWhat is the area of the intersection of two rings?Prove that in every triangle the inequality $a^3r_a + b^3r_b + c^3r_c ge 8S(2R-r)^2 $ takes place

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When do three disks in the plane intersect?


Help reducing the following expression involving Heron's formula for the area of a triangle.Radius ratio for four packed circlesTriangle Area problemGeometric inequality $fracR_a2a+b+fracR_b2b+c+fracR_c2c+ageqfrac1sqrt3$Inequality involving inradius, exradii, sides, area, semiperimeterIf $r$ is the inradius of $triangle ABC$,then prove that $frac2r=frac1r_a+frac1r_b+frac1r_c$Three tangent circlesDetermine if 3 circles intersect at a common pointWhat is the area of the intersection of two rings?Prove that in every triangle the inequality $a^3r_a + b^3r_b + c^3r_c ge 8S(2R-r)^2 $ takes place













2












$begingroup$


Suppose $ABC$ is a triangle with $|AB|=c$, $|BC|=a$, $|CA|=b$. Suppose further that $A,B,C$ are the centers of three disks with radii $r_A,r_B,r_C$, respectively.



Is there a sensible algebraic condition (inequality?) involving $a,b,c,r_A,r_B,r_C$ equivalent to the statement "these three disks have nonempty intersection"?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Seems like you should be able to do something with the law of cosines here...
    $endgroup$
    – Brian Tung
    Mar 13 at 7:23










  • $begingroup$
    One thing is clear that $r_A+r_B+r_C < frac a+b+c 2$ from the diagram.
    $endgroup$
    – Dbchatto67
    Mar 13 at 7:25











  • $begingroup$
    @Dbchatto67 I think your inequality must be reversed
    $endgroup$
    – vidyarthi
    Mar 13 at 7:51










  • $begingroup$
    I think the concept of radical centre would be much useful here
    $endgroup$
    – vidyarthi
    Mar 14 at 12:33















2












$begingroup$


Suppose $ABC$ is a triangle with $|AB|=c$, $|BC|=a$, $|CA|=b$. Suppose further that $A,B,C$ are the centers of three disks with radii $r_A,r_B,r_C$, respectively.



Is there a sensible algebraic condition (inequality?) involving $a,b,c,r_A,r_B,r_C$ equivalent to the statement "these three disks have nonempty intersection"?










share|cite|improve this question









$endgroup$







  • 1




    $begingroup$
    Seems like you should be able to do something with the law of cosines here...
    $endgroup$
    – Brian Tung
    Mar 13 at 7:23










  • $begingroup$
    One thing is clear that $r_A+r_B+r_C < frac a+b+c 2$ from the diagram.
    $endgroup$
    – Dbchatto67
    Mar 13 at 7:25











  • $begingroup$
    @Dbchatto67 I think your inequality must be reversed
    $endgroup$
    – vidyarthi
    Mar 13 at 7:51










  • $begingroup$
    I think the concept of radical centre would be much useful here
    $endgroup$
    – vidyarthi
    Mar 14 at 12:33













2












2








2





$begingroup$


Suppose $ABC$ is a triangle with $|AB|=c$, $|BC|=a$, $|CA|=b$. Suppose further that $A,B,C$ are the centers of three disks with radii $r_A,r_B,r_C$, respectively.



Is there a sensible algebraic condition (inequality?) involving $a,b,c,r_A,r_B,r_C$ equivalent to the statement "these three disks have nonempty intersection"?










share|cite|improve this question









$endgroup$




Suppose $ABC$ is a triangle with $|AB|=c$, $|BC|=a$, $|CA|=b$. Suppose further that $A,B,C$ are the centers of three disks with radii $r_A,r_B,r_C$, respectively.



Is there a sensible algebraic condition (inequality?) involving $a,b,c,r_A,r_B,r_C$ equivalent to the statement "these three disks have nonempty intersection"?







geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 13 at 7:20









Michal AdamaszekMichal Adamaszek

2,13659




2,13659







  • 1




    $begingroup$
    Seems like you should be able to do something with the law of cosines here...
    $endgroup$
    – Brian Tung
    Mar 13 at 7:23










  • $begingroup$
    One thing is clear that $r_A+r_B+r_C < frac a+b+c 2$ from the diagram.
    $endgroup$
    – Dbchatto67
    Mar 13 at 7:25











  • $begingroup$
    @Dbchatto67 I think your inequality must be reversed
    $endgroup$
    – vidyarthi
    Mar 13 at 7:51










  • $begingroup$
    I think the concept of radical centre would be much useful here
    $endgroup$
    – vidyarthi
    Mar 14 at 12:33












  • 1




    $begingroup$
    Seems like you should be able to do something with the law of cosines here...
    $endgroup$
    – Brian Tung
    Mar 13 at 7:23










  • $begingroup$
    One thing is clear that $r_A+r_B+r_C < frac a+b+c 2$ from the diagram.
    $endgroup$
    – Dbchatto67
    Mar 13 at 7:25











  • $begingroup$
    @Dbchatto67 I think your inequality must be reversed
    $endgroup$
    – vidyarthi
    Mar 13 at 7:51










  • $begingroup$
    I think the concept of radical centre would be much useful here
    $endgroup$
    – vidyarthi
    Mar 14 at 12:33







1




1




$begingroup$
Seems like you should be able to do something with the law of cosines here...
$endgroup$
– Brian Tung
Mar 13 at 7:23




$begingroup$
Seems like you should be able to do something with the law of cosines here...
$endgroup$
– Brian Tung
Mar 13 at 7:23












$begingroup$
One thing is clear that $r_A+r_B+r_C < frac a+b+c 2$ from the diagram.
$endgroup$
– Dbchatto67
Mar 13 at 7:25





$begingroup$
One thing is clear that $r_A+r_B+r_C < frac a+b+c 2$ from the diagram.
$endgroup$
– Dbchatto67
Mar 13 at 7:25













$begingroup$
@Dbchatto67 I think your inequality must be reversed
$endgroup$
– vidyarthi
Mar 13 at 7:51




$begingroup$
@Dbchatto67 I think your inequality must be reversed
$endgroup$
– vidyarthi
Mar 13 at 7:51












$begingroup$
I think the concept of radical centre would be much useful here
$endgroup$
– vidyarthi
Mar 14 at 12:33




$begingroup$
I think the concept of radical centre would be much useful here
$endgroup$
– vidyarthi
Mar 14 at 12:33










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