1
$\begingroup$

Lets say I have to draw an incircle of radius R in a triangle with side lengths a,b and c.

Can I say that no side of all the possible triangles that can contain the in circle of radius R will be less than the length of R ?

For example,

I have to draw an in circle of radius 5 units. Can I say that all the triangles that can contain the in circle of Radius 5 will have no side less than 5 units?

$\endgroup$
1
$\begingroup$

actually no side can be less than or equal to $Diameter = 2*R$

when $Diameter = 2*R$ the limiting "2 infinite sides isosceles triangle" case gives the 2 sides infinite in length and parallel with base interior angles of $\frac{\pi}{2}$

just seeing the limiting case, applying geometric intuition to other cases seems to be "intuitive proof"

$\endgroup$
  • $\begingroup$ This could be improved by a proof that each side is greater than $2R.$ Or maybe just a link to a proof... $\endgroup$ – coffeemath Jan 16 '17 at 5:44
  • $\begingroup$ @coffeemath Is there such a proof available ? $\endgroup$ – ng.newbie Jan 16 '17 at 6:01
0
$\begingroup$

For any $\triangle ABC$, the centre of the incircle $I$ is at the intersection of the angle bisectors.

Therefore in $\triangle ABI$ (and similarly in $\triangle BCI$ and $\triangle CAI$), we know that $\angle AIB$ must be obtuse, since the original angles at $A$ and $B$ total to less than $180°$ therefore the half-angles $\angle ABI$ and $\angle IAB$ total to less than $90°$.

Thus the altitude from $I$ to $AB$ - which is the radius $R$ of the incircle - is less than half of $|AB|$. To see why this is true, draw a circle with $AB$ as the diameter and note that $I$ must be inside the circle (and thus, less than the radius of that $AB$-based circle from the diameter $AB$).

Therefore the radius of the incircle of any triangle is less than half the length of any of its sides.

$\endgroup$
0
$\begingroup$

Draw line $L$ through the incenter $I$ and parallel to side $AB.$ This line cuts through sides $AC,BC$ at some points $M,N.$

Now the incircle is interior to triangle $ABC,$ and so segment $MN$ contains a diameter of the incircle somewhere in it, showing the diameter of the incircle to be at most $MN.$ However one sees that $MN<AB$ since it is parallel to side $AB$ and goes through the point $I$ which is inside triangle $ABC.$

We can concude then that the diameter $2R$ of the incircle is indeed less than side $AB$ as desired. [Note the usual notation for the incircle radius is the lower case $r,$ upper case being used for circumcircle radius.]

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.