R denotes the radius of circle(ABC) and MK=KT=TN=y.

Using power of point K with respect to the circle we have

KMxKN=R^2-OK^2 => R^2-OK^2=2y^2.

Now using power of point T with respect to the circle we have

TMxTN=R^2-OT^2 => R^2-OT^2=2y^2

Therefore OK=OT.

By Heron's formula we obtain the area S of triangle ABC

S^2=(m/2+1)(1-m/2)(m/2)(m/2) => S=((1/4)m)sqrt(4-m^2) =>R=1/sqrt(4-m^2).

Now we must find OK, but at the moment I don't have any idea!

Людмила, доброго дня.

Я випадково побачив Ваше питання.

Воно просте.

Let x=AK and y=MK=KT=TN.

Then x(1-x)=y*2y=AT*(1-AT).

So either AT=x=AK or AT=1-x.

Suppose first that AT=AK.

Then the triangles ABC and ATK are similar, so y=KT=m*AT.

Therefore 2y^2=(y/m)(1-y/m), 2ym^2=m-y, and y=m/(1+2m^2).

Suppose now that AT=1-x.

Then y^2=KT^2=

AK^2+AT^2-2*AK*AT*cos A=

x^2+(1-x)^2-2x(1-x)cos A=

2x^2-2x+1-4y^2cos A=

2x(x-1)+1-4y^2cos A=

-4y^2+1-4y^2cos A.

So y^2=1/(5+4cos A).

We have m^2=1^2+1^2-2cos A, so 2cos A=2-m^2.

Therefore y^2=1/(9-2m^2).

But we have 1/4\ge x(1-x)=2y^2, so in this case

y^2\le 1/8, that is m\le 1/sqrt{2}.

Dear Alex, Dear Liudmyla,

However something is weird here! Why we don't have unicity? How this fact can be geometrically explained ?

For example in the case of an equilateral triangle ABC( m=1) !?

Alex Ravsky Liudmyla Hetmanenko Of course, from AK(1-AK) = 2y^2=AT(1-AT) it results AK=AT or AK=1-AT, but two valid cases for MN divided in 3 parts by points K on AC and T on AB in the case m=1( ABC equilateral triangle) seems to be weird !

Dinu Teodorescu, right, two different constructions of the chord MN came from two different solutions AK=AT or AK=1-AT of the equation AK(1-AK)=AT(1-AT). But for the equilateral triangle we have only the first solution, because the second solution requires m\le 1/sqrt{2}, see the end of my answer.

Alex Ravsky Sorry for not read carefully the last part of your comment! Yes, in fact a lot of triangles with lenghts 1, 1, m do not admit the second solution! To have 2 solutions, condition 2-2cosA=3/4 is necessary! Very restrictive and cosA>=3/4 an weird limitation for angle A.

Liudmyla Hetmanenko A nice and interesting problem!