Valahia University of Targoviste
Question
Asked 4 September 2024
I have a new interesting geometric problem. Do you have any ideas on how to solve it?
Given:
In an isosceles triangle, the lengths of the two equal sides are each 1, and the base of the triangle is m.
A circle is circumscribed around the triangle.
Find the chord of the circle that intersects the two equal sides of the triangle and is divided into three equal segments by the points of intersection.

Most recent 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<=1/2 i.e. cosA>=3/4 is necessary! Very restrictive and cosA>=3/4 an weird limitation for angle A.
Liudmyla Hetmanenko A nice and interesting problem!
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Popular answers (1)
Pidstryhach Institute for Applied Problems of Mechanics and Mathematics
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}.
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All Answers (6)
Valahia University of Targoviste
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!
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Pidstryhach Institute for Applied Problems of Mechanics and Mathematics
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}.
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Valahia University of Targoviste
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) !?
Valahia University of Targoviste
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 !
Pidstryhach Institute for Applied Problems of Mechanics and Mathematics
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.
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Valahia University of Targoviste
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<=1/2 i.e. cosA>=3/4 is necessary! Very restrictive and cosA>=3/4 an weird limitation for angle A.
Liudmyla Hetmanenko A nice and interesting problem!
3 Recommendations
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