Geometric Problems: Can You Help Me Find Solutions?

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SUMMARY

This discussion focuses on solving four geometric problems involving triangles and parallelograms. The first problem asserts that if a triangle can be formed with sides an, bn, and cn for all n in natural numbers, then these triangles must be isosceles. The second problem involves finding integer side lengths of a triangle inscribed in a circle with a radius of 6.25, utilizing the Law of Sines. The third problem discusses a line that splits a triangle into two figures with equal perimeters and areas, suggesting that the inscribed circle's center lies on this line. Lastly, the fourth problem proves that the area of an octagon formed by connecting the vertices of a parallelogram to the midpoints of its sides is one-sixth of the parallelogram's area.

PREREQUISITES
  • Understanding of triangle properties and isosceles triangles
  • Familiarity with the Law of Sines and the cosine rule
  • Knowledge of geometric properties of parallelograms
  • Basic concepts of inscribed circles and their properties
NEXT STEPS
  • Study the properties of isosceles triangles and their implications in geometry
  • Learn how to apply the Law of Sines in triangle problems
  • Explore the relationship between inscribed circles and triangle bisectors
  • Investigate the area relationships in polygons, particularly in parallelograms and octagons
USEFUL FOR

Students, educators, and geometry enthusiasts seeking to deepen their understanding of geometric properties and problem-solving techniques in triangles and polygons.

Ebn_Alnafees
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Problem 1
Prove that if for {a, b, c} R and all n N there exists
a triangle with the sides an, bn and cn, then all of these triangles
are isoscoles.


Problem 2
A circle with a radius of 6.25 is circumscribed around a triangle
with the sides a, b and c. Find these sides, if {a, b, c} N.

Problem 3
A line splits a triangle into two new figures with equal perimeters
and areas. Prove that the center of the inscribed circle lies on this line.


Problem 4
The eight lines that connect the vertices of a parallelogram with the
centers of the two opposite sides form an octogon. Prove that the
octogon's area is exactly one sixth the area of the parallelogram



please guys I'm a new member here can you help me in these geometry problems?
 
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1] Why should it be isosceles?
Any three sides form a triangle if,
sum of any two sides is greater than the third side

Once such a triangle is formed i guess any integer multiple of the lengths work!
a+b<c
then na+nb<nc

2] Probably some clumsy work in the offing here,
a/sinA = b/sinB = c/sinC = 2R
find cosines using the cosine rule, convert them to sines and sub it above.
So you should find 3 bizarre equations in 3 unknowns, now the biggest trouble is solving them for integer solutions?? I am not sure if that's any easier :rolleyes:

-- AI
P.S-> The 3rd question seems interesting, must check it
 
thanx tenaliraman
 
any 1 for 3 and 4
 
problem 3
works only with equilateral triangles
 
come to think of it. doesn't work at all.
 
Ebn_Alnafees, you should give the problems a try and post as much of your work as possible here. Read the sticky thread.
Problem 3: Hint : Draw one bisector that cut the line. Call the intersection of that bisector and the line E. What do you reckon? If the line cuts 2 legs AB (at F), and AC (at G), then the bisector should go through A, so the distance between E and AB, AC is the same.
Use what the problem tells you : The two new figure have the same perimeters and area.
Hope you get it.
Viet Dao,
 
Last edited:
can you email me the answer. from how i understand the question. there's no solution. except the special cases of isosolese and equilateral triangles split congruently.
 

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