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Triangle determined by arithmetic and geometric sequences

  • Thread starter Robin04
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  • #1
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Homework Statement


Determine the triangles where the sides are consecutive elements of a geometric sequence and the angles are consecutive elements of an arithmetic sequence.

Homework Equations



The Attempt at a Solution


I don't really know how to approach this problem, what the solution would look like. I suppose it should be one or more equations that link the geometric and the arithmetic sequence.
Maybe the law of cosine but I don't know what to do with the cosines there.
 

Answers and Replies

  • #2
symbolipoint
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EDIT: Not all of this is correct, since I misunderstood part of the description.

I don't quite have it either. My attempt right now is this:
Draw triangle. Points A, B, C for the angles. Sides a, b, c which are opposite from the points. Lower case for the sides and upper case for the vertices.

Accordingly with your given conditions,
c/b=b/a
and
B=A+1 and C=A+2

Angle Sum requirement should mean that A+B+C=180 degrees;
A+(A+1)+(A+2)=180
A=59

Now, since A is found and you expected as described arithmetic progression,
A=59
and
B=60
and
C=61.


Next appears to be Law Of Sines as useful.
sin(59)/a=sin(60)/b=sin(61)/c
and this chain of equalities can be cut down by one variable
if trying a=(1/c)b^2
then
sin(59)/((b^2)/c)=sin(60)/b=sin(61)/c-----------which can be split into TWO equations in TWO variables of b and c.


I have not worked further...

(See above. Some of the work is wrong.)
 
Last edited:
  • #3
verty
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Homework Statement


Determine the triangles where the sides are consecutive elements of a geometric sequence and the angles are consecutive elements of an arithmetic sequence.

Homework Equations



The Attempt at a Solution


I don't really know how to approach this problem, what the solution would look like. I suppose it should be one or more equations that link the geometric and the arithmetic sequence.
Maybe the law of cosine but I don't know what to do with the cosines there.
Yes, use the law of cosines with the one angle you know.
 
  • #4
248
16
I found it! This can only be an equilateral triangle.

First I played with the angle as symbolipoint suggested.
Let the angles be ##\alpha_1 , \alpha_2, \alpha_3##
##\alpha_1## must be given by the arithmetic sequence, and then
##\alpha_2 = \alpha_1 + d##
##\alpha_3 = \alpha_1 + 2d##

As they must add up to 180 degrees, we get that ##\alpha_2 = 60^{\circ} , and \alpha_3 = \alpha_1 + d## , which was quite suprising for me.
Then I used the law of cosines and substituted the sides with the following notation:
Sides are ##x_1, x_2, x_3##
##x_2 =x_1 \cdot q##
##x_3 = x_1 \cdot q^2##

Got the following equations:
##q^2 + q^4 -2q^3 \cos{\alpha_1} -1 = 0##
##q^4-2q^2+1 = 0##
##1+q^2-q^4-q(\cos{d} - \sqrt{3} \sin{d}) = 0##

From the second equation we get that q can be 1 and -1, substituting those into the first one we get that ##\alpha_1## is 60 degrees too, so this has to be an equilateral triangle. I don't know any method to solve the third equation for d but d=0 works fine.
 
  • #5
verty
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Trigonometry is my favourite part of math. So thanks for the interesting question and well done for solving it.
 
  • #6
symbolipoint
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I obviously misunderstood part of the description: Arithmetic Sequence instead of Consecutive Whole Numbers. Why did I do that? Otherwise my method was good(? maybe).
 
  • #7
Ray Vickson
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I found it! This can only be an equilateral triangle.

First I played with the angle as symbolipoint suggested.
Let the angles be ##\alpha_1 , \alpha_2, \alpha_3##
##\alpha_1## must be given by the arithmetic sequence, and then
##\alpha_2 = \alpha_1 + d##
##\alpha_3 = \alpha_1 + 2d##

As they must add up to 180 degrees, we get that ##\alpha_2 = 60^{\circ} , and \alpha_3 = \alpha_1 + d## , which was quite suprising for me.
Then I used the law of cosines and substituted the sides with the following notation:
Sides are ##x_1, x_2, x_3##
##x_2 =x_1 \cdot q##
##x_3 = x_1 \cdot q^2##

Got the following equations:
##q^2 + q^4 -2q^3 \cos{\alpha_1} -1 = 0##
##q^4-2q^2+1 = 0##
##1+q^2-q^4-q(\cos{d} - \sqrt{3} \sin{d}) = 0##

From the second equation we get that q can be 1 and -1, substituting those into the first one we get that ##\alpha_1## is 60 degrees too, so this has to be an equilateral triangle. I don't know any method to solve the third equation for d but d=0 works fine.
If the angles are ##\alpha_1, \alpha_1+d## and ##\alpha_1+2d##, they need to add up to ##180^{\circ},## so you need
$$3 \alpha_1 + 3d = 180 \hspace{3ex} (1)$$ Any ##\alpha_1 > 0## and ##d \geq 0## that satisfy (1) will give the angles of a valid triangle, and that means that there will be infinitely many possibilities.
 
  • #8
248
16
If the angles are ##\alpha_1, \alpha_1+d## and ##\alpha_1+2d##, they need to add up to ##180^{\circ},## so you need
$$3 \alpha_1 + 3d = 180 \hspace{3ex} (1)$$ Any ##\alpha_1 > 0## and ##d \geq 0## that satisfy (1) will give the angles of a valid triangle, and that means that there will be infinitely many possibilities.
But there's another condition: the sides should be three consecutive elements of a geometric sequence. That cannot be satisfied only by (1).
 

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