# Triangle determined by arithmetic and geometric sequences

## 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.

## 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

symbolipoint
Homework Helper
Gold Member
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:
Robin04
verty
Homework Helper

## 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.

## 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.

Robin04
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.

Delta2
verty
Homework Helper
Trigonometry is my favourite part of math. So thanks for the interesting question and well done for solving it.

Robin04
symbolipoint
Homework Helper
Gold Member
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).

Ray Vickson
Homework Helper
Dearly Missed
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.

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).