XYZ is a triangle with angles X, Y, Z and Y = 60 degrees prove it is an arithmetic progression

A

apple12

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Hi I am slightly stuck on this proof. So before this proof I was asked to prove that if X>Y>Z and the angles form an arithmetic progression prove Y = 60 degrees. I was able to do this, but I am slightly unsure on how to make the reverse work. So this was a method I came up with but I am not sure whether this is right. Can someone please help me clarify this?
So I approached this as follows:
Assume that X < Y < Z and they do follow an arithmetic progression then:
X = a
Y = a + d
Z = a + 2d

the sum of the angles must be 180 degrees so:
a + a + d + a + 2d = 180
3a + 3d = 180
dividing by 3
a + d = 60
since a + d = 60 is Y and that this was stated in the question, it follows that they do follow an arithmetic progression given X < Y < Z
 
apple12 said:
Hi I am slightly stuck on this proof. So before this proof I was asked to prove that if X>Y>Z and the angles form an arithmetic progression prove Y = 60 degrees. I was able to do this, but I am slightly unsure on how to make the reverse work. So this was a method I came up with but I am not sure whether this is right. Can someone please help me clarify this?
So I approached this as follows:
Assume that X < Y < Z and they do follow an arithmetic progression then:
X = a
Y = a + d
Z = a + 2d

the sum of the angles must be 180 degrees so:
a + a + d + a + 2d = 180
3a + 3d = 180
dividing by 3
a + d = 60
since a + d = 60 is Y and that this was stated in the question, it follows that they do follow an arithmetic progression given X < Y < Z
Click to expand...
Correct.

the sum of the angles must be 180 degrees so:
a + a + d + a + 2d = 180
3a + 3d = 180
dividing by 3
a + d = 60 = Y ← the middle term of a three term AP
 
apple12 said:
Hi I am slightly stuck on this proof. So before this proof I was asked to prove that if X>Y>Z and the angles form an arithmetic progression prove Y = 60 degrees. I was able to do this, but I am slightly unsure on how to make the reverse work. So this was a method I came up with but I am not sure whether this is right. Can someone please help me clarify this?
So I approached this as follows:
Assume that X < Y < Z and they do follow an arithmetic progression then:
X = a
Y = a + d
Z = a + 2d

the sum of the angles must be 180 degrees so:
a + a + d + a + 2d = 180
3a + 3d = 180
dividing by 3
a + d = 60
since a + d = 60 is Y and that this was stated in the question, it follows that they do follow an arithmetic progression given X < Y < Z
Click to expand...
Do I understand correctly that the statement you are trying to prove is:

If X,Y and Z are the angles of a triangle, X < Y < Z and Y = 60 degrees than X,Y,Z form an arithmetic progression. ?​
 
blamocur said:
Do I understand correctly that the statement you are trying to prove is:

If X,Y and Z are the angles of a triangle, X < Y < Z and Y = 60 degrees than X,Y,Z form an arithmetic progression. ?​
Click to expand...
That was the first proof the second part was given Y = 60 degrees prove XYZ form an arithmetic progression
https://nrich.maths.org/10038 : the IFFY triangles section
 
BigBeachBananas said:
Have you heard of the triangle inequality?
Click to expand...
apple12 said:
yes, I have heard of it but have never worked in applying it before
Click to expand...
It's irrelevant to this problem, which is about angles, not sides. All you really need to know about triangles is what the sum of the angles is.

Please show whatever work you have done, if you need additional help.

There is a shortcut: In an arithmetic progression with an odd number of terms, the middle term is always the average of all the terms. Since we know the sum of the terms (180), we know the average is 180/3 = 60.
 
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