Prove that one of the interior angles must be 108

[bumblebee123]

can anyone help me with this?

question: the sizes of the interior angles of a non-regular pentagon ABCDE form an arithmetic sequence. Prove that one of the interior angles of ABCDE must be 108 degrees

here's what I know:

all interior angles of a pentagon add up to 540 degrees.

Sn = n/2 [ 2a + ( n - 1)d ]

S5 = 2.5 [ 2a +4d ]

S5 = 5a + 10d

540 = 5a + 10d

exterior angles of any polygon add up to 360 degrees

any suggestions about what to do next would be highly appreciated!

 

[Harry_the_cat]

Good start. Now rearrange 54 = 5a +10d to make 'a' the subject.

 

[bumblebee123]

Good start. Now rearrange 54 = 5a +10d to make 'a' the subject.

a =(540 - 10d ) / 5

a = 108 - 2d

does this prove that one of the interior angles must be 108?

 

[Harry_the_cat]

Well 'a' represents the first term, and 'd' is the common difference.
So, no matter what "a" and "d" are, the first term is 108-2d.
The second term is (108-2d)+d =108-d
What's the third term?

 

[bumblebee123]

Well 'a' represents the first term, and 'd' is the common difference.
So, no matter what "a" and "d" are, the first term is 108-2d.
The second term is (108-2d)+d =108-d
What's the third term?

ah,

third term = ( 108 - 2d ) + 2d = 108

thanks

 

[Harry_the_cat]

Just as a matter of interest, I would have done this problem a little differently.

Let x be the third term in the arithmetic sequence and d be the common difference.
So terms are x-2d, x-d, x, x+d, x+2d

Sum = (x-2d) +(x-d) +x + (x+d)+(x+2d) = 5x = 540

So x= 540/5 = 108
ie 108 has to be the middle term

 

[Steven G]

a =(540 - 10d ) / 5

a = 108 - 2d

does this prove that one of the interior angles must be 108?

Not if YOU don't see it!!!!

 

[Steven G]

Just as a matter of interest, I would have done this problem a little differently.

Let x be the third term in the arithmetic sequence and d be the common difference.
So terms are x-2d, x-d, x, x+d, x+2d

Sum = (x-2d) +(x-d) +x + (x+d)+(x+2d) = 5x = 540

So x= 540/5 = 108
ie 108 has to be the middle term

As an undergraduate student I worked in my college's math tutoring lab and use to show the students that trick (so there is no constant when you add the terms). One day I had an adjunct come into the lab and scream at me for showing her students that method. My attitude was clear--If you couldn't figure that out for yourself don't be upset with me. I guess that a student (from the lab) suggested a better way of doing the problem and she couldn't handle that. Oh poor adjunct I feel so badly for you.