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The difference in length of any two sides of an obtuse

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The difference in length of any two sides of an obtuse [#permalink] New post 25 Sep 2006, 13:01
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The difference in length of any two sides of an obtuse triangle is either 8 or 16 units. If the length of each side of the triangle is an even integer, and p is the perimeter of the triangle, then p must be one of how many distinct values?


(A) 5 (B) 6 (C) 7 (D) 8 (E) more than 8

Last edited by kevincan on 25 Sep 2006, 13:22, edited 1 time in total.
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 [#permalink] New post 25 Sep 2006, 13:08
10, 2
12,4
16,8
18,2
20,4
22,6
24,8
.....implies more than 8 possible values for 2 sides...so perimeter also can be more than 8 distinct values.
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 [#permalink] New post 25 Sep 2006, 20:43
chanmat wrote:
10, 2
12,4
16,8
18,2
20,4
22,6
24,8
.....implies more than 8 possible values for 2 sides...so perimeter also can be more than 8 distinct values.


Hey chanmat, if two sides are 2 and 10 the other side must be in between 8 and 12. In that case the difference of any two sides will not be either 8 or 16.
But anyway the answer turns out to be more than 8.

Folks, this is simple.....
Look at this......
Since it is given that the difference of any two sides is either 8 or 16 obviously the difference between the min and max side must be 16 and the difference between other possible sides must be 8.

It is also given that the sides are even.
Remember that the three sides should satisfy the two basic properites of the triangle ie.
1. Sum of any two sides must be greater than the other side
2. Differecne of any two sides must be less than the other side.

Now if we start with a,b,c as the three sides (Take a<b<c)
Now what i mean in the above explanation is
c-a=16
b-a=8
c-b=8
ie b must be the average of a and c

Then clearly c - a=16
Now let's take the least value of a ie 2 here. (remeber the sides are even)

2, , 18
Now with this combination we cannot get middle term as 9.
but 2,9,18 cannot be the sides of a triangle even though they satisfy the given condition that difference of any two sides is 8 or 16.

Proceeding along the same lines
we cannot have the smallest side as 4,6 and 8.

So the smallest side possible is 10. ie

10,18,26
12,20,28
14,22,30
16,24,32
18,26,34
20,28,36
22,30,38
24,32,40
26,34,42
............
............
............
............

Hence more than 8 are possible
_________________

Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)


Last edited by cicerone on 25 Sep 2008, 00:18, edited 1 time in total.
  [#permalink] 25 Sep 2006, 20:43
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