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anonymous

  • one year ago

In a certain pentagon, the interior angles are a,b,c,d,e degrees where a,b,c,d,e are integers strictly less than 180. ("Strictly less than 180" means they are "less than and not equal to" 180.) If the median of the interior angles is 61 degrees and there is only one mode, then what are the degree measures of all five angles?

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  1. geerky42
    • one year ago
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    I'm sure there are more than one solution to this problem.

  2. anonymous
    • one year ago
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    @geerky42 sure you can try something other than 61,61,61,177,178 please.. With some better reasoning

  3. anonymous
    • one year ago
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    The sum of exterior angles is \(360^\circ\), which means \[360=(180-a)+\cdots+(180-b)~~\implies~~a+b+c+d+e=540\] Let's say \(a\le b\le c\le d\le e\), then \(c=61\) is the median, as given. So we want to find \(a,b,d,e\) such that \[a+b+d+e=479\] I think you meant \(178\) and \(179\) as your last two angles?

  4. anonymous
    • one year ago
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    What you have adds up to \(538\), not the required \(540\).

  5. anonymous
    • one year ago
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    @ S&G Ohh... this is just guess work hit and trial.. so I am not comfortable with this.. the Ans. can be 178 and 179 for other two angles.. but you made a good attempt which is mathematical.. no guess work.. can you please try to finish it

  6. anonymous
    • one year ago
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    I don't think there are any more possibilities other than the one you mentioned. Consider the extreme case of partitioning the angles a different way. If you take 1 from any of the smaller angles above (except the median angle) and give it to a larger one, as in \[61,61,61,178,179~~\to~~60,61,61,179,179\] this forces more than one mode. On the other hand, if you take 1 from one of the smaller angles and give it to the other one of the smaller angles, as in \[61,61,61,178,179~~\to~~60,62,61,179,179~~\to~~60,61,62,178,179\] this changes the median.

  7. anonymous
    • one year ago
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    Ok.. so the answer has to be hit and trial method..only..

  8. anonymous
    • one year ago
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    It's probably not the only way. You can make a case for there being only one solution using these links: http://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle

  9. anonymous
    • one year ago
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    thanks have seen the links/sites but for my level it is too high for now

  10. geerky42
    • one year ago
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    @danyboy9169 I was just saying that there may be more than one solution, nothing else lol... Just curious, what is your motivation on this one? So far it looks like you are only trying to bait *good* solution.

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