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RMS Rigor (Posted on 2014-11-04) Difficulty: 3 of 5
A < B < C < D are four positive integers such that:

(i) B is the arithmetic mean of A and C and:
(ii) C is the rms of B and D, and:
(iii) D-A = 50

Determine all possible quadruplets satisfying the given conditions and prove that there are no others.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Almost full solution | Comment 2 of 4 |
As Charlie shows, the conditions (i),(ii),(iii) can be written a two variable relationship since C=2B-A, B²+D²=2C² and D=A+50
Combine and simplify these to get
0=7B²-8AB+A²-100A-2500
Or B = [4A±√(9A+700A+17500)]/7
The minus can be discarded since it would make B < 4A/7 < A.
[If we ignore the inequality requirement we get a weird solution: {155,5,-145,205} but a negative rms is also weird.]
Which leaves

B = [4A+√(9A+700A+17500)]/7

Searching with a calculator I quickly came upon Charlie's solution
{29,47,65,79}

I discovered something interesting I don't know how to prove:
the difference between A and B is decreasing as A increases.
For the solution given the difference is 18.  
Keep increasing A and the difference decreases slightly it isn't until A is 239 that the difference dips below 17.
We get two very near solutions if you round B:
{238, 255, 272.0009191, 288}
{239, 256, 272.9990842, 289}
As you keep going the difference between A and B appears to approach an asymptote at 16+2/3.  Meaning we won't get any near solutions or solutions, beyond those given.

If this result can be proven, then it is proven there are no others.

Edit:  I had to crack open a big calculus textbook, but I managed to do it.  Too much to enter in here though.

Edited on November 5, 2014, 2:19 pm
  Posted by Jer on 2014-11-05 13:11:15

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