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) DA = 50
Determine all possible quadruplets satisfying the given conditions and prove that there are no others.
As Charlie shows, the conditions (i),(ii),(iii) can be written a two variable relationship since C=2BA, B²+D²=2C² and D=A+50
Combine and simplify these to get
0=7B²8AB+A²100A2500
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 20141105 13:11:15 