Problem:
Let a,b,c, and d be real numbers that satisfy the system of equations
a+bab+bc+caabc+bcd+cda+dababcd=−3=−4=14=30
There exist relatively prime positive integers m and n such that
a2+b2+c2+d2=nm
Find m+n.
Solution:
We are asked to find the value of a2+b2+c2+d2 in the form nm, where m and n are relatively prime positive integers, and then compute m+n.
From the fourth equation, we can express d as d=abc30. Substituting this into the third equation gives abc+abc30(ab+bc+ca). Using the second equation, where ab+bc+ca=−4, this becomes abc−abc120. Setting this equal to 14 gives the equation abc−abc120=14. Multiplying both sides by abc leads to the quadratic (abc)2−14(abc)−120=0, whose solutions are abc=−6 or abc=20.
To determine which value is valid, we return to the first two equations. From a+b=−3, we can rewrite ab+bc+ca as ab+c(a+b)=ab−3c. Setting this equal to −4 gives ab=3c−4.
Now, if abc=−6, then substituting for ab gives c(3c−4)=−6. This leads to the quadratic 3c2−4c+6=0, which has no real roots. So we discard abc=−6.
Thus, abc=20, and from the fourth equation, d=2030=23. Substituting into abc=20 with ab=3c−4, we get c(3c−4)=20, which simplifies to 3c2−4c−20=0. Solving this yields c=−2 or c=310.
If c=310, then ab=3c−4=10−4=6, and a+b=−3, so a and b are roots of x2+3x+6=0, which has no real solutions. Therefore, c must be −2.
Now that we know c=−2 and d=23, and since a+b=−3 and ab=−10, a and b are roots of x2+3x−10=0, which gives real solutions.
Finally, compute a2+b2+c2+d2=a2+b2+(−2)2+(23)2=(a+b)2−2ab+4+49=9+20+4+49=33+49=4141. So the answer is 145.
The problems on this page are the property of the MAA's American Mathematics Competitions