An Essayed Answer I
Let , , , and be positive real numbers which satisfy the system of equations above. Find the smallest possible value of .
Note: Full credit is given to the makers of this problem.
The answer is 214.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Solution. Note that the equations expand to ac + ad + bc + bd = 143, ab + ad + bc + cd = 150, ab + ac + bd + cd = 169. Adding all these equalities together yields 2(ab+ac+ad+bc+bd+cd) = 143+150+169 = 462. As a result, we have (a+b+c+d)2 =a2 +b2 +c2 +d2 +2(ab+ac+ad+bc+bd+cd)=a2 +b2 +c2 +d2 +462. Hence in order to minimize a2 +b2 +c2 +d2 it suffices to minimize a+b+c+d. To do this, note that by AM-GM on the last equation we have a+b+c+d 2 (a+d)(b+c)≤ 2 =⇒ (a+b+c+d)2 ≥4·169=676. This is in fact sufficient to guarantee the existence of a, b, c, d which satisfy all three equations. To see this, let s = a + b + c + d, and note that the original system can be written as (a + b)(s − (a + b)) = 143, (a + c)(s − (a + c)) = 150, (a + d)(s − (a + d)) = 169. These are quadratics in a + b, a + c, and a + d respectively; as a result, whenever s ≥ 26 the values of a + b, a + c, and a + d are all real. Adding these together allows one to solve for a, from which the values of the other three variables follow. (A computer simulation ensures that a, b, c, and d are all positive.) Hence, we have 676-462=214.