2023 AIME II Problems/Problem 4
Contents
Problem
Let and be real numbers satisfying the system of equations Let be the set of possible values of Find the sum of the squares of the elements of
Solution 1
We first subtract the second equation from the first, noting that they both equal .
Case 1: Let .
The first and third equations simplify to: from which it is apparent that and are solutions.
Case 2: Let .
The first and third equations simplify to:
We subtract the following equations, yielding:
We thus have and , substituting in and solving yields and .
Then, we just add the squares of the solutions (make sure not to double count the ), and get ~SAHANWIJETUNGA
Solution 2
We index these equations as (1), (2), and (3), respectively. Taking , we get
Denote , , . Thus, the above equation can be equivalently written as
Similarly, by taking , we get
By taking , we get
From , we have the following two cases.
Case 1: .
Plugging this into and , we get . Thus, or .
Because we only need to compute all possible values of , without loss of generality, we only need to analyze one case that .
Plugging and into (1), we get a feasible solution , , .
Case 2: and .
Plugging this into and , we get .
Case 2.1: .
Thus, . Plugging and into (1), we get a feasible solution , , .
Case 2.2: and .
Thus, . Plugging these into (1), we get or .
Putting all cases together, . Therefore, the sum of the squares of the elements of is
~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 3 (Quadratic Formula and Vieta's Formulas)
We index these equations as (1), (2), and (3), respectively. Using equation (1), we get We need to solve for x, so we plug this value of z into equation (3) to get:
We use the quadratic formula to get possible values of x:
Here, we have two cases, (plus) and (minus) In the plus case, we have:
So, our first case gives us one value of x, which is 4. In the minus case, we have:
For this case, we now have values of x in terms of y. Plugging this expression for x in equation (1), we get
So we know that for this case, z = y. Using this information in equation (2), we get Multiplying both sides by y, we get a cubic expression: Here we just have to figure out the values of y that make this equation true. I used Vieta's Formulas to get a possible list, but you could also use the rational root theorem and synthetic division to find these. We call the three values of y that solve this equation: Using Vieta's Formulas, you get these three expressions:
In addition, we know that , because of our expression for x. Since the three values of y multiply to a negative number but also add to 0, we know that one value is negative and the other two are positive, and that the absolute value of the negative value is greater than both of the positive values. List of possible values for y are From a list of these values, the only values that work are because
Plugging in these values for y into our expression for x, we get:
So, now we have accounted for both cases, and we have 4 values of x = Squaring all these terms we get: 100 + 16 + 36 + 121 = 273, so our answer is
~Cardtricks
Solution 4 (even more Vieta's)
Since all three equations are in the form where , we can rearrange this to see that , , and all satisfy
Let this quadratic have roots and . Then, there are two cases to consider: two of , , are equal to and the third is equal to , or all of , , are equal to .
Case 1: WLOG let and .
Then by Vieta's,
which gives and . But we can swap , , and however we like, so this also gives as a solution. In total, this case yields and as possible values of .
Case 2: WLOG let .
Again, by Vieta's,
We can use to isolate in terms of , then plug that into to get that . This yields and as additional possible values of .
In all, can be any of , so the requested answer is
Solution 5
We index these equations as (1), (2), and (3) respectively (same as solution 2). There are two possible cases:
Case 1:
In this case, we simply plug in and . We note that is a valid case.
Case 2:
In this case, using equation (3), we get . Plugging that into equation (1), we get . Plugging that expression back into the original expression for we obtain . Now we plug these two expressions into equation (2): multiplying both sides by and factoring, we get: which simplifies to: we note that is a root. Factoring, we get the other roots, -10 and 11.
Our desired answer is the sum of the square of all these roots:
~Chupdogs
Video Solution by The Power of Logic(#3 and #4)
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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