Difference between revisions of "2023 AIME II Problems/Problem 4"
Cardtricks (talk | contribs) m (→Solution 3: Edited title) |
Cardtricks (talk | contribs) m (→Solution 3(Quadratic Formula and Vieta's Formulas): Minor Edits) |
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~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution 3(Quadratic Formula and Vieta's Formulas)== | + | ==Solution 3 (Quadratic Formula and Vieta's Formulas)== |
We index these equations as (1), (2), and (3), respectively. | We index these equations as (1), (2), and (3), respectively. | ||
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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. | 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 <math>\{-60,-30,-20,-15,-12,-10,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,10,12,15,20,30,60\}</math> | List of possible values for y are <math>\{-60,-30,-20,-15,-12,-10,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,10,12,15,20,30,60\}</math> | ||
− | From a list of these values, the only values that work are <math>y_1 = -10, y_2 = 6, | + | From a list of these values, the only values that work are <math>y_1 = -10, y_2 = 6, y_3 = 4</math> because |
<cmath> -10 + 6 + 4 = 0</cmath> | <cmath> -10 + 6 + 4 = 0</cmath> | ||
<cmath> -10 * 6 * 4 = -240</cmath> | <cmath> -10 * 6 * 4 = -240</cmath> |
Revision as of 14:56, 17 February 2023
Contents
[hide]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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.