Difference between revisions of "2023 AIME II Problems/Problem 4"
Cardtricks (talk | contribs) |
|||
Line 111: | Line 111: | ||
~ 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 | ||
+ | |||
+ | We index these equations as (1), (2), and (3), respectively. | ||
+ | Using equation (1), we get <math>z = \frac{60 - xy}{4) = 15 - \frac{xy}{4}</math> | ||
+ | We need to solve for x, so we plug this value of z into equation (3) to get: | ||
+ | <cmath> | ||
+ | <cmath>15x - \frac{x^2y}{4} - 4y = 60 </cmath> | ||
+ | <cmath>\frac{y}{4} * x^2 - 15x + (60 + 4y) = 0</cmath> | ||
+ | </cmath> | ||
+ | We use the quadratic formula to get possible values of x: | ||
+ | <math></math> | ||
+ | <cmath>x = \frac{15 \pm \sqrt{15^2 - 4(\frac{y}{4})(60 + 4y)}}{\frac{y}{2}}</cmath> | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2023|num-b=3|num-a=5|n=II}} | {{AIME box|year=2023|num-b=3|num-a=5|n=II}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 06:17, 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 $z = \frac{60 - xy}{4) = 15 - \frac{xy}{4}$ (Error compiling LaTeX. Unknown error_msg) We need to solve for x, so we plug this value of z into equation (3) to get: </cmath> We use the quadratic formula to get possible values of x: $$ (Error compiling LaTeX. Unknown error_msg)
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.