Difference between revisions of "2022 AIME I Problems/Problem 1"
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==Solution== | ==Solution== | ||
− | <b> | + | Let |
+ | <cmath>\begin{align*} | ||
+ | P(x) &= 2x^2 + ax + b, \\ | ||
+ | Q(x) &= -2x^2 + cx + d, | ||
+ | \end{align*}</cmath> | ||
+ | for some constants <math>a,b,c</math> and <math>d.</math> | ||
+ | |||
+ | We are given that | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | P(16) &= 512 + 16a + b &&= 54, \hspace{20mm}&&(1) \\ | ||
+ | Q(16) &= -512 + 16c + d &&= 54, \hspace{20mm}&&(2) \\ | ||
+ | P(20) &= 800 + 20a + b &&= 53, \hspace{20mm}&&(3) \\ | ||
+ | Q(20) &= -800 + 20c + d &&= 53, \hspace{20mm}&&(4) | ||
+ | \end{alignat*}</cmath> | ||
+ | and we wish to find <math>P(0)+Q(0)=b+d.</math> | ||
+ | |||
+ | Subtracting <math>4\cdot[(3)+(4)]</math> from <math>5\cdot[(1)+(2)],</math> we have <cmath>b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.</cmath> | ||
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 15:36, 17 February 2022
Problem 1
Quadratic polynomials and have leading coefficients and respectively. The graphs of both polynomials pass through the two points and Find
Solution
Let for some constants and
We are given that and we wish to find
Subtracting from we have
~MRENTHUSIASM
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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