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 16:36, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.