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 ![$4\cdot[(3)+(4)]$](http://latex.artofproblemsolving.com/5/4/2/542accac0ef0176c10be1a9dde3767a34ee23b2e.png)
from ![$5\cdot[(1)+(2)],$](http://latex.artofproblemsolving.com/2/3/a/23a492146185ed37807c070131331840a0ae111b.png)
we have ![\[b+d=5\cdot(54+54)-4\cdot(53+53)=\boxed{116}.\]](http://latex.artofproblemsolving.com/a/4/9/a4905e8f155a93ecc45101b300bb24601f7be0f3.png)
~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. 
