Difference between revisions of "2020 AIME I Problems/Problem 14"
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== Solution == | == Solution == | ||
+ | Either <math>P(3) = P(4)</math> or not. We first see that if <math>P(3) = P(4)</math> it's easy to obtain by Vieta's that <math>(a+b)^2 = 49</math>. Now, take <math>P(3) \neq P(4)</math> and WLOG <math>P(3) = P(a), P(4) = P(b)</math>. Now, consider the parabola formed by the graph of <math>P</math>. It has vertex <math>\frac{3+a}{2}</math>. Now, say that <math>P(x) = x^2 - (3+a)x + c</math>. We note <math>P(3)P(4) = c = P(3)(4 - 4a + \frac{8a - 1}{2}) \implies a = \frac{7P(3) + 1}{8}</math>. Now, we note <math>P(4) = \frac{7}{2}</math> by plugging in again. Now, it's easy to find that <math>a = -2.5, b = -3.5</math>, yielding a value of <math>36</math>. Finally, we add <math>49 + 36 = \boxed{085}</math>. ~awang11, charmander3333 | ||
==See Also== | ==See Also== |
Revision as of 16:28, 12 March 2020
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Problem
Solution
Either or not. We first see that if it's easy to obtain by Vieta's that . Now, take and WLOG . Now, consider the parabola formed by the graph of . It has vertex . Now, say that . We note . Now, we note by plugging in again. Now, it's easy to find that , yielding a value of . Finally, we add . ~awang11, charmander3333
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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