Difference between revisions of "2020 AIME II Problems/Problem 11"
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We could work out the system of equations, but it's pretty easy to see that <math>p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}</math>. | We could work out the system of equations, but it's pretty easy to see that <math>p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}</math>. | ||
− | <math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2} | + | <math>\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}\text{, } \text{multiplying everything together a}\text{nd taking the sqrt of both sides,}</math> |
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<cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath> | <cmath>(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)</cmath> | ||
<cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath> | <cmath>pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)} </cmath> | ||
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<cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath> | <cmath>\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}} </cmath> | ||
<math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math> | <math>\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}</math> | ||
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~quacker88 | ~quacker88 |
Revision as of 10:39, 9 June 2020
Problem
Let , and let and be two quadratic polynomials also with the coefficient of equal to . David computes each of the three sums , , and and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If , then , where and are relatively prime positive integers. Find .
Solution 1
Let and . We can write the following: Let the common root of be ; be ; and be . We then have that the roots of are , the roots of are , and the roots of are .
By Vieta's, we have:
Subtracting from , we get . Adding this to , we get . This gives us that from . Substituting these values into and , we get and . Equating these values, we get . Thus, our answer is . ~ TopNotchMath
Solution 2
Let have shared root , have shared root , and the last pair having shared root . We will now set , and . We wish to find , and now we compute . From here, we equate coefficients. This means . Now, . Finally, we know that
Solution 3
We know that .
Since , the constant term in is . Let .
Finally, let .
. Let its roots be and .
Let its roots be and .
. Let its roots be and .
By vietas,
We could work out the system of equations, but it's pretty easy to see that .
~quacker88
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.