Difference between revisions of "2020 AIME II Problems/Problem 11"
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Let <math>P(X) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | Let <math>P(X) = x^2 - 3x - 7</math>, and let <math>Q(x)</math> and <math>R(x)</math> be two quadratic polynomials also with the coefficient of <math>x^2</math> equal to <math>1</math>. David computes each of the three sums <math>P + Q</math>, <math>P + R</math>, and <math>Q + R</math> and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If <math>Q(0) = 2</math>, then <math>R(0) = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following: | Let <math>Q(x) = x^2 + ax + 2</math> and <math>R(x) = x^2 + bx + c</math>. We can write the following: | ||
<cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath> | <cmath>P + Q = 2x^2 + (a - 3)x - 5</cmath> | ||
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Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath | Subtracting <math>(3)</math> from <math>(1)</math>, we get <math>r - s = \dfrac{3 + b}{2}</math>. Adding this to <math>(2)</math>, we get <math>2r = 3 \implies r = \dfrac{3}{2}</math>. This gives us that <math>t = \dfrac{-5}{3}</math> from <math>(4)</math>. Substituting these values into <math>(5)</math> and <math>(6)</math>, we get <math>s = \dfrac{c-7}{3}</math> and <math>s = \dfrac{-3c - 6}{10}</math>. Equating these values, we get <math>\dfrac{c-7}{3} = \dfrac{-3c-6}{10} \implies c = \dfrac{52}{19} = R(0)</math>. Thus, our answer is <math>52 + 19 = \boxed{071}</math>. ~ TopNotchMath | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let <math>P+Q, Q+R</math> have shared root <math>q</math>, <math>Q+R, R+P</math> have shared root <math>r</math>, and the last pair having shared root <math>p</math>. We will now set <math>Q(x) = x^2+ax+2</math>, and <math>R(x) = x^2+bx+c</math>. We wish to find <math>c</math>, and now we compute <math>P+Q,Q+R,R+P</math>. | ||
+ | <cmath>P+Q = 2x^2+(a-3)x-5 = 2(x-p)(x-q)</cmath> | ||
+ | <cmath>Q+R = 2x^2+(a+b)x+(2+c) = 2(x-q)(x-r)</cmath> | ||
+ | <cmath>R+P = 2x^2+(b-3)x+(c-7) = 2(x-r)(x-p)</cmath> | ||
+ | From here, we equate coefficients. This means <math>p+q = \frac{3-a}{2}, p+r = \frac{3-b}{2}, q+r = \frac{-a-b}{2} \implies p = \frac{3}{2}</math>. Now, <math>pq = \frac{-5}{2} \implies q = -\frac{5}{3}</math>. Finally, we know that <math>pr = \frac{c-7}{2}, qr = \frac{c+2}{2} \implies c = \frac{52}{19} = R(0) \implies \boxed{071}.</math> | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/BQlab3vjjxw ~ CNCM | https://youtu.be/BQlab3vjjxw ~ CNCM | ||
==See Also== | ==See Also== |
Revision as of 18:46, 7 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
Video Solution
https://youtu.be/BQlab3vjjxw ~ CNCM