Difference between revisions of "2020 AIME II Problems/Problem 11"
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==Solution 4 (Official MAA)== | ==Solution 4 (Official MAA)== | ||
Let the common root of <math>P+Q</math> and <math>P+R</math> be <math>p</math>, the common root of <math>P+Q</math> and <math>Q+R</math> be <math>q</math>, and the common root of <math>Q+R</math> and <math>P+R</math> be <math>r</math>. Because <math>p</math> and <math>q</math> are both roots of <math>P+Q</math> and <math>P+Q</math> has leading coefficient <math>2</math>, it follows that <math>P(x) + Q(x) = 2(x-p)(x-q).</math> Similarly, <math>P(x) + R(x) = 2(x-p)(x-r)</math> and <math>Q(x) + R(x) = 2(x-q)(x-r)</math>. Adding these three equations together and dividing by <math>2</math> yields<cmath>P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),</cmath>so | Let the common root of <math>P+Q</math> and <math>P+R</math> be <math>p</math>, the common root of <math>P+Q</math> and <math>Q+R</math> be <math>q</math>, and the common root of <math>Q+R</math> and <math>P+R</math> be <math>r</math>. Because <math>p</math> and <math>q</math> are both roots of <math>P+Q</math> and <math>P+Q</math> has leading coefficient <math>2</math>, it follows that <math>P(x) + Q(x) = 2(x-p)(x-q).</math> Similarly, <math>P(x) + R(x) = 2(x-p)(x-r)</math> and <math>Q(x) + R(x) = 2(x-q)(x-r)</math>. Adding these three equations together and dividing by <math>2</math> yields<cmath>P(x) + Q(x) + R(x) = (x-p)(x-q) + (x-p)(x-r) + (x-q)(x-r),</cmath>so | ||
− | + | <cmath>P(x) = (P(x) + Q(x) + R(x)) - (Q(x) + R(x))</cmath> | |
− | P(x) | + | <cmath>= (x-p)(x-q) + (x-p)(x-r) - (x-q)(x-r) </cmath> |
− | + | <cmath>= x^2 - 2px + (pq + pr - qr).</cmath> | |
− | + | Similarly, | |
− | + | <cmath>Q(x) &= x^2 - 2qx + (pq + qr - pr) \text{~ and}</cmath> | |
− | + | <cmath>R(x) &= x^2 - 2rx + (pr + qr - pq).</cmath> | |
− | Q(x) &= x^2 - 2qx + (pq + qr - pr) \text{~ and} | + | Comparing the <math>x</math> coefficients yields <math>p = \tfrac32</math>, and comparing the constant coefficients yields <math>-7 = pq + pr - qr = \tfrac32(q+r) - qr</math>. The fact that <math>Q(0) = 2</math> implies that <math>\tfrac32(q-r) + qr = 2</math>. Adding these two equations yields <math>q = -\tfrac53</math>, and so substituting back in to solve for <math>r</math> gives <math>r=-\tfrac{27}{19}</math>. Finally,<cmath>R(0) = pr + qr - pq = \left(-\frac{27}{19}\right)\left(\frac32-\frac53\right) + \frac52 = \frac{9}{38} + \frac52 = \frac{52}{19}.</cmath>The requested sum is <math>52 + 19 = 71</math>. Note that <math>Q(x) = x^2 + \frac{10}3x + 2</math> and <math>R(x) = x^2 + \frac{54}{19}x + \frac{52}{19}</math>. |
− | R(x) &= x^2 - 2rx + (pr + qr - pq). | ||
− | |||
==Video Solution== | ==Video Solution== |
Revision as of 19:34, 9 June 2020
Contents
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
Solution 4 (Official MAA)
Let the common root of and be , the common root of and be , and the common root of and be . Because and are both roots of and has leading coefficient , it follows that Similarly, and . Adding these three equations together and dividing by yieldsso Similarly,
\[Q(x) &= x^2 - 2qx + (pq + qr - pr) \text{~ and}\] (Error compiling LaTeX. Unknown error_msg)
\[R(x) &= x^2 - 2rx + (pr + qr - pq).\] (Error compiling LaTeX. Unknown error_msg)
Comparing the coefficients yields , and comparing the constant coefficients yields . The fact that implies that . Adding these two equations yields , and so substituting back in to solve for gives . Finally,The requested sum is . Note that and .
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 |
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