Difference between revisions of "2010 AMC 12B Problems/Problem 23"
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x^2 - (c+d)x + cd &= a, b | x^2 - (c+d)x + cd &= a, b | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | The two possible equations are then <math>x^2 - (c+d)x + cd-a=0</math> and <math>x^2 - (c+d)x + cd-b=0</math>. The solutions are <math>-23, -21, -17, -15</math>. From Vieta's we know that the total sum <math>2(c+d) = -76 \implies c+d = -38</math> so the roots are paired <math>-23, -15</math> and <math>-21, -17</math>. | + | The two possible equations are then <math>x^2 - (c+d)x + cd-a=0</math> and <math>x^2 - (c+d)x + cd-b=0</math>. The solutions are <math>-23, -21, -17, -15</math>. From Vieta's we know that the total sum <math>2(c+d) = -76 \implies c+d = -38</math> so the roots are paired <math>-23, -15</math> and <math>-21, -17</math>. Then, <math>cd - a = 23*15</math> and <math>cd - b = 21*17</math>. |
We can similarly get that <math>ab - c = 59*49</math> and <math>ab - d = 57*51</math>, and <math>a+b = -108</math>. Add the first two equations to get <cmath>2cd - (a+b) = 23*15 + 21*17 \implies cd = \frac{23*15+21*17 - 108}{2} = 297.</cmath> This means <math>Q(x) = x^2 + 38x + 297</math>. | We can similarly get that <math>ab - c = 59*49</math> and <math>ab - d = 57*51</math>, and <math>a+b = -108</math>. Add the first two equations to get <cmath>2cd - (a+b) = 23*15 + 21*17 \implies cd = \frac{23*15+21*17 - 108}{2} = 297.</cmath> This means <math>Q(x) = x^2 + 38x + 297</math>. |
Latest revision as of 16:29, 30 June 2024
Contents
[hide]Problem
Monic quadratic polynomial and have the property that has zeros at and , and has zeros at and . What is the sum of the minimum values of and ?
Solution 1
. Notice that has roots , so that the roots of are the roots of . For each individual equation, the sum of the roots will be (symmetry or Vieta's). Thus, we have , or . Doing something similar for gives us . We now have . Since is monic, the roots of are "farther" from the axis of symmetry than the roots of . Thus, we have , or . Adding these gives us , or . Plugging this into , we get . The minimum value of is , and the minimum value of is . Thus, our answer is , or answer .
Solution 2 (Bash)
Let and .
Then is , which simplifies to:
We can find by simply doing and to get:
The sum of the zeros of is . From Vieta, the sum is . Therefore, .
The sum of the zeros of is . From Vieta, the sum is . Therefore, .
Plugging in, we get:
Let's tackle the coefficients, which is the sum of the six double-products possible. Since gives the sum of these six double products of the roots of , we have:
Similarly with , we get:
Thus, our polynomials are and .
The minimum value of happens at , and is .
The minimum value of happens at , and is .
The sum of these minimums is . -srisainandan6
Solution 3 (Mild Bash)
Let and . Notice that the roots of are and the roots of are Then we get:
The two possible equations are then and . The solutions are . From Vieta's we know that the total sum so the roots are paired and . Then, and .
We can similarly get that and , and . Add the first two equations to get This means .
Once more, we can similarly obtain Therefore .
Now we can find the minimums to be and Summing, the answer is
~Leonard_my_dude~
Solution 4
Let , .
Notice how the coefficient for has to be the same for the two quadratics that are multiplied to create , and .
, , , , ,
, , ,
,
.
Video Solution by MOP 2024
~r00tsOfUnity
See Also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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