Difference between revisions of "2020 AIME II Problems/Problem 11"

Revision as of 19:24, 7 June 2020

Problem

Let $P(X) = x^2 - 3x - 7$ , and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1$ . David computes each of the three sums $P + Q$ , $P + R$ , and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2$ , then $R(0) = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

Let $Q(x) = x^2 + ax + 2$ and $R(x) = x^2 + bx + c$ . We can write the following: $P + Q = 2x^2 + (a - 3)x - 5$ $P + R = 2x^2 + (b - 3)x + (c - 7)$ $Q + R = 2x^2 + (a + b)x + (c + 2)$ Let the common root of $P+Q,P+R$ be $r$ ; $P+R,Q+R$ be $s$ ; and $P+Q,Q+R$ be $t$ . We then have that the roots of $P+Q$ are $r,t$ , the roots of $P + R$ are $r, s$ , and the roots of $Q + R$ are $s,t$ .

Video Solution

https://youtu.be/BQlab3vjjxw ~ CNCM

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 ==Solution==
+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 + R = 2x^2 + (b - 3)x + (c - 7)</cmath>
+<cmath>Q + R = 2x^2 + (a + b)x + (c + 2)</cmath>
+Let the common root of <math>P+Q,P+R</math> be <math>r</math>; <math>P+R,Q+R</math> be <math>s</math>; and <math>P+Q,Q+R</math> be <math>t</math>.  We then have that the roots of <math>P+Q</math> are <math>r,t</math>, the roots of <math>P + R</math> are <math>r, s</math>, and the roots of <math>Q + R</math> are <math>s,t</math>.
 ==Video Solution==
 https://youtu.be/BQlab3vjjxw ~ CNCM
 ==See Also==

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Difference between revisions of "2020 AIME II Problems/Problem 11"

Revision as of 19:24, 7 June 2020

Contents

Problem

Solution

Video Solution

See Also