Difference between revisions of "2011 AIME II Problems/Problem 15"
Line 1: | Line 1: | ||
− | Problem | + | ==Problem== |
Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>. | Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>. | ||
− | + | ==Solution== | |
− | Solution | ||
− | |||
− |
Revision as of 18:17, 3 April 2011
Problem
Let . A real number is chosen at random from the interval . The probability that is equal to , where , , , , and are positive integers. Find .