2017 AIME I Problems/Problem 6

Problem 6

A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure $x$ . Two points are chosen independently and uniformly at random on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\frac{14}{25}$ . Find the difference between the largest and smallest possible values of $x$ .

Solution

The probability that the chord doesn't intersect the triangle is $\frac{11}{25}$ . The only way this can happen is if the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite one of the base angles is $\frac{x}{180}$ , and the probability that a point is chosen on the arc between the two base angles is $\frac{180-2x}{180}$ . Therefore, we can write $2\left(\frac{x}{180}\right)^2+\left(\frac{180-2x}{180}\right)^2=\frac{11}{25}$ This simplifies to $x^2-120x+3024=0$ Which factors as $(x-84)(x-36)=0$ So $x=84, 36$ . The difference between these is $\boxed{048}$ .

Note:

We actually do not need to spend time factoring $x^2 - 120x + 3024$ . Since the problem asks for $|x_1 - x_2|$ , where $x_1$ and $x_2$ are the roots of the quadratic, we can utilize Vieta's by noting that $(x_1 - x_2) ^ 2 = (x_1 + x_2) ^ 2 - 4x_1x_2$ . Vieta's gives us $x_1 + x_2 = 120,$ and $x_1x_2 = 3024.$ Plugging this into the above equation and simplifying gives us $(x_1 - x_2) ^ 2 = 2304,$ or $|x_1 - x_2| = 48$ .

Our answer is then $\boxed{048}$ .

2017 AIME I (Problems • Answer Key • Resources)
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2017 AIME I Problems/Problem 6

Problem 6

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