# 2017 AIME I Problems/Problem 6

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## 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 $\boxed{048}$.

## Solution 2 (Not Complementary Counting method)

Because we know that we have an isosceles triangle with angles of $x$ (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is $2x$. We form this same conclusion for the other angle $x$, and $180-2x$. Therefore we get $3$ arcs, namely, $2x$, $2x$, and $360-4x$. To have the chords intersect the triangle, we need the two points selected (to make a chord) to be on completely different arcs. An important idea to understand is that order matters in this case, so we have the equation $2$ * $\frac{2x}{360}$ * $\frac{2x}{360}$ + $2$ * $2$ * $\frac{2x}{360}$ * $\frac{360-4x}{360}$ = $\frac{14}{25}$ which using trivial algebra gives you $x^2-120x+3024$ and factoring gives you $(x-84)(x-36)$ and so your answer is $\boxed{048}$. ~jske25

## Solution 3 ( 3 System Algebra )

After constructing the circumscribed circle, realize that the only time when the chord does not intersect the circle is when our $2$ points fall on only one arc formed by the sides of the triangle. Thus, lets call our isosceles triangle $ABC$, where $AB=BC$. Thus, the arcs formed by $BC$ and $AB$ can be called $a$, and the arc formed by $AC$ is called $b$. So, we can create the following system $$2a+b=1$$ $$2a^2+4ab=\frac{14}{25}$$ $$2a^2+b^2=\frac{11}{25}$$

Notice we are denoting $a$ and $b$ as our probabilities, which we will be converting to degrees later. The 2 remaining systems can be calculated by using our rule about intersecting arcs and chords. So, after some hairy algebra we get: $$a=\frac{1}{5}$$ if $$b=\frac{3}{5}$$ $$a=\frac{7}{15}$$ if $$b=\frac{1}{15}$$

From here we find the absolute difference by doing $\frac{7}{15}-\frac{1}{5} = \frac{4}{15}$. Converting to degrees, since the angles of a triangle add up to $180^o$, we find that $\frac{4}{15} \cdot 180 =\boxed{048}$, which is our answer.

-Geometry285

## Video Solution

https://youtu.be/Mk-MCeVjSGc?t=690 ~Shreyas S

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 