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Difference between revisions of "2017 AIME I Problems/Problem 6"

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Our answer is then <math>\boxed{048}</math>.
 
Our answer is then <math>\boxed{048}</math>.
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==Video Solution==
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https://youtu.be/Mk-MCeVjSGc
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~Shreyas S
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2017|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:04, 17 June 2020

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 $\boxed{048}$.

Video Solution

https://youtu.be/Mk-MCeVjSGc ~Shreyas S

See Also

2017 AIME I (Problems • Answer Key • Resources)
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Problem 7
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All AIME Problems and Solutions

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