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

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m (048)
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This simplifies to <cmath>x^2-120x+3024=0</cmath>
 
This simplifies to <cmath>x^2-120x+3024=0</cmath>
 
Which factors as <cmath>(x-84)(x-36)=0</cmath>
 
Which factors as <cmath>(x-84)(x-36)=0</cmath>
So <math>x=84, 36</math>. The difference between these is <math>\boxed{48}</math>.
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So <math>x=84, 36</math>. The difference between these is <math>\boxed{048}</math>.
  
 
==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 20:22, 8 March 2017

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 $\boxed{048}$.

See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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