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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==Solution 2 (Not Complementary Counting method)==
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Because we know that we have an isosceles triangle with angles of <math>x</math> (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is <math>2x</math>. We form this same conclusion for the other angle <math>x</math>, and <math>180-2x</math>. Therefore we get <math>3</math> arcs, namely, <math>2x</math>, <math>2x</math>, and <math>360-4x</math>. 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 <math>2</math> * <math>\frac{2x}{360}</math> * <math>\frac{2x}{360}</math> + <math>2</math> * <math>2</math> * <math>\frac{2x}{360}</math> * <math>\frac{360-4x}{360}</math> = <math>\frac{14}{25}</math> which using trivial algebra gives you <math>x^2-120x+3024</math> and factoring gives you <math>(x-84)(x-36)</math> and so your answer is <math>\boxed{048}</math>.
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~jske25
 
==Video Solution==
 
==Video Solution==
  

Revision as of 18:51, 8 March 2021

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

Video Solution

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

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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