Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (048)
(→‎Solution)
Line 7: Line 7:
 
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{048}</math>.
 
So <math>x=84, 36</math>. The difference between these is <math>\boxed{048}</math>.
 +
 +
Note:
 +
 +
We actually do not need to spend time factoring <math>x^2 - 120x + 3024</math>. Since the problem asks for <math>|x_1 - x_2|</math>, where <math>x_1</math> and <math>x_2</math> are the roots of the quadratic, we can utilize Vieta's by noting that <math>(x_1 - x_2) ^ 2 = (x_1 + x_2) ^ 2 - 4x_1x_2</math>. Vieta's gives us <math>x_1 + x_2 = 120,</math> and <math>x_1x_2 = 3024.</math> Plugging this into the above equation and simplifying gives us <math>(x_1 - x_2) ^ 2 = 2304,</math> or <math>|x_1 - x_2| = 48</math>.
 +
 +
Our answer is then <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 14:27, 5 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 $\boxed{048}$.

See Also

2017 AIME I (Problems • Answer Key • Resources)
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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_I_Problems/Problem_6&oldid=123903"