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Difference between revisions of "2017 AIME I Problems/Problem 6"
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==Solution==
 
==Solution==
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The probability that the chord doesn't intersect the triangle is <math>\frac{11}{25}</math>. 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 <math>\frac{x}{180}</math>, and the probability that a point is chosen on the arc between the two base angles is <math>\frac{180-2x}{180}</math>. Therefore, we can write <cmath>2\left(\frac{x}{180}\right)^2+\left(\frac{180-2x}{180}\right)^2=\frac{11}{25}</cmath>
+
The probability that the chord doesn't intersect the triangle is <math>\frac{11}{25}</math>. 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 <math>\frac{2x}{360}=\frac{x}{180}</math> (this comes from the Central Angle Theorem, which states that the central angle from two points on a circle is always twice the inscribed angle from those two points), and the probability that a point is chosen on the arc between the two base angles is <math>\frac{180-2x}{180}</math>. Therefore, we can write <cmath>2\left(\frac{x}{180}\right)^2+\left(\frac{180-2x}{180}\right)^2=\frac{11}{25}</cmath>
 
This simplifies to <cmath>x^2-120x+3024=0</cmath>
 
This simplifies to <cmath>x^2-120x+3024=0</cmath>
 +
 +
(Note that the simplification is quite tedious)
 +
 
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>.
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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>.  
 
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>.  
 
~jske25
 
~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 <math>2</math> points fall on only one arc formed by the sides of the triangle. Thus, lets call our isosceles triangle  <math>ABC</math>, where <math>AB=BC</math>. Thus, the arcs formed by <math>BC</math> and <math>AB</math> can be called <math>a</math>, and the arc formed by <math>AC</math> is called <math>b</math>. So, we can create the following system
 +
 +
<cmath>2a+b=1</cmath>
 +
<cmath>2a^2+4ab=\frac{14}{25}</cmath>
 +
<cmath>2a^2+b^2=\frac{11}{25}</cmath>
 +
 +
Notice we are denoting <math>a</math> and <math>b</math> 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:
 +
 +
<cmath>a=\frac{1}{5}</cmath> if <cmath>b=\frac{3}{5}</cmath>
 +
<cmath>a=\frac{7}{15}</cmath> if <cmath>b=\frac{1}{15}</cmath>
 +
 +
From here we find the absolute difference by doing <math>\frac{7}{15}-\frac{1}{5} = \frac{4}{15}</math>. Converting to degrees, since the angles of a triangle add up to <math>180^o</math>, we find that <math>\frac{4}{15} \cdot 180 =\boxed{048}</math>, which is our answer.
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 +
-Geometry285
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==Video Solution==
 
==Video Solution==
  

Latest revision as of 15:18, 5 January 2024

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 $\frac{2x}{360}=\frac{x}{180}$ (this comes from the Central Angle Theorem, which states that the central angle from two points on a circle is always twice the inscribed angle from those two points), 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 \[x^2-120x+3024=0\]

(Note that the simplification is quite tedious)

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

See Also

2017 AIME I (Problems • Answer Key • Resources)
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Problem 5 		Followed by
Problem 7
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