Difference between revisions of "2017 AIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | 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>. | ||
Line 21: | Line 24: | ||
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 | 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 | + | 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 < | + | 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. |
-Geometry285 | -Geometry285 |
Latest revision as of 16:18, 5 January 2024
Contents
Problem 6
A circle is circumscribed around an isosceles triangle whose two congruent angles have degree measure . 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
. Find the difference between the largest and smallest possible values of
.
Solution
The probability that the chord doesn't intersect the triangle is . 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
(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
. Therefore, we can write
This simplifies to
(Note that the simplification is quite tedious)
Which factors as
So
. The difference between these is
.
Note:
We actually do not need to spend time factoring . Since the problem asks for
, where
and
are the roots of the quadratic, we can utilize Vieta's by noting that
. Vieta's gives us
and
Plugging this into the above equation and simplifying gives us
or
.
Our answer is then .
Solution 2 (Not Complementary Counting method)
Because we know that we have an isosceles triangle with angles of (and we know that x is an inscribed angle), that means that the arc that is intercepted by this angle is
. We form this same conclusion for the other angle
, and
. Therefore we get
arcs, namely,
,
, and
. 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
*
*
+
*
*
*
=
which using trivial algebra gives you
and factoring gives you
and so your answer is
.
~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 points fall on only one arc formed by the sides of the triangle. Thus, lets call our isosceles triangle
, where
. Thus, the arcs formed by
and
can be called
, and the arc formed by
is called
. So, we can create the following system
Notice we are denoting and
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:
if
if
From here we find the absolute difference by doing . Converting to degrees, since the angles of a triangle add up to
, we find that
, 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) | ||
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.