Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 6"
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==Solution== | ==Solution== | ||
− | + | Let <math>x</math>, <math>y</math>, and <math>z</math> be the three distances from point <math>P</math> to each of the vertices, with <math>x</math> being the longest distance. Let's consider the case in which point <math>P</math> is actually on the line: | |
<asy> | <asy> | ||
unitsize(0.75cm); | unitsize(0.75cm); | ||
Line 15: | Line 15: | ||
label("$z$", (7.5,1+4*sqrt(3)),NW); | label("$z$", (7.5,1+4*sqrt(3)),NW); | ||
</asy> | </asy> | ||
+ | |||
+ | In this case, we can use Stewart's Theorem to find the relationship between the three variables. | ||
+ | <cmath>yz(y+z)+x^2(y+z)=y(y+z)^2+z(y+z)^2</cmath> | ||
+ | <cmath>yz+x^2=y(y+z)+z(y+z)</cmath> | ||
+ | <cmath>yz+x^2=y^2+z^2+2yz</cmath> | ||
+ | <cmath>x^2=y^2+z^2+yz</cmath> | ||
+ | This can be manipulated into the Law of Cosines, with an angle of <math>120^{\circ}</math>. | ||
+ | <cmath>x^2=y^2+z^2-2yz(\cos120^{\circ})</cmath> | ||
+ | |||
+ | In order for point <math>P</math> to be inside the equilateral triangle: | ||
+ | <cmath>x^2<y^2+z^2-2yz(\cos120^{\circ})</cmath> | ||
+ | Thus, Triangle <math>ABC</math> (which has sides of <math>x,y,</math> and <math>z</math>) cannot have any angles greater than or equal to <math>120^{\circ}</math>. Next, we can fix point <math>A</math> to the <math>180^{\circ}</math> mark of a unit circle, as shown below: | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.5cm); | ||
+ | draw((-4,0)--(4,0),Arrows); | ||
+ | draw((0,-4)--(0,4),Arrows); | ||
+ | draw(circle((0,0),3),red+linewidth(1)); | ||
+ | dot((-3,0)); | ||
+ | label("$A$", (-3,0), NW); | ||
+ | </asy> | ||
+ | |||
+ | Thus, the places points <math>B</math> and <math>C</math> should be on the circle, such that Triangle <math>ABC</math> has all angles less than <math>120^{\circ}</math> is OUTSIDE of the red outline, but inside of the green outline shown in the graph below: | ||
+ | <asy> | ||
+ | unitsize(0.7cm); | ||
+ | draw((-1,0)--(7,0),Arrows); | ||
+ | draw((0,-1)--(0,7),Arrows); | ||
+ | draw((0,6)--(6,6)--(6,0)--(0,0)--cycle,green+linewidth(1)); | ||
+ | draw((1,1)--(3,1)--(5,3)--(5,5)--(3,5)--(1,3)--cycle, red); | ||
+ | label("$B$",(7,0),SE); | ||
+ | label("$C$",(0,7),NW); | ||
+ | </asy> | ||
+ | |||
+ | The red hexagon has vertices at coordinates <math>(60,60),(180,60),(300,180),(300,300),(180,300),</math> and <math>(60,180)</math>, while the green square has vertices at coordinates <math>(0,0),(0,360),(360,360),</math> and <math>(360,0)</math>. Each other these coordinates represent positions for points <math>B</math> and <math>C</math>. For example, <math>(240, 181)</math> means point <math>B</math> is at <math>(240^{\circ})</math> on the unit circle, while point <math>C</math> is at <math>(181^{\circ})</math> on the unit circle. | ||
+ | |||
+ | Therefore, the probability of points <math>A, B,</math> and <math>C</math> forming a triangle with angles less than <math>120^{\circ}</math> is: | ||
+ | <cmath>1-\frac{2\cdot120^2+2\cdot\frac{120^2}{2}}{360^2}</cmath> | ||
+ | <cmath>=1-\frac{1}{3}</cmath> | ||
+ | <cmath>=\frac{2}{3}</cmath>. | ||
+ | And the answer is <math>2+3=\boxed{005}</math> | ||
+ | - Solution by adyj | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://youtube.com/watch?v=q0lUoXmkAyk&t=794s |
Latest revision as of 03:41, 25 December 2022
Problem
Three points are chosen at random on a circle. The probability that there exists a point
inside an equilateral triangle
such that
can be expressed in the form
where
are relatively prime positive integers. Find
Solution
Let ,
, and
be the three distances from point
to each of the vertices, with
being the longest distance. Let's consider the case in which point
is actually on the line:
In this case, we can use Stewart's Theorem to find the relationship between the three variables.
This can be manipulated into the Law of Cosines, with an angle of
.
In order for point to be inside the equilateral triangle:
Thus, Triangle
(which has sides of
and
) cannot have any angles greater than or equal to
. Next, we can fix point
to the
mark of a unit circle, as shown below:
Thus, the places points and
should be on the circle, such that Triangle
has all angles less than
is OUTSIDE of the red outline, but inside of the green outline shown in the graph below:
The red hexagon has vertices at coordinates and
, while the green square has vertices at coordinates
and
. Each other these coordinates represent positions for points
and
. For example,
means point
is at
on the unit circle, while point
is at
on the unit circle.
Therefore, the probability of points and
forming a triangle with angles less than
is:
.
And the answer is
- Solution by adyj