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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 6"

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==Solution==
 
==Solution==
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The problem asks for the probability that point <math>P</math> is inside an equilateral triangle <math>A_1B_1C_1</math>. 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:
+
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 10: Line 10:
 
draw((8,1+4*sqrt(3))--(4,1));
 
draw((8,1+4*sqrt(3))--(4,1));
 
draw((6,1+4*sqrt(3))--(4,1));
 
draw((6,1+4*sqrt(3))--(4,1));
−
label("$P$",(6,1+2*sqrt(3)),NNW);
+
label("$P$",(6,1+4*sqrt(3)),NNW);
 
label("$x$",(5,1+2*sqrt(3)),NNW);
 
label("$x$",(5,1+2*sqrt(3)),NNW);
 
label("$y$", (3.5,1+4*sqrt(3)),NW);
 
label("$y$", (3.5,1+4*sqrt(3)),NW);
 
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 02:41, 25 December 2022

Problem

Three points $A,B,C$ are chosen at random on a circle. The probability that there exists a point $P$ inside an equilateral triangle $A_1B_1C_1$ such that $PA_1=BC,PB_1=AC,PC_1=AB$ can be expressed in the form $\frac{m} {n},$ where $m,n$ are relatively prime positive integers. Find $m+n.$

Solution

Let $x$, $y$, and $z$ be the three distances from point $P$ to each of the vertices, with $x$ being the longest distance. Let's consider the case in which point $P$ is actually on the line: [asy] unitsize(0.75cm); draw((0,1+4*sqrt(3))--(8,1+4*sqrt(3))); draw((0,1+4*sqrt(3))--(4,1)); draw((8,1+4*sqrt(3))--(4,1)); draw((6,1+4*sqrt(3))--(4,1)); label("$P$",(6,1+4*sqrt(3)),NNW); label("$x$",(5,1+2*sqrt(3)),NNW); label("$y$", (3.5,1+4*sqrt(3)),NW); label("$z$", (7.5,1+4*sqrt(3)),NW); [/asy]

In this case, we can use Stewart's Theorem to find the relationship between the three variables. \[yz(y+z)+x^2(y+z)=y(y+z)^2+z(y+z)^2\] \[yz+x^2=y(y+z)+z(y+z)\] \[yz+x^2=y^2+z^2+2yz\] \[x^2=y^2+z^2+yz\] This can be manipulated into the Law of Cosines, with an angle of $120^{\circ}$. \[x^2=y^2+z^2-2yz(\cos120^{\circ})\]

In order for point $P$ to be inside the equilateral triangle: \[x^2<y^2+z^2-2yz(\cos120^{\circ})\] Thus, Triangle $ABC$ (which has sides of $x,y,$ and $z$) cannot have any angles greater than or equal to $120^{\circ}$. Next, we can fix point $A$ to the $180^{\circ}$ 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 $B$ and $C$ should be on the circle, such that Triangle $ABC$ has all angles less than $120^{\circ}$ 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 $(60,60),(180,60),(300,180),(300,300),(180,300),$ and $(60,180)$, while the green square has vertices at coordinates $(0,0),(0,360),(360,360),$ and $(360,0)$. Each other these coordinates represent positions for points $B$ and $C$. For example, $(240, 181)$ means point $B$ is at $(240^{\circ})$ on the unit circle, while point $C$ is at $(181^{\circ})$ on the unit circle.

Therefore, the probability of points $A, B,$ and $C$ forming a triangle with angles less than $120^{\circ}$ is: \[1-\frac{2\cdot120^2+2\cdot\frac{120^2}{2}}{360^2}\] \[=1-\frac{1}{3}\] \[=\frac{2}{3}\]. And the answer is $2+3=\boxed{005}$ - Solution by adyj

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=q0lUoXmkAyk&t=794s

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