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

Revision as of 19:38, 7 July 2019

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

The problem asks for the probability that point $P$ is inside an equilateral triangle $A_1B_1C_1$ . 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: $[asy] unitsize(0.75cm); draw((0,1+4*sqrt(3))--(8,1+4*sqrt(3))); draw((0,1+4*sqrt(3))--(4,1)); draw((0,1+4*sqrt(3))--(4,1)); draw((6,1+4*sqrt(3))--(4,1)); label("$x$",(5,1+2*sqrt(3)),NNW); label("$y$", (4,1+4*sqrt(3)),NW); label("$z$", (8,1+4*sqrt(3)),NW); [/asy]$

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

Revision as of 19:38, 7 July 2019

Problem

Solution

@@ Line 6: / Line 6: @@
 <asy>
 unitsize(0.75cm);
-draw((0,4*sqrt(3))--(8,4*sqrt(3)));
+draw((0,1+4*sqrt(3))--(8,1+4*sqrt(3)));
-draw((0,4*sqrt(3))--(4,0));
+draw((0,1+4*sqrt(3))--(4,1));
-draw((8,4*sqrt(3))--(4,0));
+draw((0,1+4*sqrt(3))--(4,1));
-draw((6,4*sqrt(3))--(4,0));
+draw((6,1+4*sqrt(3))--(4,1));
-label("$x$",(5,2sqrt(3)),NNW);
+label("$x$",(5,1+2*sqrt(3)),NNW);
-label("$y$", (3,0),N);
+label("$y$", (4,1+4*sqrt(3)),NW);
-label("$z$", (7,0),N);
+label("$z$", (8,1+4*sqrt(3)),NW);
 </asy>