Art of Problem Solving

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I did everything related to the [[ Mock AIME 3 Pre 2005]]
<asy>
all in one edit - each solution page, the answers, and the entire aime page.


unitsize(.5cm);
pair bar(pair a) {
return (a.x,-a.y);
};
pair A[] = { (6.5,0),(1.5,12) };
pair P[] = { A[0],A[1],-bar(A[1]),-bar(A[0]),(2A[0]-bar(A[0]))/3,(2A[0]+A[1])/3,
(5A[0]-4bar(A[0]))/9,(5A[0]+4A[1])/9
};
pair D[] = { (2P[4]+P[5])/3,(P[4]+2P[5])/3 };
draw( D[0]--P[6] ,blue+1mm);
draw( D[0]--P[4] ,blue+1mm);
draw( P[6]--P[4], black+1mm);
pair E[] = {(D[0]+P[6])/2,D[0] + P[6] - P[4]};
dot(E[0]);dot(E[1]);
draw(P[6]--E[1],red+1mm);
draw(P[4]--E[1],red+1mm);
draw(P[4]--P[5]);
draw( D[0] -- E[1]);
draw(D[1]--P[7]);
draw( P[3] -- P[0] -- P[1]) ;
for(int i=0;i<10;++i) {
if(i<4) {


Hmmm... I think dollars signs convert automatically to math symbols
} else {
 
dot( (i<8) ? P[i] : D[i-8] );
<pre> <math>ab</math> </pre>
}
<math>ab</math>
};


<asy>import graph;
import markers;
unitsize(3cm);
int n=18;
real a=360/n;
for (int i = -3; i < n-4; ++i)
{
  draw(dir(i*a)--dir((i+1)*a),.5+black);
}
for (int i = -2; i < n-4; ++i)
{
  draw(dir(i*a)--2*dir(i*a)-dir((i-1)*a),blue);
  markangle(scale(.7)*"<math>\theta</math>",radius=15,2*dir(i*a)-dir((i-1)*a),dir(i*a),dir((i+1)*a),ArcArrow(1mm),.5+blue);
}
markangle(scale(.7)*"<math>\theta</math>",radius=15,2*dir(15*a)-dir(14*a),dir(15*a),dir(16*a),ArcArrow(1mm),.5+blue);
draw(dir(14*a)--6*dir(14*a)-5*dir(13*a));
label("<math>\ell</math>",(7*dir(14*a)-5*dir(13*a))/2,S);
label("<math>m</math>",(7*dir(15*a)-5*dir(14*a))/2,dir(3*a));
label(scale(.8)*"<math>180-\theta</math>",.95*dir(14*a),dir(5*a),red);
draw(dir(14*a)--6*dir(15*a)-5*dir(14*a));
markangle("<math>\theta</math>",radius=60,2*dir(14*a)-dir(13*a),dir(14*a),dir(15*a),ArcArrow(1mm),.5+red);
</asy>
</asy>

Latest revision as of 00:21, 15 September 2008

[asy]  unitsize(.5cm); pair bar(pair a) {  return (a.x,-a.y); }; pair A[] = { (6.5,0),(1.5,12) }; pair P[] = { A[0],A[1],-bar(A[1]),-bar(A[0]),(2A[0]-bar(A[0]))/3,(2A[0]+A[1])/3, (5A[0]-4bar(A[0]))/9,(5A[0]+4A[1])/9  }; pair D[] = { (2P[4]+P[5])/3,(P[4]+2P[5])/3 }; draw( D[0]--P[6] ,blue+1mm); draw( D[0]--P[4] ,blue+1mm); draw( P[6]--P[4], black+1mm); pair E[] = {(D[0]+P[6])/2,D[0] + P[6] - P[4]}; dot(E[0]);dot(E[1]); draw(P[6]--E[1],red+1mm); draw(P[4]--E[1],red+1mm); draw(P[4]--P[5]); draw( D[0] -- E[1]); draw(D[1]--P[7]); draw( P[3] -- P[0] -- P[1]) ; for(int i=0;i<10;++i) { if(i<4) {  } else {  dot( (i<8) ? P[i] : D[i-8] ); } };  [/asy]