2008 AMC 12A Problems/Problem 24

Problem

Triangle has and . Point is the midpoint of . What is the largest possible value of ?

Solution

unitsize(12mm);
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
pair E=(1,0), F=(2,0);

draw(C--B--A--C);
draw(A--D);
draw(D--E);
draw(B--F);

dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);

label("C",C,SW);
label("B",B,N);
label("A",A,SE);
label("D",D,NW);
label("E",E,S);
label("F",F,S);
label("<math>60^\circ</math>",C,NE);
label("2",1*dir(60),NW);
label("2",3*dir(60),NW);
label("<math>\theta</math>",(7,.4));
label("1",(.5,0),S);
label("1",(1.5,0),S);
label("x-2",(5,0),S);
 (Error making remote request. Unknown error_msg)

Where

Since and , we have

Multiplying numerator and denominator by

If you know calculus, you can use that right here to max , but if you don't:

By AM-GM

Which means that the minimum is