2008 AMC 12A Problems/Problem 24: Difference between revisions
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<math>\textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} - 3} \qquad \textbf{(E)}\ 1</math> | <math>\textbf{(A)} \ \frac {\sqrt {3}}{6} \qquad \textbf{(B)} \ \frac {\sqrt {3}}{3} \qquad \textbf{(C)} \ \frac {\sqrt {3}}{2\sqrt {2}} \qquad \textbf{(D)} \ \frac {\sqrt {3}}{4\sqrt {2} - 3} \qquad \textbf{(E)}\ 1</math> | ||
==Solution== | ==Solution== | ||
<asy> | |||
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); | |||
</asy> | |||
Where <math>CA = x</math> | |||
<math>\tan\theta = \tan(\angle BAF - \angle DAE)</math> | |||
Since <math>\tan\angle BAF = \frac{2\sqrt{3}}{x-2}</math> and <math>\tan\angle DAE = \frac{\sqrt{3}}{x-1}</math>, we have | |||
<math>\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}</math> | |||
Multiplying numerator and denominator by <math>(x-2)(x-1)</math> | |||
<math>\tan\theta = \frac{x\sqrt{3}}{x^2-3x+8}</math> | |||
If you know calculus, you can use that right here to max <math>\tan\theta</math>, but if you don't: | |||
By AM-GM | |||
<math>\frac{x + \frac{8}{x}}{2} \geq \sqrt{8}</math> | |||
<math>x + \frac{8}{x} \geq 4\sqrt{2}</math> | |||
<math>x + \frac{8}{x} -3 \geq 4\sqrt{2} - 3</math> | |||
<math>\frac{x^2 - 3x + 8}{x} \geq 4\sqrt{2} - 3</math> | |||
<math>\frac{x}{x^2 - 3x + 8} \leq \frac{1}{4\sqrt{2}-3}</math> | |||
<math>\frac{x\sqrt{3}}{x^2 - 3x + 8} \leq \frac{\sqrt{3}}{4\sqrt{2}-3}</math> | |||
<math>\tan\theta \leq \frac{\sqrt{3}}{4\sqrt{2}-3}</math> | |||
Which means that the minimum is | |||
<math>\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}</math> | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}} | {{AMC12 box|year=2008|ab=A|num-b=23|num-a=25}} | ||
Revision as of 11:05, 24 February 2008
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
See Also
| 2008 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 23 |
Followed by Problem 25 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
