Difference between revisions of "2008 AMC 12A Problems/Problem 24"
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− | ==Problem== | + | == Problem == |
− | Triangle <math>ABC</math> has <math>\angle C = 60^{\circ}</math> and <math>BC = 4</math>. Point <math>D</math> is the midpoint of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>? | + | [[Triangle]] <math>ABC</math> has <math>\angle C = 60^{\circ}</math> and <math>BC = 4</math>. Point <math>D</math> is the [[midpoint]] of <math>BC</math>. What is the largest possible value of <math>\tan{\angle BAD}</math>? |
− | <math>\ | + | <math>\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\qquad\mathrm{(E)}\ 1</math> |
− | ==Solution== | + | == Solution 1 == |
+ | <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("\(60^\circ\)",C+(.1,.1),ENE); | ||
+ | label("\(2\)",1*dir(60),NW); | ||
+ | label("\(2\)",3*dir(60),NW); | ||
+ | label("\(\theta\)",(7,.4)); | ||
+ | label("\(1\)",(.5,0),S); | ||
+ | label("\(1\)",(1.5,0),S); | ||
+ | label("\(x-2\)",(5,0),S);</asy> | ||
− | ==See | + | Let <math>x = CA</math>. Then <math>\tan\theta = \tan(\angle BAF - \angle DAE)</math>, and 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 |
+ | |||
+ | <cmath>\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}}= \frac{x\sqrt{3}}{x^2-3x+8}</cmath> | ||
+ | |||
+ | With calculus, taking the [[derivative]] and setting equal to zero will give the maximum value of <math>\tan \theta</math>. Otherwise, we can apply [[AM-GM]]: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \frac{x^2 - 3x + 8}{x} = \left(x + \frac{8}{x}\right) -3 &\geq 2\sqrt{x \cdot \frac 8x} - 3 = 4\sqrt{2} - 3\\ | ||
+ | \frac{x}{x^2 - 3x + 8} &\leq \frac{1}{4\sqrt{2}-3}\\ | ||
+ | \frac{x\sqrt{3}}{x^2 - 3x + 8} = \tan \theta &\leq \frac{\sqrt{3}}{4\sqrt{2}-3}\end{align*}</cmath> | ||
+ | |||
+ | Thus, the maximum is at | ||
+ | <math>\frac{\sqrt{3}}{4\sqrt{2}-3} \Rightarrow \mathbf{(D)}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | We notice that <math>\tan(x)</math> is strictly increasing on the interval <math>[0, \frac{\pi}{2})</math> (if <math>\angle BAD\ge 90^\circ</math>, then it is impossible for <math>\angle C=60^\circ</math>), so we want to maximize <math>\angle BAD</math>. | ||
+ | |||
+ | Consider the circumcircle of <math>BAD</math> and let it meet <math>AC</math> again at <math>F</math>. Any point <math>P</math> between <math>A</math> and <math>F</math> on line <math>AC</math> is inside this circle, so it follows that <math>\angle BPD>\angle BAD</math>. Therefore to maximize <math>\angle BAD</math>, the circumcircle of <math>BAD</math> must be tangent to <math>AC</math> at <math>A</math>. By PoP we find that <math>CA^2=CD\cdot CB \Rightarrow AC = 2\sqrt{2}</math>. | ||
+ | |||
+ | Now our computations are straightforward: | ||
+ | <cmath>\tan\angle BAD = \frac{\sin \angle BAD}{\cos \angle BAD} = \frac{\frac{2\sin\angle ABD}{AD}}{\frac{AB^2+AD^2-BD^2}{2AB\cdot AD}}</cmath> | ||
+ | <cmath>=\frac{4\sin \angle ABD\cdot AB}{AB^2+AD^2-4} = \frac{4 AC \sin \angle ACB}{AB^2 + AD^2 - 4}</cmath> | ||
+ | <cmath>=\frac{4\sqrt{6}}{(4^2+(2\sqrt{2})^2-4\cdot 2\sqrt{2}) + (2^2+(2\sqrt{2})^2 - 2\cdot 2\sqrt{2}) - 4} = \frac{4\sqrt{6}}{32-12\sqrt{2}}</cmath> | ||
+ | <cmath>=\boxed{\frac{\sqrt{3}}{4\sqrt{2}-3}}</cmath> | ||
+ | |||
+ | == 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}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 14:35, 7 June 2018
Contents
Problem
Triangle has
and
. Point
is the midpoint of
. What is the largest possible value of
?
Solution 1
Let . Then
, and since
and
, we have
With calculus, taking the derivative and setting equal to zero will give the maximum value of . Otherwise, we can apply AM-GM:
Thus, the maximum is at
.
Solution 2
We notice that is strictly increasing on the interval
(if
, then it is impossible for
), so we want to maximize
.
Consider the circumcircle of and let it meet
again at
. Any point
between
and
on line
is inside this circle, so it follows that
. Therefore to maximize
, the circumcircle of
must be tangent to
at
. By PoP we find that
.
Now our computations are straightforward:
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 |
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