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Difference between revisions of "2006 AIME II Problems/Problem 6"
Line 1: Line 1:
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== Problem ==
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== Solution 1 ==
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[[Square]] <math> ABCD </math> has sides of length 1. Points <math> E </math> and <math> F </math> are on <math> \overline{BC} </math> and <math> \overline{CD}, </math> respectively, so that <math> \triangle AEF </math> is [[equilateral]]. A square with vertex <math> B </math> has sides that are [[parallel]] to those of <math> ABCD </math> and a vertex on <math> \overline{AE}. </math> The length of a side of this smaller square is <math>\frac{a-\sqrt{b}}{c}, </math> where <math> a, b, </math> and <math> c </math> are positive integers and <math> b</math> is not divisible by the square of any prime. Find <math> a+b+c. </math>
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<asy>
 +
unitsize(32mm);
 +
defaultpen(linewidth(.8pt)+fontsize(10pt));
 +
dotfactor=3;
 +
 
 +
pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1);
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pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1);
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pair Ap = (0, (3 - sqrt(3))/6);
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pair Cp = ((3 - sqrt(3))/6, 0);
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pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6);
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pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp};
 +
 
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draw(A--B--C--D--cycle);
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draw(A--F--Ep--cycle);
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draw(Ap--B--Cp--Dp--cycle);
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dot(dots);
  
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== Solution 1 ==
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label("$A$", A, NW);
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[[Image:2006_II_AIME-6.png|left|300px]]
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label("$B$", B, SW);
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label("$C$", C, SE);
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label("$D$", D, NE);
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label("$E$", Ep, SE);
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label("$F$", F, E);
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label("$A'$", Ap, W);
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label("$C'$", Cp, SW);
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label("$D'$", Dp, E);
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label("$s$", Ap--B, W);
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label("$1$", A--D, N);
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</asy>
 
Call the vertices of the new square A', B', C', and D', in relation to the vertices of <math>ABCD</math>, and define <math>s</math> to be one of the sides of that square. Since the sides are [[parallel]], by [[corresponding angles]] and AA~ we know that triangles <math>AA'D'</math> and <math>D'C'E</math> are similar. Thus, the sides are proportional: <math>\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}</math>. Simplifying, we get that <math>s^2 = (1 - s)(1 - s - CE)</math>.
 
Call the vertices of the new square A', B', C', and D', in relation to the vertices of <math>ABCD</math>, and define <math>s</math> to be one of the sides of that square. Since the sides are [[parallel]], by [[corresponding angles]] and AA~ we know that triangles <math>AA'D'</math> and <math>D'C'E</math> are similar. Thus, the sides are proportional: <math>\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}</math>. Simplifying, we get that <math>s^2 = (1 - s)(1 - s - CE)</math>.
  
Line 11: Line 36:
  
 
Here's an alternative geometric way to calculate <math>AE</math> (as opposed to [[trigonometry|trigonometric]]): The diagonal <math>\overline{AC}</math> is made of the [[altitude]] of the equilateral triangle and the altitude of the <math>45-45-90 \triangle</math>. The former is <math>\frac{AE\sqrt{3}}{2}</math>, and the latter is <math>\frac{AE}{2}</math>; thus <math>\frac{AE\sqrt{3} + AE}{2} = AC = \sqrt{2} \Longrightarrow AE= \sqrt{6}-\sqrt{2}</math>. The solution continues as above.
 
Here's an alternative geometric way to calculate <math>AE</math> (as opposed to [[trigonometry|trigonometric]]): The diagonal <math>\overline{AC}</math> is made of the [[altitude]] of the equilateral triangle and the altitude of the <math>45-45-90 \triangle</math>. The former is <math>\frac{AE\sqrt{3}}{2}</math>, and the latter is <math>\frac{AE}{2}</math>; thus <math>\frac{AE\sqrt{3} + AE}{2} = AC = \sqrt{2} \Longrightarrow AE= \sqrt{6}-\sqrt{2}</math>. The solution continues as above.
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== Solution 2 ==
 
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Since <math>\triangle AFE</math> is equilateral, <math>\overline{AE} = \overline{AF}</math>. It follows that <math>\overline{FC} = \overline{EC}</math>. Let <math>\overline{FC} = x</math>. Then, <math>\overline{EF} = x\sqrt{2}</math> and <math>\overline{DF} = 1-x</math>.
 
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<math>\overline{AF} = \sqrt{1+(1-x)^2} = x\sqrt{2}</math>.
 
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Square both sides and combine/move terms to get <math>x^2+2x-2 = 0</math>.
 
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Therefore <math>x = -1 + \sqrt{3}</math> and <math>x = -1 - \sqrt{3}</math>. The second solution is obviously extraneous, so <math>x = -1 + \sqrt{3}</math>.
 
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Now, consider the square ABCD to be on the Cartesian Coordinate Plane with <math>A = (0,0)</math>. Then, the line containing <math>\overline{AF}</math> has slope <math>\frac{1}{2-\sqrt{3}}</math> and equation <math>y = \frac{1}{2-\sqrt{3}}x</math>.
 
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The distance from <math>\overline{DC}</math> to <math>\overline{AF}</math> is the distance from <math>y = 1</math> to <math>y = \frac{1}{2-\sqrt{3}}x</math>.
 
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Similarly, the distance from <math>\overline{AD}</math> to <math>\overline{AF}</math> is the distance from <math>x = 0</math> to <math>y = \frac{1}{2-\sqrt{3}}x</math>.
 
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For some value <math>x = s</math>, these two distances are equal.
 
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<math>(s-0) = (1 - (\frac{1}{2-\sqrt{3}})s)</math>
 
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Solving for s, <math>s = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 012</math>.
 
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== See also ==
 
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{{AIME box|year=2006|n=II|num-b=5|num-a=7}}
 
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[[Category:Intermediate Geometry Problems]]
 
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{{MAA Notice}}
 

Revision as of 18:11, 9 December 2013

Solution 1

[asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp}; draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$E$", Ep, SE); label("$F$", F, E); label("$A'$", Ap, W); label("$C'$", Cp, SW); label("$D'$", Dp, E); label("$s$", Ap--B, W); label("$1$", A--D, N); [/asy] Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$, and define $s$ to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$. Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$.

$\angle EAF$ is $60$ degrees, so $\angle BAE = \frac{90 - 60}{2} = 15$. Thus, $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}$, so $AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}$. Since $\triangle AEF$ is equilateral, $EF = AE = \sqrt{6} - \sqrt{2}$. $\triangle CEF$ is a $45-45-90 \triangle$, so $CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1$. Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \sqrt{3} - s)$, so $(3 - \sqrt{3})s = 2 - \sqrt{3}$. Therefore, $s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}$, and $a + b + c = 3 + 3 + 6 = 012$.

Here's an alternative geometric way to calculate $AE$ (as opposed to trigonometric): The diagonal $\overline{AC}$ is made of the altitude of the equilateral triangle and the altitude of the $45-45-90 \triangle$. The former is $\frac{AE\sqrt{3}}{2}$, and the latter is $\frac{AE}{2}$; thus $\frac{AE\sqrt{3} + AE}{2} = AC = \sqrt{2} \Longrightarrow AE= \sqrt{6}-\sqrt{2}$. The solution continues as above.

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