Difference between revisions of "2006 AIME II Problems/Problem 6"
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− | + | == Problem == | |
+ | [[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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+ | == Solution == | ||
+ | [[Image:2006_II_AIME-6.png|thumb|left|300px]] | ||
+ | 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>. | ||
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+ | <math>\angle EAF</math> is <math>60</math> degrees, so <math>\angle BAE = \frac{90 - 60}{2} = 15</math>. Thus, <math>\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}</math>, so <math>AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}</math>. Since <math>\triangle AEF</math> is [[equilateral]], <math>EF = AE = \sqrt{6} - \sqrt{2}</math>. <math>\triangle CEF</math> is a <math>45-45-90 \triangle</math>, so <math>CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1</math>. Substituting back into the equation from the beginning, we get <math>s^2 = (1 - s)(2 - \sqrt{3} - s)</math>, so <math>(3 - \sqrt{3})s = 2 - \sqrt{3}</math>. Therefore, <math>s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 012</math>. | ||
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+ | ---- | ||
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+ | Here's an alternative geometric way to calculate <math>CE</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{CE\sqrt{3}}{2}</math>, and the latter is <math>\frac{CE}{2}</math>; thus <math>\frac{CE\sqrt{3} + CE}{2} = AC = \sqrt{2} \Longrightarrow CE = \sqrt{6}-\sqrt{2}</math>. The solution continues as above. | ||
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+ | == See also == | ||
+ | {{AIME box|year=2006|n=II|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Revision as of 18:06, 25 September 2007
Problem
Square has sides of length 1. Points and are on and respectively, so that is equilateral. A square with vertex has sides that are parallel to those of and a vertex on The length of a side of this smaller square is where and are positive integers and is not divisible by the square of any prime. Find
Solution
Call the vertices of the new square A', B', C', and D', in relation to the vertices of , and define to be one of the sides of that square. Since the sides are parallel, by corresponding angles and AA~ we know that triangles and are similar. Thus, the sides are proportional: . Simplifying, we get that .
is degrees, so . Thus, , so . Since is equilateral, . is a , so . Substituting back into the equation from the beginning, we get , so . Therefore, , and .
Here's an alternative geometric way to calculate (as opposed to trigonometric): The diagonal is made of the altitude of the equilateral triangle and the altitude of the . The former is , and the latter is ; thus . The solution continues as above.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |