Difference between revisions of "2006 AIME I Problems/Problem 6"
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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>. | <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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+ | 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. | ||
== See also == | == See also == |
Revision as of 17:34, 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 I (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 |