Difference between revisions of "2006 AIME II Problems/Problem 6"
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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> | [[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 1 == | == Solution 1 == | ||
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Revision as of 18:12, 9 December 2013
Contents
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 1
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.
Solution 2
Since is equilateral,
. It follows that
. Let
. Then,
and
.
.
Square both sides and combine/move terms to get .
Therefore
and
. The second solution is obviously extraneous, so
.
Now, consider the square ABCD to be on the Cartesian Coordinate Plane with . Then, the line containing
has slope
and equation
.
The distance from to
is the distance from
to
.
Similarly, the distance from to
is the distance from
to
.
For some value , these two distances are equal.
Solving for s, , and
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.