Difference between revisions of "2006 AIME II Problems/Problem 6"
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Solving for s, <math>s = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 12</math>. | Solving for s, <math>s = \frac{3 - \sqrt{3}}{6}</math>, and <math>a + b + c = 3 + 3 + 6 = 12</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | Suppose <math>\overline{AB} = \overline{AD} = x.</math> Note that <math>\angle EAF = 60</math> since the triangle is equilateral, and by symmetry, <math>\angle BAE = \angle DAF = 15.</math> Note that if <math>\overline{AD} = x</math> and <math>\angle BAE = 15</math>, then <math>\overline{AA'}=\frac{x}{\tan(15)}.</math> Also note that <cmath>AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x</cmath> | ||
+ | Using the fact <math>\tan(15) = 2-\sqrt{3}</math>, this yields <cmath>x = \frac{1}{3+\sqrt{3}} = \frac{3-\sqrt{3}}{6} \rightarrow 3 + 3 + 6 = \boxed{12}</cmath> | ||
==Elegant Solution== | ==Elegant Solution== |
Revision as of 17:42, 4 June 2021
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
.
Solution 3
Suppose Note that
since the triangle is equilateral, and by symmetry,
Note that if
and
, then
Also note that
Using the fact
, this yields
Elegant Solution
Why not solve in terms of the side only (single-variable beauty)? By similar triangles we obtain that
, therefore
. Then
. Using Pythagorean Theorem on
yields
. This means
, and it's clear we take the smaller root:
. Answer:
.
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.