Difference between revisions of "2006 AIME II Problems/Problem 6"
Expilncalc (talk | contribs) (Added solution) |
Expilncalc (talk | contribs) (→Elegant Solution: Changed variable name.) |
||
Line 62: | Line 62: | ||
==Elegant Solution== | ==Elegant Solution== | ||
− | Why not solve in terms of <math> | + | Why not solve in terms of the side <math>x</math> only (single-variable beauty)? By similar triangles we obtain that <math>BE=\frac{x}{1-x}</math>, therefore <math>CE=\frac{1-2x}{1-x}</math>. Then <math>AE=\sqrt{2}*\frac{1-2x}{1-x}</math>. Using Pythagorean Theorem on <math>\triangle{ABE}</math> yields <math>\frac{x^2}{(1-x)^2} + 1 = 2 * \frac{(1-2x)^2}{(1-x)^2}</math>. This means <math>6x^2-6x+1=0</math>, and it's clear we take the smaller root: <math>x=\frac{3-\sqrt{3}}{3}</math>. Answer: <math>\boxed{012}</math>. |
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2006|n=II|num-b=5|num-a=7}} | {{AIME box|year=2006|n=II|num-b=5|num-a=7}} |
Revision as of 19:31, 1 January 2018
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
.
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.