Difference between revisions of "2006 AIME II Problems/Problem 6"
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− | == Problem == 1+ | + | == 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> | ||
== Solution 1 == | == Solution 1 == | ||
Line 66: | Line 67: | ||
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}}{6}</math>. Answer: <math>\boxed{12}</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}}{6}</math>. Answer: <math>\boxed{12}</math>. | ||
+ | == Solution 5 (First part is similar to Solution 2) == | ||
+ | Since <math>AEF</math> is equilateral, <math>AE=EF</math>. Let <math>BE=x</math>. By the [[Pythagorean theorem]], <math>1+x^2=2(1-x)^2</math>. Simplifying, we get <math>x^2-4x+1=0</math>. By the quadratic formula, the roots are <math>2 \pm \sqrt{3}</math>. Since <math>x<1</math>, we discard the root with the "+", giving <math>x=2-\sqrt{3}</math>. | ||
+ | <asy> | ||
+ | real n; | ||
+ | n=0.26794919243; | ||
+ | real m; | ||
+ | m=0.2113248654; | ||
+ | draw((0,0)--(0,n)--(1,0)--(0,0)); | ||
+ | draw((0,m)--(m,m)--(m,0)); | ||
+ | label((0,0), "$B$",SW); | ||
+ | label((0,n), "$E$",SW); | ||
+ | label((0,m), "$M$",SW); | ||
+ | label((1,0), "$A$",SW); | ||
+ | label((m,0), "$N$",SW); | ||
+ | label((m,m), "$K$",NE); | ||
+ | </asy> | ||
+ | Let the side length of the square be s. Since <math>MEK</math> is similar to <math>ABE</math>, <math>s=\frac{2-\sqrt{3}-s}{2-\sqrt{3}}</math>. Solving, we get <math>s=\frac{3-\sqrt{3}}{6}</math> and the final answer is <math>\boxed{012}</math>. | ||
== See also == | == See also == |
Latest revision as of 12:23, 8 April 2024
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
.
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:
.
Solution 5 (First part is similar to Solution 2)
Since is equilateral,
. Let
. By the Pythagorean theorem,
. Simplifying, we get
. By the quadratic formula, the roots are
. Since
, we discard the root with the "+", giving
.
Let the side length of the square be s. Since
is similar to
,
. Solving, we get
and the final answer is
.
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.