Difference between revisions of "2013 AIME I Problems/Problem 3"
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== Solution 5 (Fast) == | == Solution 5 (Fast) == | ||
Let <math>AE = x</math> and <math>BE = y</math>. From this, we get <math>AB = x + y</math>. The problem is asking for <math>\frac{x}{y} + \frac{y}{x}</math>, which can be rearranged to give <math>\frac{x^2 + y^2}{xy}</math>. The problem tells us that <math>x^2 + y^2 = \frac{9(x+y)^2}{10}</math>. We simplify to get <math>x^2 + y^2 = 18xy</math>. Finally, we divide both sides by <math>xy</math> to get <math>\frac{x^2 + y^2}{xy} = \boxed{018}</math>. - Spacesam | Let <math>AE = x</math> and <math>BE = y</math>. From this, we get <math>AB = x + y</math>. The problem is asking for <math>\frac{x}{y} + \frac{y}{x}</math>, which can be rearranged to give <math>\frac{x^2 + y^2}{xy}</math>. The problem tells us that <math>x^2 + y^2 = \frac{9(x+y)^2}{10}</math>. We simplify to get <math>x^2 + y^2 = 18xy</math>. Finally, we divide both sides by <math>xy</math> to get <math>\frac{x^2 + y^2}{xy} = \boxed{018}</math>. - Spacesam | ||
+ | |||
+ | == A faster Vieta's == | ||
+ | |||
+ | After we get the polynomial <math>x^2 - 18x + 1,</math> we want to find <math>x + \frax 1 {x}.</math> Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply <math>x, \frac 1 {x}.</math> Hence <math>x + \frax 1 {x}</math> is just <math>18</math> by Vieta's formula. | ||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=2|num-a=4}} | {{AIME box|year=2013|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:45, 20 February 2019
Contents
[hide]Problem 3
Let be a square, and let
and
be points on
and
respectively. The line through
parallel to
and the line through
parallel to
divide
into two squares and two nonsquare rectangles. The sum of the areas of the two squares is
of the area of square
Find
Solution
It's important to note that is equivalent to
We define as the length of the side of larger inner square, which is also
,
as the length of the side of the smaller inner square which is also
, and
as the side length of
. Since we are given that the sum of the areas of the two squares is
of the the area of ABCD, we can represent that as
. The sum of the two nonsquare rectangles can then be represented as
.
Looking back at what we need to find, we can represent as
. We have the numerator, and dividing
by two gives us the denominator
. Dividing
gives us an answer of
.
Solution 2
Let the side of the square be . Therefore the area of the square is also
.
We label
as
and
as
. Notice that what we need to find is equivalent to:
.
Since the sum of the two squares (
) is
(as stated in the problem) the area of the whole square, it is clear that the
sum of the two rectangles is
. Since these two rectangles are congruent, they
each have area:
. Also note that the area of this is
. Plugging this into our equation we get:
Solution 3
Let be
, and
be
. Then we are looking for the value
. The areas of the smaller squares add up to
of the area of the large square,
. Cross multiplying and simplifying we get
. Rearranging, we get
Solution 4 (Vieta)
As before, is equivalent to
. Let
represent the value of
. Since
the area of the two rectangles is
. Adding
to both sides and dividing by
gives
Note that the two possible values of
in the quadratic both sum to
like how
and
does. Therefore,
must be the other root of the quadratic that
isn't. Applying Vietas and manipulating the numerator, we get
.
Solution 5 (Fast)
Let and
. From this, we get
. The problem is asking for
, which can be rearranged to give
. The problem tells us that
. We simplify to get
. Finally, we divide both sides by
to get
. - Spacesam
A faster Vieta's
After we get the polynomial we want to find $x + \frax 1 {x}.$ (Error compiling LaTeX. Unknown error_msg) Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply
Hence $x + \frax 1 {x}$ (Error compiling LaTeX. Unknown error_msg) is just
by Vieta's formula.
See also
2013 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.