Difference between revisions of "2013 AIME I Problems/Problem 3"
Neeneemath (talk | contribs) (→Solution 5 (Fast)) |
Pi is 3.14 (talk | contribs) (→Video Solution) |
||
(8 intermediate revisions by 5 users not shown) | |||
Line 1: | Line 1: | ||
− | == Problem | + | == Problem == |
Let <math>ABCD</math> be a square, and let <math>E</math> and <math>F</math> be points on <math>\overline{AB}</math> and <math>\overline{BC},</math> respectively. The line through <math>E</math> parallel to <math>\overline{BC}</math> and the line through <math>F</math> parallel to <math>\overline{AB}</math> divide <math>ABCD</math> into two squares and two nonsquare rectangles. The sum of the areas of the two squares is <math>\frac{9}{10}</math> of the area of square <math>ABCD.</math> Find <math>\frac{AE}{EB} + \frac{EB}{AE}.</math> | Let <math>ABCD</math> be a square, and let <math>E</math> and <math>F</math> be points on <math>\overline{AB}</math> and <math>\overline{BC},</math> respectively. The line through <math>E</math> parallel to <math>\overline{BC}</math> and the line through <math>F</math> parallel to <math>\overline{AB}</math> divide <math>ABCD</math> into two squares and two nonsquare rectangles. The sum of the areas of the two squares is <math>\frac{9}{10}</math> of the area of square <math>ABCD.</math> Find <math>\frac{AE}{EB} + \frac{EB}{AE}.</math> | ||
Line 26: | Line 26: | ||
Let <math>AE</math> be <math>x</math>, and <math>EB</math> be <math>1</math>. Then we are looking for the value <math>x+\frac{1}{x}</math>. The areas of the smaller squares add up to <math>9/10</math> of the area of the large square, <math>(x+1)^2</math>. Cross multiplying and simplifying we get <math>x^2-18x+1=0</math>. Rearranging, we get <math>x+\frac{1}{x}=\boxed{018}</math> | Let <math>AE</math> be <math>x</math>, and <math>EB</math> be <math>1</math>. Then we are looking for the value <math>x+\frac{1}{x}</math>. The areas of the smaller squares add up to <math>9/10</math> of the area of the large square, <math>(x+1)^2</math>. Cross multiplying and simplifying we get <math>x^2-18x+1=0</math>. Rearranging, we get <math>x+\frac{1}{x}=\boxed{018}</math> | ||
− | == Solution 4 (Vieta)== | + | == Solution 4 (Vieta's)== |
As before, <math>\dfrac{AE}{EB} + \dfrac{EB}{AE}</math> is equivalent to <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math>. Let <math>x</math> represent the value of <math>AE=CF</math>. Since <math>EB=FB=1-x,</math> the area of the two rectangles is <math>2x(1-x)=-2x^2+2x=\frac1{10}</math>. Adding <math>2x^2-2x</math> to both sides and dividing by <math>2</math> gives <math>x^2-x+\frac1{20}=0.</math> Note that the two possible values of <math>x</math> in the quadratic both sum to <math>1,</math> like how <math>AE</math> and <math>EB</math> does. Therefore, <math>EB</math> must be the other root of the quadratic that <math>AE</math> isn't. Applying Vietas and manipulating the numerator, we get <math>\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{\frac{1}{20}}=\frac{1^2-\frac1{10}}{\frac1{20}}=\frac{\frac9{10}}{\frac{1}{20}}=\boxed{018}</math>. | As before, <math>\dfrac{AE}{EB} + \dfrac{EB}{AE}</math> is equivalent to <math>\dfrac{AE^2 + EB^2}{(AE)(EB)}</math>. Let <math>x</math> represent the value of <math>AE=CF</math>. Since <math>EB=FB=1-x,</math> the area of the two rectangles is <math>2x(1-x)=-2x^2+2x=\frac1{10}</math>. Adding <math>2x^2-2x</math> to both sides and dividing by <math>2</math> gives <math>x^2-x+\frac1{20}=0.</math> Note that the two possible values of <math>x</math> in the quadratic both sum to <math>1,</math> like how <math>AE</math> and <math>EB</math> does. Therefore, <math>EB</math> must be the other root of the quadratic that <math>AE</math> isn't. Applying Vietas and manipulating the numerator, we get <math>\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{\frac{1}{20}}=\frac{1^2-\frac1{10}}{\frac1{20}}=\frac{\frac9{10}}{\frac{1}{20}}=\boxed{018}</math>. | ||
Line 33: | Line 33: | ||
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 == | + | == Solution 5 (A faster Vieta's) == |
− | After we get the polynomial <math>x^2 - 18x + 1,</math> we want to find <math>x + \ | + | After we get the polynomial <math>x^2 - 18x + 1,</math> we want to find <math>x + \frac 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 + \frac 1 {x}</math> is just <math>18</math> by Vieta's formula, or <math>\boxed{018}</math> |
+ | |||
+ | == Solution 6== | ||
+ | We have the equation <math>x^2 + y^2</math> = <math>\frac {9}{10} \cdot (x+y)^2</math>. We get <math>x^2 + y^2 = 18xy</math>. We rearrange to get <math>x^2 + y^2 - 18xy = 0</math>. Since the problem only asks us for a ratio, we assume <math>x</math> = <math>1</math>. We have <math>y^2 - 18y + 1</math> = <math>0</math>. Solving the quadratic yields <math>9 + 4 \sqrt 5</math> and <math>9 - 4 \sqrt 5</math>. It doesn't really matter which one it is, since both of them are positive. We will use <math>9 + 4 \sqrt 5</math>. | ||
+ | |||
+ | We have <math>9 + 4 \sqrt 5 + \frac {1}{9+4 \sqrt 5}</math>. Rationalizing the denominator gives us <math>9 + 4 \sqrt 5 + \frac {9 - 4 \sqrt 5}{81-80} = (9 + 4 \sqrt 5) + (9 - 4 \sqrt 5) = 18</math>. Our answer is <math>\boxed {018}</math> | ||
+ | |||
+ | ~Arcticturn | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/FWmrHV1dWPM?t=39 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=kz3ZX4PT-_0 | ||
+ | ~Shreyas S | ||
== 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 03:51, 23 January 2023
Contents
Problem
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's)
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
Solution 5 (A faster Vieta's)
After we get the polynomial we want to find
Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply
Hence
is just
by Vieta's formula, or
Solution 6
We have the equation =
. We get
. We rearrange to get
. Since the problem only asks us for a ratio, we assume
=
. We have
=
. Solving the quadratic yields
and
. It doesn't really matter which one it is, since both of them are positive. We will use
.
We have . Rationalizing the denominator gives us
. Our answer is
~Arcticturn
Video Solution by OmegaLearn
https://youtu.be/FWmrHV1dWPM?t=39
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=kz3ZX4PT-_0 ~Shreyas S
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.