Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AIME I Problems/Problem 3"
(→‎Solution 7)
 
(13 intermediate revisions by 8 users not shown)
Line 1: Line 1:
−
== Problem 3 ==
+
== 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>.
  
 
== 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>. 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
 +
 
 +
== Solution 5 (A faster Vieta's) ==
 +
 
 +
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
 +
 
 +
== Solution 7==
 +
 
 +
Set side length of square to be <math>10</math>, <math>AE = x</math> and <math>EB = y</math>. From this, we get <math>y+x=10</math>, and since the area of the square will be 100, the area of the two rectangles will be <math>2xy = 10</math>. We can substitute and say that <math>2xy = x+y</math>, and subtract <math>y</math> from both sides, and then divide by <math>y</math>, getting the equation <math>\frac {x}{y} = 2x-1</math>, and doing the same thing with <math>x</math> to get <math>\frac {y}{x} = 2y-1</math>. Adding these equations, we get the desired sum to be <math>2(x+y) - 2</math>, or <math>20-2</math> which is equal to <math>\boxed {018}</math>.
 +
 
 +
~ E___
 +
 
 +
== 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}}

Latest revision as of 17:46, 3 July 2024

Problem

Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$


Solution

It's important to note that $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$

We define $a$ as the length of the side of larger inner square, which is also $EB$, $b$ as the length of the side of the smaller inner square which is also $AE$, and $s$ as the side length of $ABCD$. Since we are given that the sum of the areas of the two squares is$\frac{9}{10}$ of the the area of ABCD, we can represent that as $a^2 + b^2 = \frac{9s^2}{10}$. The sum of the two nonsquare rectangles can then be represented as $2ab = \frac{s^2}{10}$.

Looking back at what we need to find, we can represent $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ as $\dfrac{a^2 + b^2}{ab}$. We have the numerator, and dividing$\frac{s^2}{10}$ by two gives us the denominator $\frac{s^2}{20}$. Dividing $\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}$ gives us an answer of $\boxed{018}$.

Solution 2

Let the side of the square be $1$. Therefore the area of the square is also $1$. We label $AE$ as $a$ and $EB$ as $b$. Notice that what we need to find is equivalent to: $\frac{a^2+b^2}{ab}$. Since the sum of the two squares ($a^2+b^2$) is $\frac{9}{10}$ (as stated in the problem) the area of the whole square, it is clear that the sum of the two rectangles is $1-\frac{9}{10} \implies \frac{1}{10}$. Since these two rectangles are congruent, they each have area: $\frac{1}{20}$. Also note that the area of this is $ab$. Plugging this into our equation we get:

$\frac{\frac{9}{10}}{\frac{1}{20}} \implies \boxed{018}$


Solution 3

Let $AE$ be $x$, and $EB$ be $1$. Then we are looking for the value $x+\frac{1}{x}$. The areas of the smaller squares add up to $9/10$ of the area of the large square, $(x+1)^2$. Cross multiplying and simplifying we get $x^2-18x+1=0$. Rearranging, we get $x+\frac{1}{x}=\boxed{018}$

Solution 4 (Vieta's)

As before, $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$. Let $x$ represent the value of $AE=CF$. Since $EB=FB=1-x,$ the area of the two rectangles is $2x(1-x)=-2x^2+2x=\frac1{10}$. Adding $2x^2-2x$ to both sides and dividing by $2$ gives $x^2-x+\frac1{20}=0.$ Note that the two possible values of $x$ in the quadratic both sum to $1,$ like how $AE$ and $EB$ does. Therefore, $EB$ must be the other root of the quadratic that $AE$ isn't. Applying Vietas and manipulating the numerator, we get $\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{\frac{1}{20}}=\frac{1^2-\frac1{10}}{\frac1{20}}=\frac{\frac9{10}}{\frac{1}{20}}=\boxed{018}$.

Solution 5 (Fast)

Let $AE = x$ and $BE = y$. From this, we get $AB = x + y$. The problem is asking for $\frac{x}{y} + \frac{y}{x}$, which can be rearranged to give $\frac{x^2 + y^2}{xy}$. The problem tells us that $x^2 + y^2 = \frac{9(x+y)^2}{10}$. We simplify to get $x^2 + y^2 = 18xy$. Finally, we divide both sides by $xy$ to get $\frac{x^2 + y^2}{xy} = \boxed{018}$. - Spacesam

Solution 5 (A faster Vieta's)

After we get the polynomial $x^2 - 18x + 1,$ we want to find $x + \frac 1 {x}.$ Since the product of the roots of the polynomial is 1, the roots of the polynomial are simply $x, \frac 1 {x}.$ Hence $x + \frac 1 {x}$ is just $18$ by Vieta's formula, or $\boxed{018}$

Solution 6

We have the equation $x^2 + y^2$ = $\frac {9}{10} \cdot (x+y)^2$. We get $x^2 + y^2 = 18xy$. We rearrange to get $x^2 + y^2 - 18xy = 0$. Since the problem only asks us for a ratio, we assume $x$ = $1$. We have $y^2 - 18y + 1$ = $0$. Solving the quadratic yields $9 + 4 \sqrt 5$ and $9 - 4 \sqrt 5$. It doesn't really matter which one it is, since both of them are positive. We will use $9 + 4 \sqrt 5$.

We have $9 + 4 \sqrt 5 + \frac {1}{9+4 \sqrt 5}$. Rationalizing the denominator gives us $9 + 4 \sqrt 5 + \frac {9 - 4 \sqrt 5}{81-80} = (9 + 4 \sqrt 5) + (9 - 4 \sqrt 5) = 18$. Our answer is $\boxed {018}$

~Arcticturn

Solution 7

Set side length of square to be $10$, $AE = x$ and $EB = y$. From this, we get $y+x=10$, and since the area of the square will be 100, the area of the two rectangles will be $2xy = 10$. We can substitute and say that $2xy = x+y$, and subtract $y$ from both sides, and then divide by $y$, getting the equation $\frac {x}{y} = 2x-1$, and doing the same thing with $x$ to get $\frac {y}{x} = 2y-1$. Adding these equations, we get the desired sum to be $2(x+y) - 2$, or $20-2$ which is equal to $\boxed {018}$.

~ E___

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_3&oldid=221622"