Difference between revisions of "2013 AIME I Problems/Problem 3"
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− | == 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> | ||
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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>. | ||
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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 + \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> | 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> |
Latest revision as of 19:04, 24 January 2021
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
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.