Difference between revisions of "2014 AMC 12B Problems/Problem 21"
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− | ==Problem | + | == Problem == |
In the figure, <math> ABCD </math> is a square of side length <math> 1 </math>. The rectangles <math> JKHG </math> and <math> EBCF </math> are congruent. What is <math> BE </math>? | In the figure, <math> ABCD </math> is a square of side length <math> 1 </math>. The rectangles <math> JKHG </math> and <math> EBCF </math> are congruent. What is <math> BE </math>? | ||
+ | |||
<asy> | <asy> | ||
pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); | pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); | ||
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label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W); | label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W); | ||
</asy> | </asy> | ||
+ | |||
<math> \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}</math> | <math> \textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}</math> | ||
− | |||
− | ==Solution 1== | + | == Solution 1 == |
Draw the altitude from <math>H</math> to <math>AB</math> and call the foot <math>L</math>. Then <math>HL=1</math>. Consider <math>HJ</math>. It is the hypotenuse of both right triangles <math>\triangle HGJ</math> and <math>\triangle HLJ</math>, and we know <math>JG=HL=1</math>, so we must have <math>\triangle HGJ\cong\triangle JLH</math> by Hypotenuse-Leg congruence. From this congruence we have <math>LJ=HG=BE</math>. | Draw the altitude from <math>H</math> to <math>AB</math> and call the foot <math>L</math>. Then <math>HL=1</math>. Consider <math>HJ</math>. It is the hypotenuse of both right triangles <math>\triangle HGJ</math> and <math>\triangle HLJ</math>, and we know <math>JG=HL=1</math>, so we must have <math>\triangle HGJ\cong\triangle JLH</math> by Hypotenuse-Leg congruence. From this congruence we have <math>LJ=HG=BE</math>. | ||
Notice that all four triangles in this picture are similar. Also, we have <math>LA=HD=EJ</math>. So set <math>x=LJ=HG=BE</math> and <math>y=LA=HD=EJ</math>. Now <math>BE+EJ+JL+LA=2(x+y)=1</math>. This means <math>x+y=\frac{1}{2}=BE+EJ=BJ</math>, so <math>J</math> is the midpoint of <math>AB</math>. So <math>\triangle AJG</math>, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have <math>AG=\sqrt{3} \cdot AJ=\sqrt{3}/2</math> and subsequently <math>GD=\frac{2-\sqrt{3}}{2}=KE</math>. This means <math>EJ=\sqrt{3} \cdot KE=\frac{2\sqrt{3}-3}{2}</math>, which gives <math>BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}</math>, so the answer is <math>\textbf{(C)}</math>. | Notice that all four triangles in this picture are similar. Also, we have <math>LA=HD=EJ</math>. So set <math>x=LJ=HG=BE</math> and <math>y=LA=HD=EJ</math>. Now <math>BE+EJ+JL+LA=2(x+y)=1</math>. This means <math>x+y=\frac{1}{2}=BE+EJ=BJ</math>, so <math>J</math> is the midpoint of <math>AB</math>. So <math>\triangle AJG</math>, along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have <math>AG=\sqrt{3} \cdot AJ=\sqrt{3}/2</math> and subsequently <math>GD=\frac{2-\sqrt{3}}{2}=KE</math>. This means <math>EJ=\sqrt{3} \cdot KE=\frac{2\sqrt{3}-3}{2}</math>, which gives <math>BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}</math>, so the answer is <math>\textbf{(C)}</math>. | ||
− | ==Solution 2== | + | == Solution 2 == |
Let <math>BE = x</math>. Let <math>JA = y</math>. Because <math>\angle FKH = \angle EJK = \angle AGJ = \angle DHG</math> and <math>\angle FHK = \angle EKJ = \angle AJG = \angle DGH</math>, <math>\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK</math> are all similar. Using proportions and the pythagorean theorem, we find | Let <math>BE = x</math>. Let <math>JA = y</math>. Because <math>\angle FKH = \angle EJK = \angle AGJ = \angle DHG</math> and <math>\angle FHK = \angle EKJ = \angle AJG = \angle DGH</math>, <math>\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK</math> are all similar. Using proportions and the pythagorean theorem, we find | ||
<cmath>EK = xy</cmath> | <cmath>EK = xy</cmath> | ||
<cmath>FK = \sqrt{1-y^2}</cmath> | <cmath>FK = \sqrt{1-y^2}</cmath> | ||
<cmath>EJ = x\sqrt{1-y^2}</cmath> | <cmath>EJ = x\sqrt{1-y^2}</cmath> | ||
− | Because we know that <math>BE+EJ+AJ = EK + FK = 1</math>, we can set up a systems of equations | + | Because we know that <math>BE+EJ+AJ = EK + FK = 1</math>, we can set up a systems of equations and solving for <math>x</math>, we get |
− | <cmath>x + x\sqrt{1-y^2} + y = 1</cmath> | + | <cmath>x + x\sqrt{1-y^2} + y = 1 \implies x= \frac{1-y}{1+\sqrt{1-y^2}}</cmath> |
− | <cmath>xy + \sqrt{1-y^2} = 1 | + | <cmath>xy + \sqrt{1-y^2} = 1 \implies x= \frac{1-\sqrt{1-y^2}}{y}</cmath> |
− | + | Now solving for <math>y</math>, we get | |
− | + | <cmath>\frac{1-y}{1+\sqrt{1-y^2}}=\frac{1-\sqrt{1-y^2}}{y} \implies y(1-y)=(1-\sqrt{1-y^2})(1+\sqrt{1-y^2}) \implies y-y^2=y^2 \implies y=\frac{1}{2}</cmath> | |
− | + | Plugging into the second equations with <math>x</math>, we get | |
− | <cmath>\frac | ||
− | Plugging into the | ||
<cmath>x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</cmath> | <cmath>x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</cmath> | ||
− | ==Solution 3== | + | == Solution 3 == |
Let <math>BE = x</math>, <math>EK = a</math>, and <math>EJ = b</math>. Then <math>x^2 = a^2 + b^2</math> and because <math>\triangle KEJ \cong \triangle GDH</math> and <math>\triangle KEJ \sim \triangle JAG</math>, <math>\frac{GA}{1} = 1 - a = \frac{b}{x}</math>. Furthermore, the area of the four triangles and the two rectangles sums to 1: | Let <math>BE = x</math>, <math>EK = a</math>, and <math>EJ = b</math>. Then <math>x^2 = a^2 + b^2</math> and because <math>\triangle KEJ \cong \triangle GDH</math> and <math>\triangle KEJ \sim \triangle JAG</math>, <math>\frac{GA}{1} = 1 - a = \frac{b}{x}</math>. Furthermore, the area of the four triangles and the two rectangles sums to 1: | ||
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Then by the rational root theorem, this has roots <math>\frac{1}{4}</math>, <math>2 - \sqrt{3}</math>, and <math>2 + \sqrt{3}</math>. The first and last roots are extraneous because they imply <math>a = 0</math> and <math>x > 1</math>, respectively, thus <math>x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</math>. | Then by the rational root theorem, this has roots <math>\frac{1}{4}</math>, <math>2 - \sqrt{3}</math>, and <math>2 + \sqrt{3}</math>. The first and last roots are extraneous because they imply <math>a = 0</math> and <math>x > 1</math>, respectively, thus <math>x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}</math>. | ||
− | ==Solution 4== | + | == Solution 4 == |
Let <math>\angle FKH</math> = <math>k</math> and <math>CF</math> = <math>a</math>. It is shown that all four triangles in the picture are similar. From the square side lengths: | Let <math>\angle FKH</math> = <math>k</math> and <math>CF</math> = <math>a</math>. It is shown that all four triangles in the picture are similar. From the square side lengths: | ||
− | |||
<cmath>a + \sin(k) \cdot 1 + \cos(k) \cdot a = 1</cmath> | <cmath>a + \sin(k) \cdot 1 + \cos(k) \cdot a = 1</cmath> | ||
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<cmath>\sin(k)a + \cos(k) = 1</cmath> | <cmath>\sin(k)a + \cos(k) = 1</cmath> | ||
Solving for <math>a</math> we get: | Solving for <math>a</math> we get: | ||
− | |||
<cmath>a = \frac{1-\sin(k)}{\cos(k) + 1} = \frac{1 - \cos(k)}{\sin(k)}</cmath> | <cmath>a = \frac{1-\sin(k)}{\cos(k) + 1} = \frac{1 - \cos(k)}{\sin(k)}</cmath> | ||
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<cmath>a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3</cmath> | <cmath>a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3</cmath> | ||
− | ==Solution 5== | + | == Solution 5 == |
Note that <math>HJ</math> is a diagonal of <math>JKHG</math>, so it must be equal in length to <math>FB</math>. Therefore, quadrilateral <math>FHJB</math> has <math>FH\parallel BJ</math>, and <math>FB=HJ</math>, so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of <math>FB</math> and <math>HJ</math>, we see that it must be a parallelogram. Therefore, <math>FH=BJ</math>. But by the symmetry in rectangle <math>FEAD</math>, we see that <math>FH=JA</math>. Therefore, <math>BJ=FH=JA</math>. We also know that <math>BJ+JA=1</math>, hence <math>BJ=JA=\frac12</math>. | Note that <math>HJ</math> is a diagonal of <math>JKHG</math>, so it must be equal in length to <math>FB</math>. Therefore, quadrilateral <math>FHJB</math> has <math>FH\parallel BJ</math>, and <math>FB=HJ</math>, so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of <math>FB</math> and <math>HJ</math>, we see that it must be a parallelogram. Therefore, <math>FH=BJ</math>. But by the symmetry in rectangle <math>FEAD</math>, we see that <math>FH=JA</math>. Therefore, <math>BJ=FH=JA</math>. We also know that <math>BJ+JA=1</math>, hence <math>BJ=JA=\frac12</math>. | ||
As <math>JG=1</math> and <math>JA=\frac12</math>, and as <math>\triangle GJA</math> is right, we know that <math>\triangle GJA</math> must be a 30-60-90 triangle. Therefore, <math>GA=\sqrt{3}/2</math> and <math>DG=1-\sqrt{3}/2</math>. But by similarity, <math>\triangle DHG</math> is also a 30-60-90 triangle, hence <math>DH=\sqrt{3}-3/2</math>. But <math>\triangle DHG\cong\triangle EJK</math>, hence <math>EJ=\sqrt{3}-3/2</math>. As <math>BJ=1/2</math>, this implies that <math>BE=BJ-EJ=1/2-\sqrt{3}+3/2=2-\sqrt{3}</math>. Thus the answer is <math>\boxed{\textbf{(C)}}</math>. | As <math>JG=1</math> and <math>JA=\frac12</math>, and as <math>\triangle GJA</math> is right, we know that <math>\triangle GJA</math> must be a 30-60-90 triangle. Therefore, <math>GA=\sqrt{3}/2</math> and <math>DG=1-\sqrt{3}/2</math>. But by similarity, <math>\triangle DHG</math> is also a 30-60-90 triangle, hence <math>DH=\sqrt{3}-3/2</math>. But <math>\triangle DHG\cong\triangle EJK</math>, hence <math>EJ=\sqrt{3}-3/2</math>. As <math>BJ=1/2</math>, this implies that <math>BE=BJ-EJ=1/2-\sqrt{3}+3/2=2-\sqrt{3}</math>. Thus the answer is <math>\boxed{\textbf{(C)}}</math>. | ||
− | == See | + | == Solution 6 == |
+ | |||
+ | Let <math>BE = x</math>, <math>KE = y</math> | ||
+ | |||
+ | <math>DG = y</math>, <math>AG = 1-y</math>, <math>KJ = x</math>, <math>JG = 1</math> | ||
+ | |||
+ | <math>\frac{AJ}{JG} = \frac{KE}{JK}</math>, <math>AJ = \frac{y}{x}</math>, <math>EJ = 1 - x - \frac{y}{x}</math>, | ||
+ | |||
+ | <math>\frac{AG}{JG} = \frac{EJ}{JK}</math>, <math>1-y = \frac{1 - x - \frac{y}{x}}{x}</math>, <math>x - xy = 1 - x - \frac{y}{x}</math>, <math>2x^2 - x^2y + y - x = 0</math> <math>(1)</math> | ||
+ | |||
+ | <math>AJ^2 + AG^2 = JG^2</math>, <math>\left( \frac{y}{x} \right) ^2 + (1-y)^2 = 1</math>, <math>\frac{y^2}{x^2} + 1 - 2y + y^2 = 1</math>, <math>x^2 y -2x^2 + y = 0</math> <math>(2)</math> | ||
+ | |||
+ | <math>(1) + (2)</math>, <math>2y - x = 0</math>, <math>y = \frac{x}{2}</math> | ||
+ | |||
+ | Substitute <math>y</math> into <math>(2)</math> we get <math>x^2 \cdot \frac{x}{2} - 2 x^2 + \frac{x}{2} = 0</math>, <math>x^2 - 4x + 1 = 0</math> | ||
+ | |||
+ | <math>x = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}</math>, as <math>x < 1</math>, <math>x = \boxed{\textbf{(C)}2 - \sqrt{3}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == See Also == | ||
{{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category: Introductory Geometry Problems]] |
Latest revision as of 01:07, 1 January 2023
Contents
Problem
In the figure, is a square of side length . The rectangles and are congruent. What is ?
Solution 1
Draw the altitude from to and call the foot . Then . Consider . It is the hypotenuse of both right triangles and , and we know , so we must have by Hypotenuse-Leg congruence. From this congruence we have .
Notice that all four triangles in this picture are similar. Also, we have . So set and . Now . This means , so is the midpoint of . So , along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have and subsequently . This means , which gives , so the answer is .
Solution 2
Let . Let . Because and , are all similar. Using proportions and the pythagorean theorem, we find Because we know that , we can set up a systems of equations and solving for , we get Now solving for , we get Plugging into the second equations with , we get
Solution 3
Let , , and . Then and because and , . Furthermore, the area of the four triangles and the two rectangles sums to 1:
By the Pythagorean theorem:
Then by the rational root theorem, this has roots , , and . The first and last roots are extraneous because they imply and , respectively, thus .
Solution 4
Let = and = . It is shown that all four triangles in the picture are similar. From the square side lengths:
Solving for we get:
Solution 5
Note that is a diagonal of , so it must be equal in length to . Therefore, quadrilateral has , and , so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of and , we see that it must be a parallelogram. Therefore, . But by the symmetry in rectangle , we see that . Therefore, . We also know that , hence .
As and , and as is right, we know that must be a 30-60-90 triangle. Therefore, and . But by similarity, is also a 30-60-90 triangle, hence . But , hence . As , this implies that . Thus the answer is .
Solution 6
Let ,
, , ,
, , ,
, , ,
, , ,
, ,
Substitute into we get ,
, as ,
See Also
2014 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.