Difference between revisions of "2014 AMC 12B Problems/Problem 21"

Revision as of 11:56, 29 November 2016

Problem 21

In the figure, $ABCD$ is a square of side length $1$ . The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$ ? $[asy] pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); draw(A--B--C--D--cycle); draw(K--H--G--J--cycle); draw(F--E); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N); label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W); [/asy]$ $\textbf{(A) }\frac{1}{2}(\sqrt{6}-2)\qquad\textbf{(B) }\frac{1}{4}\qquad\textbf{(C) }2-\sqrt{3}\qquad\textbf{(D) }\frac{\sqrt{3}}{6}\qquad\textbf{(E) } 1-\frac{\sqrt{2}}{2}$

Solution 1

Draw the altitude from $H$ to $AB$ and call the foot $L$ . Then $HL=1$ . Consider $HJ$ . It is the hypotenuse of both right triangles $\triangle HGJ$ and $\triangle HLJ$ , and we know $JG=HL=1$ , so we must have $\triangle HGJ\cong\triangle JLH$ by Hypotenuse-Leg congruence. From this congruence we have $LJ=HG=BE$ .

Notice that all four triangles in this picture are similar. Also, we have $LA=HD=EJ$ . So set $x=LJ=HG=BE$ and $y=LA=HD=EJ$ . Now $BE+EJ+JL+LA=2(x+y)=1$ . This means $x+y=\frac{1}{2}=BE+EJ=BJ$ , so $J$ is the midpoint of $AB$ . So $\triangle AJG$ , along with all other similar triangles in the picture, is a 30-60-90 triangle, and we have $AG=\sqrt{3} \cdot AJ=\sqrt{3}/2$ and subsequently $GD=\frac{2-\sqrt{3}}{2}=KE$ . This means $EJ=\sqrt{3} \cdot KE=\frac{2\sqrt{3}-3}{2}$ , which gives $BE=\frac{1}{2}-EJ=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}$ , so the answer is $\textbf{(C)}$ .

Solution 2

Let $BE = x$ . Let $JA = y$ . Because $\angle FKH = \angle EJK = \angle AGJ = \angle DHG$ and $\angle FHK = \angle EKJ = \angle AJG = \angle DGH$ , $\triangle KEJ, \triangle JAG, \triangle GDH, \triangle HFK$ are all similar. Using proportions and the pythagorean theorem, we find $EK = xy$ $FK = \sqrt{1-y^2}$ $EJ = x\sqrt{1-y^2}$ Because we know that $BE+EJ+AJ = EK + FK = 1$ , we can set up a systems of equations $x + x\sqrt{1-y^2} + y = 1$ $xy + \sqrt{1-y^2} = 1$ Solving for $x$ in the second equation, we get $x= \frac{1-\sqrt{1-y^2}}{y}$ Plugging this into the first equation, we get $\frac{1-\sqrt{1-y^2}}{y} + (\sqrt{1-y^2})\frac{1-\sqrt{1-y^2}}{y} + y = 1 \implies \frac{2y^2}{y}=1 \implies y=\frac{1}{2}$ Plugging into the previous equation with $x$ , we get $x= 2\left(1-\sqrt{1-\frac{1}{4}}\right) = 2\left(\frac{2-\sqrt{3}}{2} \right) = \boxed{\textbf{(C)}\ 2-\sqrt{3}}$

Solution 3

Let $BE = x$ , $EK = a$ , and $EJ = b$ . Then $x^2 = a^2 + b^2$ and because $\triangle KEJ \cong \triangle GDH$ and $\triangle KEJ \sim \triangle JAG$ , $\frac{GA}{1} = 1 - a = \frac{b}{x}$ . Furthermore, the area of the four triangles and the two rectangles sums to 1:

$1 = 2x + GA\cdot JA + ab$

$1 = 2x + (1 - a)(1 - (x + b)) + ab$

$1 = 2x + \frac{b}{x}(1 - x - b) + \left(1 - \frac{b}{x}\right)b$

$1 = 2x + \frac{b}{x} - b - \frac{b^2}{x} + b - \frac{b^2}{x}$

$x = 2x^2 + b - 2b^2$

$x - b = 2(x - b)(x + b)$

$x + b = \frac{1}{2}$

$b = \frac{1}{2} - x$

$a = 1 - \frac{b}{x} = 2 - \frac{1}{2x}$

By the Pythagorean theorem: $x^2 = a^2 + b^2$

$x^2 = \left(2 - \frac{1}{2x}\right)^2 + \left(\frac{1}{2} - x\right)^2$

$x^2 = 4 - \frac{2}{x} + \frac{1}{4x^2} + \frac{1}{4} - x + x^2$

$0 = \frac{1}{4x^2} - \frac{2}{x} + \frac{17}{4} - x$

$0 = 1 - 8x + 17x^2 - 4x^3.$

Then by the rational root theorem, this has roots $\frac{1}{4}$ , $2 - \sqrt{3}$ , and $2 + \sqrt{3}$ . The first and last roots are extraneous because they imply $a = 0$ and $x > 1$ , respectively, thus $x = \boxed{\textbf{(C)}\ 2-\sqrt{3}}$ .

Solution 4

Let $\angle FKH$ = $k$ and $CF$ = $a$ . It is shown that all four triangles in the picture are similar. From the square side lengths:

[center] $a + sin(k)*1 + cos(k)a = 1$

$sin(k)a + cos(k) = 1$ [/center] Solving for $a$ we get:

[center] $a = \frac{1-sin(k)}{cos(k) + 1} = \frac{1 - cos(k)}{sin(k)}$

$(1-sin(k))*sin(k) = (1 - cos(k))*(cos(k) + 1)$

$sin(k)-sin(k)^2 = cos(k) + 1 - cos(k)^2 - cos(k)$

$sin(k)-sin(k)^2 = sin(k)^2$

$1-sin(k) = sin(k)$

$sin(k) = \frac{1}{2}, cos(k) = \frac{\sqrt 3}{2}$

$a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3$ [/center]

@@ Line 64: / Line 64: @@
 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==
+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:
+[center]<math>a + sin(k)*1 + cos(k)a = 1</math>
+<math>sin(k)a + cos(k) = 1</math>[/center]
+Solving for <math>a</math> we get:
+[center]<math>a = \frac{1-sin(k)}{cos(k) + 1} = \frac{1 - cos(k)}{sin(k)}</math>
+<math>(1-sin(k))*sin(k) = (1 - cos(k))*(cos(k) + 1)</math>
+<math>sin(k)-sin(k)^2 = cos(k) + 1 - cos(k)^2 - cos(k)</math>
+<math>sin(k)-sin(k)^2 = sin(k)^2</math>
+<math>1-sin(k) = sin(k)</math>
+<math>sin(k) = \frac{1}{2}, cos(k) = \frac{\sqrt 3}{2}</math>
+<math>a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3</math>
+[/center]
 == See also ==
 {{AMC12 box|year=2014|ab=B|num-b=20|num-a=22}}
 {{MAA Notice}}

2014 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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