2014 AMC 12B Problems/Problem 21

Problem

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 and solving for $x$ , we get $x + x\sqrt{1-y^2} + y = 1 \implies x= \frac{1-y}{1+\sqrt{1-y^2}}$ $xy + \sqrt{1-y^2} = 1 \implies x= \frac{1-\sqrt{1-y^2}}{y}$ Now solving for $y$ , we get $\frac{1-y}{1+\sqrt{1-y^2}}=\frac{1-\sqrt{1-y^2}}{y} \implies y(1-y)=(1-\sqrt{1-y^2})(1+\sqrt{1-y^2}) \implies y-y^2=y^2 \implies y=\frac{1}{2}$ Plugging into the second equations 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:

$a + \sin(k) \cdot 1 + \cos(k) \cdot a = 1$

$\sin(k)a + \cos(k) = 1$ Solving for $a$ we get:

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

$(1-\sin(k)) \cdot sin(k) = (1 - \cos(k))\cdot(\cos(k) + 1)$

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

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

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

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

$a = \frac{1 - \frac{\sqrt 3}{2}}{\frac{1}{2}} = 2 - \sqrt 3$

Solution 5

Note that $HJ$ is a diagonal of $JKHG$ , so it must be equal in length to $FB$ . Therefore, quadrilateral $FHJB$ has $FH\parallel BJ$ , and $FB=HJ$ , so it must be either an isosceles trapezoid or a parallelogram. But due to the slope of $FB$ and $HJ$ , we see that it must be a parallelogram. Therefore, $FH=BJ$ . But by the symmetry in rectangle $FEAD$ , we see that $FH=JA$ . Therefore, $BJ=FH=JA$ . We also know that $BJ+JA=1$ , hence $BJ=JA=\frac12$ .

As $JG=1$ and $JA=\frac12$ , and as $\triangle GJA$ is right, we know that $\triangle GJA$ must be a 30-60-90 triangle. Therefore, $GA=\sqrt{3}/2$ and $DG=1-\sqrt{3}/2$ . But by similarity, $\triangle DHG$ is also a 30-60-90 triangle, hence $DH=\sqrt{3}-3/2$ . But $\triangle DHG\cong\triangle EJK$ , hence $EJ=\sqrt{3}-3/2$ . As $BJ=1/2$ , this implies that $BE=BJ-EJ=1/2-\sqrt{3}+3/2=2-\sqrt{3}$ . Thus the answer is $\boxed{\textbf{(C)}}$ .

Solution 6

Let $BE = x$ , $KE = y$

$DG = y$ , $AG = 1-y$ , $KJ = x$ , $JG = 1$

$\frac{AJ}{JG} = \frac{KE}{JK}$ , $AJ = \frac{y}{x}$ , $EJ = 1 - x - \frac{y}{x}$ ,

$\frac{AG}{JG} = \frac{EJ}{JK}$ , $1-y = \frac{1 - x - \frac{y}{x}}{x}$ , $x - xy = 1 - x - \frac{y}{x}$ , $2x^2 - x^2y + y - x = 0$ $(1)$

$AJ^2 + AG^2 = JG^2$ , $\left( \frac{y}{x} \right) ^2 + (1-y)^2 = 1$ , $\frac{y^2}{x^2} + 1 - 2y + y^2 = 1$ , $x^2 y -2x^2 + y = 0$ $(2)$

$(1) + (2)$ , $2y - x = 0$ , $y = \frac{x}{2}$

Substitute $y$ into $(2)$ we get $x^2 \cdot \frac{x}{2} - 2 x^2 + \frac{x}{2} = 0$ , $x^2 - 4x + 1 = 0$

$x = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$ , as $x < 1$ , $x = \boxed{\textbf{(C)}2 - \sqrt{3}}$

~isabelchen

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