2014 AMC 12B Problems/Problem 21

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:


\[a + \sin(k) \cdot 1 + \cos(k) \cdot a = 1\]

\[\sin(k)a + \cos(k) = 1\] 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 $\boxed{\textbf{(C)}}$.

See also

2014 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AMC 12 Problems and Solutions

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