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2005 Indonesia MO Problems/Problem 7

Problem

Let $ABCD$ be a convex quadrilateral. Square $AB_1A_2B$ is constructed such that the two vertices $A_2,B_1$ is located outside $ABCD$. Similarly, we construct squares $BC_1B_2C$, $CD_1C_2D$, $DA_1D_2A$. Let $K$ be the intersection of $AA_2$ and $BB_1$, $L$ be the intersection of $BB_2$ and $CC_1$, $M$ be the intersection of $CC_2$ and $DD_1$, and $N$ be the intersection of $DD_2$ and $AA_1$. Prove that $KM$ is perpendicular to $LN$.

Solution

Let the coordinates of $A$ be $(2a,0)$, the coordinates of $B$ be $(2b_1, 2b_2)$, the coordinates of $C$ be $(-2c,0)$, and the coordinates of $D$ be $(2d_1, -2d_2)$, where all variables are rational and $a, b_2, c, d_2 \ge 0$.

[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4); draw(A--B--C--D--A); dot(A); label("$(2a,0)$",A,NE); dot(B); label("$(2b_1,2b_2)$",B,N); dot(C); label("$(-2c,0)$",C,NW); dot(D); label("$(2d_1,-2d_2)$",D,S); [/asy]

Let $X$ be the midpoint of $AB$, which is point $(a+b_1,b_2)$. Additionally, mark points $Y = (2b_1,b_2)$, $K = (x,y)$, and $Z = (x,b_2)$.

[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-0.2,xmax=10.2,ymin=-0.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8); draw(A--B); dot(A); label("$(2a,0)$",A,NE); dot(B); label("$(2b_1,2b_2)$",B,N); dot((5,4)); label("$(a+b_1,b_2)$",(5,4),NE); dot((9,7)); label("$(x,y)$",(9,7),NE); dot((2,4)); label("$(2b_1,b_2)$",(2,4),SW); dot((9,4)); label("$(x,b_2)$",(9,4),SE); [/asy]

Note that since $K$ is the center of square $AB_1A_2B$, $BX \perp KX$ and $BX = KX$. Additionally, $BY \parallel KZ$ and $BY \perp YZ$.


$YZ$ is a line, so $\angle BXY + \angle KXZ + \angle BXK = 180^\circ$. Since $BX \perp KX$, $\angle BXK = 90^\circ$, so $\angle BXY + \angle KXZ = 90^\circ$. Additionally, because $BXY$ is a right triangle, $\angle YBX + \angle YXB = 90^\circ$. Rearranging and substituting results in $\angle KXZ = \angle YBX$.


Since both $BYX$ and $XZK$ are right triangles, by AAS Congruency, $\triangle BYX \cong \triangle XZK$. Therefore $BY = XZ = b_2$ and $YX = ZK = a - b_1$. From this information, the coordinates of $K$ are $(a+b_1+b_2, b_2 + a - b_1)$.

[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=2,gy=2; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); draw((-10,0)--(10,0),Arrows); draw((0,10)--(0,-10),Arrows); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); pair A=(8,0), B=(2,8), C=(-6,0), D=(-4,-4), K=(9,7), L=(-6,8), M=(-7,-3), N=(4,-8); draw(A--B--C--D--A); dot(A); label("$(2a,0)$",A,NE); dot(B); label("$(2b_1,2b_2)$",B,NE); dot(C); label("$(-2c,0)$",C,NW); dot(D); label("$(2d_1,-2d_2)$",D,S); dot(K); label("$K$",K,NE); dot(L); label("$L$",L,NW); dot(M); label("$M$",M,SW); dot(N); label("$N$",N,SE); [/asy]


By using similar reasoning, the coordinates of $L$ are $(b_1-b_2-c,b_2+b_1+c)$, the coordinates of $M$ are $(-c+d_1-d_2,-d_2-c-d_1)$, and the coordinates of $N$ are $(a+d_1+d_2,d_1-d_2-a)$.


The slope of $KM$ is $\frac{a+b_2-b_1+c+d_2+d_1}{a+b_1+b_2+c+d_2-d_1}$. The slope of $LN$ is $\frac{b_1+b_2+c+a+d_2-d_1}{b_1-b_2-c-a-d_1-d_2}$. The product of the two slopes is $-1$, so $KM \perp LN$.

See Also

2005 Indonesia MO (Problems)
Preceded by
Problem 6 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 8
All Indonesia MO Problems and Solutions
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