Find area of triangle MND (ToT 2007-Spring-A2)

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Find area of triangle MND (ToT 2007-Spring-A2)

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#1 h Sep 2, 2011, 5:57 PM • 1 Y

Y by Adventure10

$K, L, M$ and $N$ are points on sides $AB, BC, CD$ and $DA$ , respectively, of the unit square $ABCD$ such that $KM$ is parallel to $BC$ and $LN$ is parallel to $AB$ . The perimeter of triangle $KLB$ is equal to $1$ . What is the area of triangle $MND$ ?

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#2 h Sep 4, 2011, 9:28 PM • 2 Y

Y by Amir Hossein, Adventure10

amparvardi wrote:

$K, L, M$ and $N$ are points on sides $AB, BC, CD$ and $DA$ , respectively, of the unit square $ABCD$ such that $KM$ is parallel to $BC$ and $LN$ is parallel to $AB$ . The perimeter of triangle $KBL$ is equal to $1$ . What is the area of triangle $MND$ ?

Proof. Denote the midpoints $X$ , $Y$ of the sides $[AB]$ , $[BC]$ respectively. Since $BX=BY=\frac 12$ - semiperimeter of $\triangle KBL$ obtain that the incircle $w$ of the square is the $B$ -exincircle of $\triangle KBL$ , i.e. $KL$ is tangent to $w$ . Denote $T\in KL\cap w$ and $KT=KX=x$ , $LT=LY=y$ . Thus, $BK=\frac 12-x$ , $BL=\frac 12-y$ and $KL=x+y$ . Since $KL^2=BK^2+BL^2\iff$ $\frac 12-(x+y)=2xy\iff$ $4xy+2(x+y)+1=2\iff$ $\boxed{(1+2x)(1+2y)=2}\ (*)$ . On other hand, $AK=DM=\frac 12+x$ , $CL=DN=\frac 12+y$ and $[MDN]=\frac 12\cdot\left(\frac 12+x\right)\left(\frac 12+y\right)$ , i.e. $[MDN]=\frac 18\cdot (1+2x)(1+2y)$ . In conclusion, using the relation $(*)$ obtain that $[MDN]=\frac 14$ .

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#3 h Sep 4, 2011, 10:24 PM • 2 Y

Y by Amir Hossein, Adventure10

Let the length of side $KB$ be $x$ and the length of side $BL$ be $y$ . Since we're given that $KM$ is parallel to $BC$ and $NL$ is parallel to $AB$ , we can conclude that side $DM=1-x$ and side $ND=1-y$ . We are given that the perimeter of triangle $KBL$ is 1, thus $x+y+ \sqrt (x^2+y^2) = 1$ . By rewriting this equation, we see that $2xy+1-2x-2y=0$ . Now, we focus on the area of triangle $MND$ . Since the length of the legs are $1-x$ and $1-y$ , the area is $(1-x)(1-y)/2$ which is equal to $(1+xy-x-y)/2$ . Aha, this looks very similar to the equation $2xy+1-2x-2y=0$ , so we subtract $1$ on both sides, divide by $2$ , and add $1$ . This gives us $1+xy-x-y=1/2$ . Finally, we divide by $2$ and get $\boxed {1/4}$ .

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#4 h Sep 5, 2011, 2:21 PM • 2 Y

Y by Adventure10, Mango247

See here PP7 and an easy its extension.

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