Points $K, L, M, N$ lie on the sides $AB, BC, CD, DA$ of a square $ABCD$, respectively, such that the area of $KLMN$ is equal to one half of the area of $ABCD$. Prove that some diagonal of $KLMN$ is parallel to some side of $ABCD$. Proposed by Josef Tkadlec - Czech Republic
Problem
Source: Iranian Geometry Olympiad 2021 IGO Elementary p2
Tags: inscribed, geometry, areas, square
qwertyboyfromalotoftime
25.01.2022 19:02
parmenides51 wrote: Points $K, L, M, N$ lie on the sides $AB, BC, CD, DA$ of a square $ABCD$, respectively, such that the area of $KLMN$ is equal to one half of the area of $ABCD$. Prove that some diagonal of $KLMN$ is parallel to some side of $ABCD$. Proposed by Josef Tkadlec - Czech Republic
$\boxed{\text{Solution 1.}}$ Let $[P]$ denote the area of a polygon $P$ . Suppose that $LN \nparallel AB$ and let $N' \ne N$ be the point on $AD$ such that $LN' \parallel AB.$ Note that $[KLMN'] = \frac12 [ABCD].$ Thus $[KLMN] = [KLMN']$, hence $[KMN] = [KMN']$ implying that $KM \parallel NN' = AD.$
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qwertyboyfromalotoftime
25.01.2022 19:07
$\boxed{\text{Solution 2.}}$ Denote the length as on the figure and let the side of $ABCD$ be $a.$ The sum of the areas of the right triangles in the corners is $\frac {a^2}2$ and so $$\frac 12wx + \frac 12(1-x)y +\frac 12 (1-y)(1-z) + \frac 12z(1-w) = \frac 12a^2$$which can be modifier into $(w-y)(x-z),$ and so $x=z$ or $y=w,$ which gives the desired conclusion.
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lifeismathematics
18.11.2022 21:11
so consider the markups i did in the diagram
notice $[DMN]+[MCL]+[LBK]+[KAN]=\frac{a^2}{2}$
$\implies (a-w)x+(a-x)y+w(a-z)+z(a-y)=a^2\implies (a-x-z)(a-y-w)=0\implies a=x+z$ or $a=y+w$
WLOG $a=x+z$
then we have $MC\parallel KB$ and $MC=KB$ which makes $MCBK$ a parallelogram and hence $MK\parallel BC$
which proves the desired result $\blacksquare$
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Laan
20.07.2023 04:20
Case 1: ${MK}\parallel{BC}\parallel{AD}(true)$ Case 2 : ${MK}\nparallel{AD,BC}$ Construct $E,F$ lie on line $CD$ such that ${NF}\parallel{LE}\parallel{MK}$ ${NF}\parallel{MK} \Rightarrow S_{MFK}=S_{MNK}$ ${LE }\parallel{MK} \Rightarrow S_{MEK}=S_{MLK}$ Therefore: $S_{KLMN}=S_{KEF} \Rightarrow S_{KEF}={\frac{1}{2}}\cdot{S_{ABCD}}=S_{DKC}$ $\Rightarrow EF=CD \Rightarrow DF=CE$ Now we can proof $\Delta{DNF}=\Delta{CLE} $ Thus, $FN=LE$ and ${FN}\parallel{LE}$ so $NLEF$ is parallelogram. $\Rightarrow {NL}\parallel{CD}(true)$
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