PAMO 2016 Q5

Art of Problem Solving

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PAMO 2016 Q5

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Source: PAMO 2016

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410 posts

kron

#1 h Apr 29, 2016, 5:30 PM • 3 Y

Y by JoelBinu, Adventure10, Mango247

Let $ABCD$ be a trapezium such that the sides $AB$ and $CD$ are parallel and the side $AB$ is longer than the side $CD$ . Let $M$ and $N$ be on the segments $AB$ and $BC$ respectively, such that each of the segments $CM$ and $AN$ divides the trapezium in two parts of equal area.
Prove that the segment $MN$ intersects the segment $BD$ at its midpoint.

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2016 posts

PROF65

#2 h Apr 29, 2016, 10:30 PM • 2 Y

Y by Adventure10, Mango247

Let $K$ a point on the segment $AB$ s.t $KB=DC$ since $CM$ bisects the area of the trapezium then it 's clear that $M$ is the midpoint of $AK$ ,Let $d\equiv AM,d'\equiv DC ,h$ the
height of $A$ from $DC$ and $h',h''$ the heights of N from $AB,DC$ resp..since $AN$ bisects the area of the trapezium then

$(2d+d')h'=(d+d')h\implies \frac{2d+d'}{h}=\frac{d+d'}{h'}=\frac{d}{h''}\implies \frac{h'}{h''}=\frac{d+d'}{d}$ thus $\frac{NB}{NC}=\frac{d+d'}{d}$ ;

consider $G$ the intersection of $MN$ and $CK$ ,applying Menelaus to the triangle $CKB$ and the points $M-G-N$ yields

$\frac{NB}{NC}\cdot \frac{MK}{MG}\cdot \frac{GC}{GK}=1\implies \frac{d+d'}{d}\cdot \frac{d}{d+d'}\cdot \frac{GC}{GK}=1$

thus $G$ is the midpoint of $CK$ but $KBCD$ is a parallelogram hence $G$ is the midpoint of $BD$ .
R HAS

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#3 h May 6, 2016, 4:53 PM • 1 Y

Y by Adventure10

Let h be the trapezium’s height.
(AM + CD)h/2 = (BM)h/2
AM + CD = BM.
Let [XYZ] denote area of a polygon XYZ.
[CBM] = [ANB]
[BMN] + [CMN] = [BMN] + [AMN]
[CMN] = [AMN]
Therefore, MN//AC (i.e. because triangles between parallel lines with common base have equal areas and vice versa).
Let MN intersect DC at N’
Hence AMN’C is a parallelogram because MN//AC therefore MN’//AC.
Hence CN’ = AM.
AM + CD = CN’ + CD = BM
Hence BM = DN’.
BM//DC//DN ’and BM = DN’, hence BMDN’ is a parallelogram.
In a parallelogram the diagonals bisect each other at the point of intersection. Hence MN’ and MN pass through the midpoint of BD.
Q.E.D.

This post has been edited 1 time. Last edited by Mmesomachi, Apr 4, 2019, 9:14 AM
Reason: For clarification.

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square_root_of_3

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square_root_of_3

#4 h Jun 9, 2017, 11:06 PM • 2 Y

Y by Adventure10, Mango247

It can be done analitically in 5 minutes. It's ugly so I won't post the solution.

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LOCUS

#5 h Jun 9, 2017, 11:13 PM • 2 Y

Y by Wtysuw, Adventure10

I am in middle school AND what is a cyclic series the sigma with cyc on the bottom and what does it mean.

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288 posts

#6 h May 10, 2023, 10:13 AM

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LOCUS wrote:

I am in middle school AND what is a cyclic series the sigma with cyc on the bottom and what does it mean.

For example, given $3$ variables $a, b, c$ , $\sum_{cyc} a^2b = a^2b + b^2c + c^2a$
A more rigorous definition can be found here.

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288 posts

#7 h May 10, 2023, 10:31 AM

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Mmesomachi wrote:

Let $h$ be the trapezium’s height.
Since $(AM + CD) \cdot \frac{h}{2} = (BM) \cdot \frac{h}{2}$ , we have that $AM + CD = BM$ .
Let $[XYZ]$ denote area of a polygon $XYZ$ .
$[CBM] = [ANB] \Rightarrow [BMN] + [CMN] = [BMN] + [AMN] \Rightarrow [CMN] = [AMN]$
Therefore, $MN \parallel AC$ (i.e. because triangles between parallel lines with common base have equal areas and vice versa).
Let $MN$ intersect $DC$ at $N’$ .
Hence $AMN’C$ is a parallelogram because $MN \parallel AC$ therefore $MN’ \parallel AC$ .
Hence $CN’ = AM$ .
Then, $AM + CD = CN’ + CD = BM$
Hence $BM = DN’$ .
$BM \parallel DC \parallel DN’$ and $BM = DN’$ , hence $BMDN’$ is a parallelogram.
In a parallelogram the diagonals bisect each other at the point of intersection. Hence $MN’$ and $MN$ pass through the midpoint of $BD$ . $\square$

This post has been edited 2 times. Last edited by Rukevwe, May 10, 2023, 10:32 AM

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