Straightedge to draw midpoints

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anantmudgal09

1979 posts

anantmudgal09

#1 h Mar 7, 2021, 10:31 AM • 9 Y

Y by nikaaryu, Rg230403, meet18, Aritra12, p_square, NumberX, OlympusHero, sotpidot, bobjoe123

Betal marks $2021$ points on the plane such that no three are collinear, and draws all possible segments joining these. He then chooses any $1011$ of these segments, and marks their midpoints. Finally, he chooses a segment whose midpoint is not marked yet, and challenges Vikram to construct its midpoint using only a straightedge. Can Vikram always complete this challenge?

Note. A straightedge is an infinitely long ruler without markings, which can only be used to draw the line joining any two given distinct points.

Proposed by Prithwijit De and Sutanay Bhattacharya

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anantmudgal09

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anantmudgal09

#2 h Mar 7, 2021, 11:11 AM • 11 Y

Y by nikaaryu, Pluto1708, RudraRockstar, Dr_Vex, Aimingformygoal, cowcow, Euler1728, sotpidot, Siddharth03, hakN, BVKRB-

My solution (terser than the official):

Solution

This post has been edited 1 time. Last edited by anantmudgal09, Mar 7, 2021, 11:17 AM

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amar_04

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amar_04

#3 h Mar 7, 2021, 11:29 AM • 7 Y

Y by Dr_Vex, Aimingformygoal, Mathematicsislovely, gowtham_doma, sanyalarnab, Bumblebee60, PRMOisTheHardestExam

Kinda straightforward. But I liked this one. We will be using our weapons Parallelo456 and Midsegmento789 from here. Notice that if there $n$ odd points in the plane, and there are $\frac{(n+1)}{2}$ line segments joining some of the $n$ points, then atleast two of the line segments have a common vertex. So we are just left to prove this famous fact.

$\textbf{FACT:}$ $ABC$ be a triangle with midpoints $BC,CA,AB$ marked, then for any segment $PQ$ in the plane we can find its midpoint only using a straightedge.

Using Parallelo456 we can draw a line through $P$ parallel to $AB$ and a line through $Q$ parallel to $AC$ . Suppose both those lines meet at $V$ . Using Midsegmento789 we can get the midpoints $X,Y$ of $PV,QV$ . Now let $QX\cap PY=M$ . Thus $KM\cap PQ$ is the midpoint of the segment $PQ$ . $\quad\blacksquare$

This post has been edited 4 times. Last edited by amar_04, Mar 7, 2021, 11:32 AM

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Dr_Vex

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Dr_Vex

#4 h Mar 7, 2021, 11:32 AM • 1 Y

Y by amar_04

More (similar) ruler constructions can be found here

Edit: A similar idea was used in solving Sharygin P21 this year as pointed by amar_04

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554183

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554183

#5 h Mar 7, 2021, 11:35 AM

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$\text{[math]}$

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a2048

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a2048

#6 h Mar 7, 2021, 11:47 AM • 2 Y

Y by ChiefEditorQuibbler, spicemax

How to draw parallel line using only straightedge ?

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Sumitrajput0271

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Sumitrajput0271

#7 h Mar 7, 2021, 11:50 AM

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I think you didn't considered the case when four points form a trapezium with midpoint marked on opposite sides .
With use of straightedge you can find midpoint of other 2 sides and diagonals of that figure

This post has been edited 1 time. Last edited by Sumitrajput0271, Mar 7, 2021, 11:51 AM
Reason: Spelling mistake

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NumberX

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NumberX

#8 h Mar 7, 2021, 1:23 PM • 1 Y

Y by Dem0nfang

Here's another (a little complicated for no reason at all) sol. We know we have a triangle and we can prove that we can convert any connected component into a clique. Now we prove that Given a triangle ABC, with midpoints DEF, we can connect BP for any P. To see this let PC meet EF at G. Then notice that the line connecting $CF \cap DE$ and $BE \cap DF$ is parallel and midway between EF and BC. let this intersect DG at M, then if M intersects DG at K and K intersect EF at S, then SGCD is a parallelogram and so just extend DS to meet PB at its midpoint. Now if we connect everything to B, its all connected so boom.

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508669

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508669

#9 h Mar 7, 2021, 1:38 PM • 1 Y

Y by RudraRockstar

anantmudgal09 wrote:

My solution (terser than the official):

Solution

I did prove the lemma that you mentioned and showed and used php, can I expect at least 3 marks?

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spicemax

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spicemax

#10 h Mar 7, 2021, 3:12 PM

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How can one draw a parallel line using only a straightedge? :huh:

Thanks @below

This post has been edited 1 time. Last edited by spicemax, Mar 7, 2021, 3:45 PM
Reason: (Question was answered)

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amar_04

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amar_04

#11 h Mar 7, 2021, 3:28 PM • 7 Y

Y by lneis1, Mathematicsislovely, Aimingformygoal, Siddharth03, spicemax, Bumblebee60, Aritra12

spicemax wrote:

How can one draw a parallel line using only a straightedge? :huh:

It is mentioned in the problem that the line segments have their midpoints marked. So your question should be "how can one draw a line parallel to a line segment which has its midpoint marked". Refer to the diagram here
This is how we do it. The red segment $AB$ and its midpoint $M$ are given and we want to draw a line through $P$ parallel to $AB$ . Take $X$ to be a point on $AP$ . Let $XM\cap BP=V$ and let $AV\cap XB=Q$ . By Ceva's Theorem we get $\frac{XP}{PA}\cdot\frac{AM}{MB}\cdot\frac{BQ}{QA}=1$ and as $AM=MB$ we have $\frac{XP}{PA}=\frac{XQ}{QB}$ . Hence, by Thales we have $PQ\parallel AB$ .

This post has been edited 3 times. Last edited by amar_04, Mar 7, 2021, 3:31 PM

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p_square

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p_square

#12 h Mar 7, 2021, 3:59 PM • 3 Y

Y by lneis1, Mathematicsislovely, spicemax

The statement of this theorem felt slightly scam, since all we really need is that there are points $A,B,C$ such that the midpoint of $AB = M$ and midpoint of $AC = N$ . With the above configuration, one can find the midpoint of any segment $UV$ . Observe by PHP that we indeed do have a structure similar to what is above described. Even so, I really liked the problem.

The following two claims are crucial. [having seen them before helped :p]
Claim 1: Given segment $AB$ with midpoint $M$ , once can construct a line parallel to $AB$
Proof: Pick arbitrary $C$ , $X$ on $CM$ and let $AX, BX$ meet $BC, AC$ at $Y, Z$ . By Ceva's $YZ$ is the desired line.

Claim 2: Given segment $AB$ and parallel line $YZ$ , one can construct the midpoint of $BC$
Proof: Let $AY$ meet $BZ$ at $X$ and let $AZ$ meet $BY$ at $C$ . By Ceva's $CX$ passes through the midpoint of $AB$ .

Main proof:
We use this rewording:
The statement of this theorem felt slightly scam, since all we really need is that there are points $A,B,C$ such that the midpoint of $AB = M$ and midpoint of $AC = N$ . With the above configuration, find the midpoint of any segment $UV$ .

Let $X, Y$ be the midpoints of $MB$ and $NC$ . By claim 1 we can draw line parallel to $AB\AC$ and by claim 2 we can draw in $X, Y$ . Let $BC$ , $XY$ and $MN$ meet $UV$ at $I, J$ and $K$ . Observe that $J$ is the midpoint of $IK$ . By claim 1 draw a line parallel to $IK$ , and by claim $2$ draw in the midpoint of $UV$ .

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guptaamitu1

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guptaamitu1

#13 h Mar 7, 2021, 5:35 PM • 2 Y

Y by amar_04, Mathematicsislovely

How can we draw parallel lines using a straight edge only (we don't even have a compass) ??

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biomathematics

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biomathematics

#14 h Mar 7, 2021, 5:37 PM • 1 Y

Y by p_square

guptaamitu1 wrote:

How can we draw parallel lines using a straight edge only (we don't even have a compass) ??

read the solution immediately above this post of yours.

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Mathematicsislovely

245 posts

Mathematicsislovely

#15 h Mar 7, 2021, 5:49 PM • 5 Y

Y by RAMUGAUSS, amar_04, Euler1728, Aritra12, Mango247

Lemma: Given a segment $AB$ .And another segment $CD$ parallel to $AB$ and midpoint $M$ of $CD$ .Then it is possible to construct midpoint of $AB$ using straightedge.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2., xmax = 10., ymin = -1., ymax = 3.5; /* image dimensions */ /* draw figures */ draw((2.66,2.28)--(9.1,2.28), linewidth(0.8)); draw((3.12,0.12)--(8.48,0.12), linewidth(0.8)); draw((2.66,2.28)--(8.48,0.12), linewidth(0.8)); draw((9.1,2.28)--(3.12,0.12), linewidth(0.8)); draw((xmin, 26.999999999999893*xmin-156.4799999999994)--(xmax, 26.999999999999893*xmax-156.4799999999994), linewidth(0.8)); /* line */ /* dots and labels */ dot((2.66,2.28),linewidth(2.pt) + dotstyle); label("$A$", (2.74,2.36), NE * labelscalefactor); dot((9.1,2.28),linewidth(2.pt) + dotstyle); label("$B$", (9.18,2.36), NE * labelscalefactor); dot((3.12,0.12),linewidth(2.pt) + dotstyle); label("$C$", (3.2,0.2), NE * labelscalefactor); dot((8.48,0.12),linewidth(2.pt) + dotstyle); label("$D$", (8.56,0.2), NE * labelscalefactor); dot((5.8,0.12),linewidth(2.pt) + dotstyle); label("$M$", (5.88,0.2), NE * labelscalefactor); dot((5.836338983050848,1.1011525423728812),linewidth(2.pt) + dotstyle); label("$X$", (5.92,1.18), NE * labelscalefactor); dot((5.88,2.28),linewidth(2.pt) + dotstyle); label("$F$", (5.96,2.36), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Proof. At least one of $AC\cap BD$ , $AD\cap BC$ will not be a point at infinity.Wlog, $AD\cap BC=X\ne P_{\infty}$ .Then $XM$ bisects $AB$ . $\quad\square$

Now given $n$ points [ $n$ odd] and $\frac{n+1}{2}$ segments each with end point from the $n$ points. Then we can find 2 segments which are not parallel. Indeed, among $n$ points there can be at most $\frac{n-1}{2}$ segments parallel to a particular direction.
Let We want to bisect $AB$ .
Suppose there is a segment parallel to $AB$ whose midpoint is marked. Then using the lemma we are done.Otherwise, assume no segment with marked midpoint is parallel to $AB$ .Suppose $XY$ , $MN$ are 2 segment such that midpoint of $XY$ and $MN$ is marked and $XY$ and $MN$ are not parallel.

Given any segment $XY$ and its midpoint and given any other point $P$ . It is possible to draw a parallel line of $XY$ through $CD$ .proof
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 2., xmax = 14., ymin = -3., ymax = 4.; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); /* draw figures */ draw((4.6,2.7)--(9.02,2.72), linewidth(0.8)); draw((4.24,-0.14)--(8.86,-2.12), linewidth(0.8) + blue); draw((11.56,2.02)--(12.62,-0.44), linewidth(0.8) + ffqqff); draw((6.106445534619243,4.342777344185527)--(8.06642323558925,-0.20585015051845357), linewidth(0.8) + linetype("4 4") + ffqqff); draw((8.346582673653696,4.282836436615007)--(10.994514438621918,-1.8623636971791737), linewidth(0.8) + ffqqff); draw((4.6,2.7)--(10.031680911680908,0.37213675213675307), linewidth(0.8) + blue); /* dots and labels */ dot((4.6,2.7),linewidth(2.pt) + dotstyle); label("$A$", (4.7066,2.7994), NE * labelscalefactor); dot((9.02,2.72),linewidth(2.pt) + dotstyle); label("$B$", (9.111,2.8236), NE * labelscalefactor); dot((4.24,-0.14),linewidth(2.pt) + dotstyle); label("$X$", (4.3436,-0.032), NE * labelscalefactor); dot((8.86,-2.12),linewidth(2.pt) + dotstyle); label("$Y$", (8.9658,-2.0164), NE * labelscalefactor); dot((11.56,2.02),linewidth(2.pt) + dotstyle); label("$M$", (11.652,2.1218), NE * labelscalefactor); dot((12.62,-0.44),linewidth(2.pt) + dotstyle); label("$N$", (12.7168,-0.3466), NE * labelscalefactor); dot((10.031680911680908,0.37213675213675307),linewidth(2.pt) + dotstyle); label("$K$", (10.1274,0.4762), NE * labelscalefactor); dot((7.315840455840454,1.5360683760683767),linewidth(2.pt) + dotstyle); label("$G$", (7.417,1.6378), NE * labelscalefactor); dot((9.525840455840454,1.5460683760683767),linewidth(2.pt) + dotstyle); label("$R$", (9.6192,1.6378), NE * labelscalefactor); dot((6.81,2.71),linewidth(2.pt) + dotstyle); label("$C$", (6.9088,2.7994), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
So draw a parallel line of $XY$ through $A$ ,draw a parallel line of $MN$ through $B$ .[This can be done since mid-point of $XY$ , $MN$ is marked.] Let them intersects at $K$ .By the lemma it is possible to find mid-point $G$ of $KA$ ,midpoint $R$ of $KB$ [Since, $KA||XY$ and mid-point of $XY$ is given, $KB||MN$ ,midpoint of $MN$ is given.
Now again it is possible to draw a parallel line to $BK$ through $G$ since midpoint of $BK$ is $R$ .This line passes through midpoint of $AB$ .

This post has been edited 1 time. Last edited by Mathematicsislovely, Mar 7, 2021, 5:56 PM

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