Prove $L$, $M$, $N$ are collinear Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Proving a triangle is isoceles given two points on the sidesBisector of angle formed at the orthocentre passes through the circumcentreProve that L,M,N are collinear.Are lines which pass respectively through vertices $A,B,C$ and incenter, circumcenter and orthocenter of $Delta ABC$ concurrent?Collinearity of points in a projective settingWhy are auxiliary lines valid in geometric proofs?Finding the 'three points'Circumcenter of $Delta ABC$ is the centroid of the antipedal triangle of symmedian pointA conjecture related to any triangleShow that three point $G,H,G_1$ are collinear.
Time to Settle Down!
Why do we need to use the builder design pattern when we can do the same thing with setters?
How to tell that you are a giant?
How would a mousetrap for use in space work?
What does it mean that physics no longer uses mechanical models to describe phenomena?
Drawing without replacement: why is the order of draw irrelevant?
What is the font for "b" letter?
How could we fake a moon landing now?
ArcGIS Pro Python arcpy.CreatePersonalGDB_management
When a candle burns, why does the top of wick glow if bottom of flame is hottest?
A term for a woman complaining about things/begging in a cute/childish way
As a beginner, should I get a Squier Strat with a SSS config or a HSS?
How do I use the new nonlinear finite element in Mathematica 12 for this equation?
Does the Weapon Master feat grant you a fighting style?
Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?
Is grep documentation about ignoring case wrong, since it doesn't ignore case in filenames?
Putting class ranking in CV, but against dept guidelines
Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?
Chinese Seal on silk painting - what does it mean?
How does light 'choose' between wave and particle behaviour?
Why is Nikon 1.4g better when Nikon 1.8g is sharper?
Can the Great Weapon Master feat's "Power Attack" apply to attacks from the Spiritual Weapon spell?
How fail-safe is nr as stop bytes?
Illegal assignment from sObject to Id
Prove $L$, $M$, $N$ are collinear
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Proving a triangle is isoceles given two points on the sidesBisector of angle formed at the orthocentre passes through the circumcentreProve that L,M,N are collinear.Are lines which pass respectively through vertices $A,B,C$ and incenter, circumcenter and orthocenter of $Delta ABC$ concurrent?Collinearity of points in a projective settingWhy are auxiliary lines valid in geometric proofs?Finding the 'three points'Circumcenter of $Delta ABC$ is the centroid of the antipedal triangle of symmedian pointA conjecture related to any triangleShow that three point $G,H,G_1$ are collinear.
$begingroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
edited Mar 27 at 19:04
Anirban Niloy
8611419
asked Mar 27 at 15:45
StompStomp
183
$endgroup$
add a comment |
$begingroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
edited Mar 27 at 19:04
Anirban Niloy
8611419
asked Mar 27 at 15:45
StompStomp
183
$endgroup$
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
Mar 27 at 16:01
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
Mar 27 at 16:06
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
Mar 27 at 17:37
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
Mar 27 at 18:46
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
Mar 28 at 1:18
add a comment |
$begingroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
edited Mar 27 at 19:04
Anirban Niloy
8611419
asked Mar 27 at 15:45
StompStomp
183
$endgroup$
$G$ is the centroid of triangle $ABC$; $AG$ is produced to $X$ such that $GX=AC$, if we draw parallels through $X$ to $CA$, $AB$, $BC$ meeting $BC$, $CA$, $AB$ at $L$, $M$, $N$ respectively, prove that $L$, $M$, $N$ are collinear.
I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
geometry contest-math
geometry contest-math
edited Mar 27 at 19:04
Anirban Niloy
8611419
asked Mar 27 at 15:45
StompStomp
183
edited Mar 27 at 19:04
Anirban Niloy
8611419
asked Mar 27 at 15:45
StompStomp
183
edited Mar 27 at 19:04
Anirban Niloy
8611419
edited Mar 27 at 19:04
Anirban Niloy
8611419
edited Mar 27 at 19:04
Anirban Niloy
8611419
8611419
asked Mar 27 at 15:45
StompStomp
183
asked Mar 27 at 15:45
StompStomp
183
asked Mar 27 at 15:45
StompStomp
183
183
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
Mar 27 at 16:01
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
Mar 27 at 16:06
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
Mar 27 at 17:37
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
Mar 27 at 18:46
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
Mar 28 at 1:18
add a comment |
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
Mar 27 at 16:01
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
Mar 27 at 16:06
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
Mar 27 at 17:37
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
Mar 27 at 18:46
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
Mar 28 at 1:18
1
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
Mar 27 at 16:01
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
Mar 27 at 16:01
3
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
Mar 27 at 16:06
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
Mar 27 at 16:06
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
Mar 27 at 17:37
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
Mar 27 at 17:37
2
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
Mar 27 at 18:46
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
Mar 27 at 18:46
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
Mar 28 at 1:18
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
Mar 28 at 1:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
$endgroup$
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164673%2fprove-l-m-n-are-collinear%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
$endgroup$
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
add a comment |
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
$endgroup$
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
add a comment |
$begingroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
$endgroup$
I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $vecA = vec0$, and $vecB$ and $vecC$ are linearly independent. The centroid $G$ is given by a vector
$$ vecG = frac13(vecA+vecB+vecC) = frac13(vecB+vecC) $$
Point $X$ lies on the line $overlineAG$, which means that
$$ exists lambdainmathbbR : vecX = lambda (vecB+vecC) $$
For point $L$ we have
beginalign big(Lin overlineBCbig) &Rightarrow big(exists alpha_LinmathbbR : vecL = alpha_L vecB + (1-alpha_L) vecC big) \
big(overlineXL parallel overlineAC big) &Rightarrow big(exists beta_LinmathbbR : vecL = vecX + beta_L vecC = lambda vecB + (lambda + beta_L)vecCbig) endalign
since vectors $vecB=vecC$ the only solution for these conditions is
$$ vecL = lambda vecB + (1-lambda)vecC$$
For point $M$ we have
beginalign big(Min overlineACbig) &Rightarrow big(exists alpha_MinmathbbR : vecM = alpha_M vecC big) \
big(overlineXMparallel overlineABbig) &Rightarrow big(exists beta_MinmathbbR : vecM = vecX + beta_M vecB = (lambda + beta_M) vecB + lambda vecCbig) endalign
The only solution for these conditions is
$$ vecM = lambda vecC$$
Finally, for point $N$ we have
beginalign big(Nin overlineABbig) &Rightarrow big(exists alpha_NinmathbbR : vecN = alpha_N vecB big) \
big(overlineXNparallel overlineBCbig) &Rightarrow big(exists beta_NinmathbbR : vecN = vecX + beta_N (vecB-vecC) = (lambda + beta_N) vecB + (lambda-beta_N) vecCbig) endalign
and the solution for these conditions is
$$ vecN = 2lambda vecB$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
beginalign existsgammainmathbbR &: (vecL-vecM) = gamma (vecN-vecL) \
existsgammainmathbbR &: lambdavecB+(1-2lambda)vecC = gamma (lambdavecB+(lambda-1)vecC)endalign
This can only be true if $lambda=frac23$, or $lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $vecX=frac23(vecB+vecC) = 2vecG$
If you choose any other point $X$ on line $overlineAG$, points $L$, $M$, $N$ won't be colinear.
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
answered Mar 27 at 22:43
Adam LatosińskiAdam Latosiński
8759
8759
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
add a comment |
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
$begingroup$
Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that.
$endgroup$
– Stomp
Mar 28 at 1:20
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3164673%2fprove-l-m-n-are-collinear%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
$endgroup$
– Matteo
Mar 27 at 16:01
3
$begingroup$
Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true.
$endgroup$
– Adam Latosiński
Mar 27 at 16:06
$begingroup$
Could you provide a picture??
$endgroup$
– Dr. Mathva
Mar 27 at 17:37
2
$begingroup$
@Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary.
$endgroup$
– FredH
Mar 27 at 18:46
$begingroup$
Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways
$endgroup$
– Stomp
Mar 28 at 1:18