The Instigator
baggins
Pro (for)
Losing
6 Points
The Contender
royalpaladin
Con (against)
Winning
21 Points

ODT2: All triangles are isosceles

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Post Voting Period
The voting period for this debate has ended.
after 8 votes the winner is...
royalpaladin
Voting Style: Open Point System: 7 Point
Started: 11/10/2012 Category: Science
Updated: 4 years ago Status: Post Voting Period
Viewed: 4,395 times Debate No: 27092
Debate Rounds (5)
Comments (15)
Votes (8)

 

baggins

Pro


This debate is part of Official Debate Tournament 2 (ODT2).


I will present a proof that triangles are isosceles. Clearly all triangles are not isosceles. There must be a flaw in my proof. The task of my opponent is to spot the flaw.


== Structure of Debate ==


R1 is for acceptance. In R2 I will present my proof. My esteemed opponent can explain the flaw in R2, R3 or R4. If she manages to spot the flaw, she wins the debate. In case she fails in her task, I will explain the flaw in R5 and claim victory. Pro need not do anything in R2, R3 and R4. Con does not have to do anything in R5.


Con has promised to work on the problem alone. More specifically, she has agreed to not to search internet or to ask friends for help.


Voters will not vote based on truth of resolution (it is false). Voters will vote Pro if my opponent fails to spot the mistake in my proof. Otherwise they will vote Con.


royalpaladin

Con

I accept the terms of the debate.
Debate Round No. 1
baggins

Pro

I would like to thank Royalpaladin for prompt acceptance of the challenge (within few minutes).

== To Prove ==

All Triangles are isosceles

Consider an arbitrary triangle ΔABC, shown in the figure below. Our goal is to show that AB = AC. This should be sufficient to establish that all triangles are isosceles.

Fig: A general Triangle

== Construction ==

  1. Draw angle bisector of A (AO) and Perpendicular bisector of side BC (DO). Let them intersect at O.
  2. Join point O to vertex of the triangle B and C (OB and OC respectively).
  3. Drop perpendicular from O on sides AB and AC (OE and OF respectively)

The resulting diagram should look as follows.

== Claim 1 ==

ΔAOE ΔAOF

Proof:

∠OEA = ∠OFA ( = 900­­ )

EAO = ∠FAO (AO is angle bisector of ∠A)

AO = AO (Common)

By AAS congruency, ΔAOE ΔAOF

== Claim 2 ==

ΔBOD ΔCOD

Proof:
BD = CD (OD is perpendicular bisector of AD)

BDO = ∠CDO ( =900 )

DO = DO (Common)

By SAS congruency, ΔBOD ΔCOD

== Claim 3 ==

ΔEOB ΔFOC

Proof:

∠OEB = ∠OFC( =900 )

BO = CO (ΔBOD ΔCOD as per claim 2)

EO = FO (ΔAOE ΔAOF as per claim 1)

By RHS congruency, ΔEOB ΔFOC

== Claim 4 ==

AB = AC

Proof:

AE = AF(ΔAOE ΔAOF as per claim 1)

EB = FC (ΔEOB ΔFOC as per claim 3)

AE + EB = AF + FC(Adding last two lines)

AB = AC

Claim 4 establishes most of the proof. There are few more scenarios we need to consider before we can call the proof to be complete.

<<<< Alternative Scenario 1 >>>>

In construction step 1, we assumed that angle bisector of A and perpendicular bisector of side BC intersect each other. Consider the situation where angle bisector of side A is perpendicular to side BC and intersects BC and D’. On other hand, perpendicular bisector of side BC is DA’.

Fig: Scenario when angle bisector is perpendicular to opposite side.

Consider ΔAD’B and ΔAD’C
∠BAD’ = CAD’(AD’ is the angle bisector)

AD’ = AD’ (Common)

AD’B = AD’C ( =900 )

From ASA congruency, ΔAD’B ΔAD’C

This means, AD’ = D’C. This implies D’ is midpoint of BC. However midpoint of BC was assumed to be D (as A’D was perpendicular bisector). This means D and D’ are identical. This also means AD’ and AD are identical lines.

Most important of all, AB = AC. This triangle is isosceles.

<<<< Alternative Scenario 2 >>>>

What if the angle bisector and perpendicular bisector are almost parallel? In this case point O will lie below the line BC. I could have asked reader to complete this part of the proof, however I will do it myself for sake of completeness.

== Construction ==

  1. Draw angle bisector of A (AO) and Perpendicular bisector of side BC (DO). Let them intersect at O.
  2. Join point O to vertex of the triangle B and C (OB and OC respectively).
  3. Drop perpendicular from O on sides AB and AC (OE and OF respectively)

Scenario when O lies below AB

In this case the perpendicular dropped from O on line AB and BC will lie on extended part of the respective line. Apart from that, all the constructions are identical to previous cases.

== Claim 1 ==

ΔAOE ΔAOF

Proof:

∠OEA = ∠OFA ( = 900­­ )

EAO = ∠FAO (AO is angle bisector of ∠A)

AO = AO (Common)

By AAS congruency, ΔAOE ΔAOF

== Claim 2 ==

ΔBOD ΔCOD

Proof:

BD = CD(OD is perpendicular bisector of AD)

BDO = ∠CDO ( =900 )

DO = DO (Common)

By SAS congruency, ΔBOD ΔCOD

== Claim 3 ==

ΔEOB ΔFOC

Proof:

∠OEB = ∠OFC ( =900 )

BO = CO(ΔBOD ΔCOD as per claim 2)

EO = FO (ΔAOE ΔAOF as per claim 1)

By RHS congruency, ΔEOB ΔFOC

== Claim 4 ==

AB = AC

Proof:

AE = AF(ΔAOE ΔAOF as per claim 1)

EB = FC(ΔEOB ΔFOC as per claim 3)

AE - EB = AF - FC (From last two lines)

AB = AC

This completes the proof.

== Corollary ==

All triangles are equilateral.

We have proved AB = AC. Using similar procedures we can show that AB = BC. This would mean AB = BC = CA. Thus all triangles are equilateral.

Over to my esteemed opponent.

(My apologies for poor formatting. I was having several problems with DDO text editor while copyinh it from word.)

royalpaladin

Con

I have not had much time to work on this. I will wait to post the flaw until next round.
Debate Round No. 2
baggins

Pro

My oppnent has got 2 more rounds...
royalpaladin

Con

Quick Note
My opponent must prove that *all* triangles are isosceles. If I can show that even one triangle is isosceles, I win the round because then not *all* triangles are isosceles.

Definition of Isosceles Triangle and Impact
An isosceles triangle is a triangle with at least two sides that are equal. The 3-4-5 right triangle is not an isosceles triangle, so we know that my opponent’s case is flawed from the outset.

Opponent’s Case
My opponent attempts to prove that all triangles are isosceles through an examination of three different cases that center around the intersection of an angle bisector and the perpendicular bisector of the side across from it. It is a given that if the perpendicular bisector and the angle bisector are the same, the triangle will be isosceles, so he needs to prove the three other cases.

Scenario 1: The perpendicular bisector and the angle bisector intersect in the middle of the triangle.
This one gave me a great deal of trouble, but after staring at the proof for two hours, I noticed that my opponent’s diagram is actually misleading and then did some investigation of my own through drawings on paper. I found that Scenario 1 does not even exist since the perpendicular bisector and the angle bisector will always intersect outside the triangle. This proof hinges on an internal intersection, but that does not happen. Scenario 1 thus turns into Scenario 3.

Scenario 2: No intersection
This scenario doesn’t actually make sense; if you look at my opponent’s drawings, the triangles cannot possibly be congruent because one has larger sides than the other. The flaw in this scenario is that he assumes that the angle bisector will intersect the side at a 90 angle, which is not possible if it is not equal to the perpendicular bisector. In fact, the two subtriangles could not ever be congruent (which requires all sides and angles to be the same) if the angle bisector and perpendicular bisector attached to the side at two different points since the angle bisector would not divide the triangle into equal halves.

Scenario 3: External Intersection
This is where my opponent does not actually prove his case because he does not consider a scenario that actually is part of this scenario. My opponent is correct that lines can be drawn at OB and OC, but he then tells you that he wants to connect O and AC and AB. He assumes that a ray from O will intersect these lines outside of the triangle, but it is possible to have a situation in which E, for example, is between A and B. In this case, OE and OF would not be congruent, and neither would AOE and AOF. This was claim one in his proof, so in this scenario, the triangle is not isosceles and the resolution is negated.
Debate Round No. 3
baggins

Pro

== Quick notes on quick notes ==


This one gave me a great deal of trouble, but after staring at the proof for two hours…


Glad to know that my opponent enjoyed the problem.


I am glad to note that she has made some progress. In fact she has spotted where the flaw lies. However her analysis of the problem is still incomplete.


An isosceles triangle is a triangle with at least two sides that are equal. The 3-4-5 right triangle is not an isosceles triangle, so we know that my opponent’s case is flawed from the outset.


No doubt the proof is flawed. However the task of Con is to find the flaw.


Inside Intersection: My opponent thinks this scenario can never happen because…


…did some investigation of my own through drawings on paper. I found that Scenario 1 does not even exist since the perpendicular bisector and the angle bisector will always intersect outside the triangle.


There are two obvious problems with the approach.



  1. If this scenario does not occur in few test diagrams, it does not mean it will never occur. Unless she has some proof that intersections will never take place inside the triangle, this scenario cannot be ignored.

  2. Assuming that this scenario does occur, does my opponent accept that my proof is correct for this case?


No intersection: My opponent has argued that this case can never occur. I agree. She should read my proof again. I have also proved the same case by showing D and D’ must coincide. If they coincide this is the case for isosceles triangle.


External Intersection: She has correctly spotted that I have skipped few cases. Her analysis of the missing case is wrong. I guess she will do a more thorough (and formal) job in next round. To win this debate, the least we can expect from her is to post the correct analysis of the missing scenario and explain why the proof breaks down in that case.


It shall be unfortunate if she restricts herself to doing the minimum for winning the debate. It will be intellectually satisfying if she can also explain the flaw in scenario where both E and F lie outside the triangle. (Apart from the internal intersection scenario).


Wishing my opponent a few more hours of staring at the proof…

royalpaladin

Con

Notes
I am going to go ahead and protest my opponent’s claim that I need to provide a formal proof for my claims. In Round 1, he explicitly said that I just needed to find the flaw in his proof, and in the previous round, he admits that I found the flaw. This means that I have already won the debate. However, in case some of you disagree, I will continue. Note, however, that by the rules he established in the first round, you should vote Con without reading the rest of this debate.

Scenario 1
He asks me to prove that this scenario cannot occur and notes that my test cases don’t prove anything. Fine. I will ask the readers to construct the diagrams at home since I cannot do so for them. A theorem known as the Internal Angle Bisector Theorem notes that if there is a triangle ABC with an internal angle bisector AD of the angle BAC, AB/AC=BD/DC (the ratios will be the same). Now, suppose we have a triangle ABC with an internal angle bisector D and a perpendicular bisector of BC at point M. In order to prove my claim true, I need to demonstrate that BD<BM (since if BD was greater, they would intersect inside the triangle).

BC=BM+MC (since M is the midpoint)
BC=BD+DC
BD<DC
Proof that BD<BC:
By the Internal Angle Bisector Theorem
AB/AC=BD/DC
However, we know that AB is less than AC
Thus, AB/AC <1
If AB/AC<1 and BD/DC=AB/BC, BD/DC<1 and BD<DC
2BM=BM+MC=BD+DC>BD+BD
BD<BM

Thus, Scenario 1 never occurs.

Scenario 2
I already conceded that Scenario 2 only occurs if D and D’ are the same, in which case the triangle is isosceles. Otherwise, the scenario does not occur.

Scenario 3
Note that I only need to prove that in the case in which E is not outside the triangle, the triangle is not isosceles in order to win.

So, if E is between A and B and F is outside the triangle and on line AC, the proof breaks down. He was right; my previous argument was incorrect. His proof actually breaks down at Claim 4 in this scenario because he says that AE - EB = AF - FC. This is false in the case I presented; AE-EB not equal AF-FC when E is between A and B. This means that AB will not equal AC and the triangle is not isosceles.

I don’t need to prove the other scenario false since I have shown that not all triangles are isosceles.
Debate Round No. 4
baggins

Pro

My opponent has correctly spotted the flaw. On other hand her insistence that the readers draw there own diagram at home appears to be weird.

VOTE CON

This problem appears in 'Fallacies in Mathematics' by EA Maxwel [1]. The proof is often attributed to Charles Dodgeson who is better known for his novel Alice Adventures in Wonderland by Lewis Carrol).

[1] http://www.scribd.com...
royalpaladin

Con

I would like to thank my opponent for a good debate. As he said, vote Con.
Debate Round No. 5
15 comments have been posted on this debate. Showing 1 through 10 records.
Posted by Cometflash 4 years ago
Cometflash
I guess everyone that voted, was undecided, if all triangles are isosceles or not, since everyone vote as tie.

You guys probably kept the confusing to the voters, because after, they were still undecided, since again they all voted tie.
Posted by royalpaladin 4 years ago
royalpaladin
My challenge was to spot the flaw in the proof.
Posted by Luggs 4 years ago
Luggs
You do realize that all that needs to be done to prove Pro wrong is to measure the sides. By definition, an isosceles triangle has at least 2 sides that are equivalent, and the angles opposite those sides have equivalent are equivalent as well. Therefore, to prove that not all triangles are isosceles, you only need to measure sides and angles (better to measure them all to make sure).
Posted by staraa 4 years ago
staraa
though the topic is quite interesting but you can't prove it true for scalene triangle. can you?????
if yes then you are really a genius
Posted by baggins 4 years ago
baggins
@ Costas

Redefining the metric would have been a nice trick. However this proof is constrained to high school geometry (i.e R2 with the usual metric).
Posted by costas 4 years ago
costas
There is a specific step of the proof that is false. So I think Con should focus on spotting this flaw instead of trying to make some general argument against the proof.

By the way, since there is no reference to the structure of the space where the triangles lie, one could assume they are on R^2 with the trivial metric d((x,x'),(y,y'))=1, for all real x,x',y,y'. Hence all triangles are isosceles and furthermore they are equilateral.

I guess Pro assumes they are in R^2 with the usual metric d((x,x'),(y,y'))=sqrt((x-y)^2+(x'-y')^2).
Posted by baggins 4 years ago
baggins
The proof is intentionally verbose. The idea was to create a bit of panic...
Posted by Zaradi 4 years ago
Zaradi
*looks at debate*
*looks at comments*
*repeats Maikuru's comment*
Posted by bossyburrito 4 years ago
bossyburrito
@ Maikuru
+ 1000
Posted by vmpire321 4 years ago
vmpire321
measure the line.

done.
8 votes have been placed for this debate. Showing 1 through 8 records.
Vote Placed by Maikuru 4 years ago
Maikuru
bagginsroyalpaladinTied
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Total points awarded:03 
Reasons for voting decision: Pro conceded, which is good because I wasn't going to read this otherwise.
Vote Placed by RationalMadman 4 years ago
RationalMadman
bagginsroyalpaladinTied
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Reasons for voting decision: Such a nerdy debate lol. Baggins get conduct for con's rude round 2 skipping. Obviously arguments to con but I really think pro used more reliable sources... it's a tie for me.
Vote Placed by Zaradi 4 years ago
Zaradi
bagginsroyalpaladinTied
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Reasons for voting decision: Dat flaw was Waldo, and Royal found Waldo.
Vote Placed by Muted 4 years ago
Muted
bagginsroyalpaladinTied
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Reasons for voting decision: TrasguTravieso, could you explain why you're voting for baggins? Anyway, flaw spotted.
Vote Placed by tulle 4 years ago
tulle
bagginsroyalpaladinTied
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Reasons for voting decision: This hurt my head :(
Vote Placed by TrasguTravieso 4 years ago
TrasguTravieso
bagginsroyalpaladinTied
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Reasons for voting decision: Good job Royal, and kudos for the interesting idea to Baggins
Vote Placed by RyuuKyuzo 4 years ago
RyuuKyuzo
bagginsroyalpaladinTied
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Reasons for voting decision: con succeeded in finding the flaw in baggins' argument.
Vote Placed by AlwaysMoreThanYou 4 years ago
AlwaysMoreThanYou
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Reasons for voting decision: Con spotted the flaw.