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by dtweah » Wed Jun 24, 2009 1:55 am
rah_pandey wrote:Dtweah,
Does your proof say that all inscribed triangles in a circle are equilateral.I Hope NOT. If answer is NO then what specific info you have used to prove that the triangle that is drawn is equilateral. If i understand your proof right then you have assumed that the circumcenter and centroid coincide(this is true for equilateral triangles). Also how can you say that triangle ADC is right angled. I know any bisector of the chord passing through the center will be perpendicular to the chord but any line from vertex C on the circle passing through the mid point of the chord may not pass through the center of the circle.

I think your proof is wrong and as dumdoofus said you assumed one thing and later on proved the same.

Regards,
Rahul
So I told you reverse the order. Prove the equilateral triangle first, which proof is given. Since the proof is valid, then the centroid of the triangle and the center of the circle must coinside. The proof given for the equilateral triangle is incontestable. What problem do you have with the proof. Demonstrate.
A triangle is drawn in the circle.
Radia are connected to three vertices of the triangle.
Three isoceles tringles are formed.
the rest of the proof follows.

Do you have problem with this proof. This proof makes no assumption about anything. After that proceed with the rest of the argument.
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by dtweah » Wed Jun 24, 2009 2:07 am
Proof that triangle drawn in the circle of the problem is Equilateral.

Let O be center of circle.
Draw AC and BC and form triangle ACB with vertex C opposite segment AB
Draw diameter CE which bisects AB. Let D be midpoint of AB. Since AB is bisected O must be the center of the circle and the diameter and the center must meet at 90.

Connect radii with all vertices of triangle ACB
Get isoceles triangles: AOC, COB, and BOA. Label the base angles in each as x, y z. derive
2x +2y +2z=180 or
x + y +z =90

Now take right triangle ADC. summing its angles you have
2x +z =90, since angle ADC is 90. Equate the two 90 degrees we have
2x +z =x + y +z
X=Y.

Proceeding X=y=X=30 so triangle is EQualater.

Having established this, center must be centroid and follow from there.

What is the assumption made in this proof?
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Re: tunnel entrance

by Naruto » Wed Jun 24, 2009 2:43 am
uptowngirl92 wrote:The figure above shows the shape of a tunnel entrance. If the curved portion is of a circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'?



(A) 9π

(B) 12π

(C) 9 pie root2
(D) 18π

(E)9 pie/root2
Guys, Wow many proofs and all, it just went above my head.
Now I am assuming what smilles mentioned that 3/4 part is missing and i agree IMO is C, Just I arrived at the answer by a very different process,let me know if it makes sense.

Now clearly diameter is more than 12 and so if we assume 12 to be the dia then length of 3/4 th circle would be 12pie *3/4= 9pie. So clearly the tunnel arc is greater than 9 pie, that eliminates A and E,
We also know that chord or the base 12 forms an angle of 90 at the centre (since 3/4) is mentioned. Now working from the answers, if we assume length to be 18pie then dia would be 18*(4/3) i.e 24, now if we draw a perpendicular bisector from the centre to the chord and check for angles is does not make an angle of 90, instead it makes a 60 so the option is out, similarly if the answer is 12pie then the dia would be 16 and again it does not make an angle of 90, option C however does.

I know its crude but it was the fastest i could get to the answer.
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by rah_pandey » Wed Jun 24, 2009 2:45 am
So just answer my simple question, by your proof(assuming it to be correct) you can prove any triangle inside the circle is equilateral. Now Obviously this is not true that any triangle in a circle is equilateral. Then it means it will be true under special circumstances. What are those special circumstances?
Let O be center of circle.
Draw AC and BC and form triangle ACB with vertex C opposite segment AB
Draw diameter CE which bisects AB
I could have chosen C anywhere on the circle(remember this)
.
No special assumption. It is possible for any triangle inscribed in a circle.
Let D be midpoint of AB.
Nothing wrong upto this point
Since AB is bisected O must be the center of the circle and the diameter and the center must meet at 90.
Now here you assume that any line from O which bisects AB if extended backward will pass through C

Not necessary. where is the problem. Problem lies in the fact that you have assumed that any line from C that bisects AB will also be perpendicular and hence will pass through O. This will only be true for equilateral triangle for which orthocenter,circumcenter(center of the circle in question) and centre of gravity(centroid) all lie at the same point.

I hope you understand the flaw....use the basic fundamental that any arbitrary triangle inside a circle will NOT necessarily be equilateral and then check your proof
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by dtweah » Wed Jun 24, 2009 3:24 am
rah_pandey wrote:Dtweah,
Does your proof say that all inscribed triangles in a circle are equilateral.I Hope NOT. If answer is NO then what specific info you have used to prove that the triangle that is drawn is equilateral. If i understand your proof right then you have assumed that the circumcenter and centroid coincide(this is true for equilateral triangles). Also how can you say that triangle ADC is right angled. I know any bisector of the chord passing through the center will be perpendicular to the chord but any line from vertex C on the circle passing through the mid point of the chord may not pass through the center of the circle.

I think your proof is wrong and as dumdoofus said you assumed one thing and later on proved the same.

Regards,
Rahul
Also how can you say that triangle ADC is right angled?

Because CD which passes thru the center O bisects AB. Any line from one vertex that bisects the opposite side is a median of a triangle. so CD is a median of ACB. since you agree that CD is perpendicular to AB then it follows that ADC is right angled. Draw radii from the center to all vertices. My proof follows from the 3 isoceles triangles fromed by these radii. You have to demonstrate the invalidity of the proof. I make no assumption about centroid here. Except you have a problem with me labeling O as the center and running diameter CE thru that center bisecting AB. This diameter coincides with the Median of the triangle.

I stand by my proof. You have to demonstrate the mathematical invalidity of the proof. I MAKE no assumptions in the proof about centroid. I define the center of a circle and run perpendiculars thru that center.
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by dtweah » Wed Jun 24, 2009 3:34 am
rah_pandey wrote:So just answer my simple question, by your proof(assuming it to be correct) you can prove any triangle inside the circle is equilateral. Now Obviously this is not true that any triangle in a circle is equilateral. Then it means it will be true under special circumstances. What are those special circumstances?
Let O be center of circle.
Draw AC and BC and form triangle ACB with vertex C opposite segment AB
Draw diameter CE which bisects AB
I could have chosen C anywhere on the circle(remember this)

Draw C from any where else and try to bisect AB. There is only one position C can be drawn such that the median of the triangle is a bisector of AB. That position is had when The median coincides with the diameter. When that happens you have an equalateral triange. Try it.

Shift C leftward and run a segment frum C to AB. You will not bisect AB. The only bisector of AB is perdicular to AB and so C must be on that
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by rah_pandey » Wed Jun 24, 2009 4:09 am
i disagree with you that for given AB you can always find a equilateral triangle such that C always lies on the Circle. For ABC to be a equilateral traingle AB=BC=CA=radius*sqrt(3). How do you know that ABC and circle satisfy this relationship. Your proof has problem of assumption.
Draw C from any where else and try to bisect AB. There is only one position C can be drawn such that the median of the triangle is a bisector of AB. That position is had when The median coincides with the diameter. When that happens you have an equalateral triange. Try it.
Here AB is not arbitrary but of a fixed given length.
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