By your response I thought what you have thought about me. So
I request you not to judge me on my age coz I m still 14 although it is showing 22.
I do hope there will not be any clashes in b/w us in the future.
I m also sorry for being rude to you and I too hope you feeling the same way.

Oh, I was being literal because you asked for a positive response :)

No worries. On Brilliant, because the problems are much harder than what one would normally encounter, it is very common for mistakes / misconceptions to occur. The important thing is to learn from the mistakes and understand why you had made them, and how to avoid them in future.

No one is exempt from making mistakes, me included. I have had posted numerous problems where I got corrected by the community. For example, I had a problem in which I made numerous mistakes which let to all kinds of wrong conclusions. @brian charlesworth came along and helped me fix the problem, which led to this gem.

In fact, I do not know of any avid problem creator who has not had at least one valid report submitted against his problem. Eventually, they get fixed, and the community appreciates their efforts.

Skanda Prasad please follow the guide lines below to write a link.

1) Firstly open the square brackets i.e, [ . . . ]

2) Next, write something in it. example [here]

3) Next go to the page where you want to connect this link. For example if you want to connect the link to you Sound problem, go to that page

4) Next, copy the URL written on the page. For example, in this page the URL can be found above(i.e, https://brilliant.org/discussions/thread/skanda-prasads-message-board/).

5) Now, open curved brackets i.e, ( . . . )(i.e, [here] and beside it, without space( . . . ))

6) Finally copy the URL in to the curved brackets and click post or preview to see it in GREENS! For example, I've made a link to your Sound-2 problem, click here

Oh Skanda , I happened to see this problem click here. The answer to this problem is wrong , infact Jon Hausmann has given proof to prove it false. I saw that you are one of its solvers.I am very curious to know its method. Can you explain to me how you arrived at this answer? That would be helpful .. thanks!

With reference to your problem and problem, here is the proof that in any cyclic hexagon, $\angle A + \angle C + \angle E = 360 ^ \circ$. It is independent of BDF being an equilateral triangle, which is not stated in the question, and cannot be used as an assumption in the problem.

Solution: Say the (not necessarily regular) hexagon is ABCDEF. Then, $\angle A = 180 ^ \circ - \angle FDB$, $\angle C = 180 ^ \circ - \angle BFD$, $\angle E = 180 ^ \circ - \angle FDB$.

I do not abuse the power of being staff. Note that Jon Haussmann's dispute echoes mine. I have stated out the reasons for why your problem was deleted. Specifically, the 2 issues are:

In your question you did not state that $DEF$ is an equilateral triangle, as such that cannot be used in the solution. It is an unwarranted assumption, and hence a solution of "Let's assume this fact" is not a correct solution.

There are multiple ways to prove a fact. It is not reasonable to use MCQ to say "You can only prove this problem in this certain way". For example, I could have used a lot of parallel lines to show that the sum of those angles is 180, instead of using triangle $BFD$.

As such, the answer of "By construction proving triangle is equilateral and then by cyclic quadilateral property" is not correct because it is a false statement.

If you do not know to solve in this way then please see the solution given below.

By construction, OB=OD=OF.

If you use a little of your brains, you might understand that BDF is now an equilateral triangle,
by the isosceles triangle property of triangles OFD, OBD and OBF.

Hence all angles are equal in the triangle BDF as OB, OF, OD are the radii of the circle.

Now consider the quadrilateral DBAF, As angle B = 60 degrees, angle A = 120 degrees.

In the same way, it can be proved for the angles, C and E

Are you claiming that if $OB = OD = OF$ then $BDF$ must be an equilateral triangle? This would imply that any 3 points on a circle always form an equilateral triangle.

@Calvin Lin
–
Did you understand another concept of isosceles triangle property?? Sir??
Consider triangle OBD
OB=OD
hence angle OBD=ODB

Consider traingle ODF
OD=OF
Hence angle ODF=OFD

But angle ODB=ODF
and angle ODB+ODF= angle BDF

In the same way we can take the triangles OBF
OB=OF
Hence angle OBF=OFB

angle OFD+OFB=angle DFB

HENCE angles BDF = BFD =
DBF (in the same way)
AS ALL ANGLES IN A TRIANGLE ARE EQUAL, THEIR SIDES ARE EQUAL AND HENCEFORTH CONSIDERED TO BE AN EQUILATERAL TRIANGLE.

Come on!! you must know that the equal radius bring equal base angles.

Hence, angle ODB in the triangle OBD and angle ODF in triangle OFD have the same measures.

in the same way, if you add angles OFD and OFB you will get Angle DFB
if you add angles OBF and OBD you will get angle DBF
when all the angles in a triangle are equal, it is an equilateral triangle

IT IS TRUE FOR ALL SUCH ANGLES IN THE FIG. HENCE IT IS AN EQUILATERAL TRIANGLE.

I do not agree that angle ODB = OFD, unless there is some other reason which you have not stated. What does "equal radius bring equal base angles" mean?

Consider the following diagram. OB, OD, OF are radii of the circle marked in red. Is BDF an equilateral triangle (marked in blue)?

@Calvin Lin
–
Sir pls donot make and post meaningless figures, just talk about mine.
there was a hexagon in which the triangle is enclosed.

I m completely sure that my solution was correct and "meaningful"
I donot wish continue this arguement, or whatever you might call it.....

THANK YOU FOR YOUR CO-OPERATION AND SUGGESTION,
YOU ARE VERY BRAINY, INTELLIGENT, AND I AM AN ILLITERATE FELLOW WHO DOES'NT KNOWS TO READ............ HAPPY ??
PLEASE LEAVE ME ALONE AND I DO NOT EXPECT ANY MORE COMMENTS ON THE PROBLEMS I POST.
PLEASE STOP DELETING PROBLEMS WHICH "YOU" DONOT LIKE,
PEOPLE MIGHT HAVE POSTED WITH A MEANING ONLY AND NOT JUST FOR FUN.........

I dont give a damn about Jon Houseman or whoever he is... that is very unessential for me....

THANK YOU !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

@Sravanth Chebrolu
–
Oh! This one! I tried to solve it long ago.....
But unfortunately I got it wrong.... Ha Ha !! Never mind.......
I thought that in 2 and a 1/2 hours, there are two COMPLETE hours so I clicked '2'.

@Calvin Lin
–
I was just surfing in your message board.
You had changed a person's age.....
Is it possible for mine too, Sir??
Could you please change it as 14 from 22 ??

@Skanda Prasad
–
B, D, F are points on the circumference, so OB, OD, OF are all radii of the circle. So yes, OD is a radii. It is just not drawn in as yet.

@Calvin Lin
–
Sir its 12:00 am in the night, my parents are asking me to sleep now and could we please continue our conversation at round 15 hrs later, it would be afternoon here by then........

@Calvin Lin
–
If you have any confusion with the fig. then pls check my problem known as " Geometrically Brain teasing Problem by Skanda Prasad" where you will get the fig. and clear the confusions, if you had had any....
Thank You Sir !!!!

@Skanda Prasad
–
I have seen the figure. It has not implied that BDF is an equilateral triangle.

Unless, the figure is trying to say that Angles A , C ,E are equal, in which case that should be clearly stated in the problem. And if so, then ABCDEF is a regular hexagon, and so there are 2 possible answers to your problem.

@Calvin Lin
–
You have asked me for the proof and I have given it to you.
Consider the quadrilateral DBAF after proving BDF is an equilateral traingle.
as angle BDF = 60 degrees, angle BAF = 120 degrees

Consider the quadrilateral DBFE,
as angle DBF = 60 degrees, angle DEF = 120 degrees

Consider the quadrilateral DCBF
as angle DFB = 60 degrees, angle DCB = 120 degrees

I am 100 % sure that it is accurate question and with a valid answer.
As you would have noticed, I have posted my solution.
If you prove that it is wrong, I will also stop posting this question again and again.

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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You've given the literal response sir, +++.

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By your response I thought what you have thought about me. So I request you not to judge me on my age coz I m still 14 although it is showing 22. I do hope there will not be any clashes in b/w us in the future. I m also sorry for being rude to you and I too hope you feeling the same way.

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Oh, I was being literal because you asked for a positive response :)

No worries. On Brilliant, because the problems are much harder than what one would normally encounter, it is very common for mistakes / misconceptions to occur. The important thing is to learn from the mistakes and understand why you had made them, and how to avoid them in future.

No one is exempt from making mistakes, me included. I have had posted numerous problems where I got corrected by the community. For example, I had a problem in which I made numerous mistakes which let to all kinds of wrong conclusions. @brian charlesworth came along and helped me fix the problem, which led to this gem.

In fact, I do not know of any avid problem creator who has not had at least one valid report submitted against his problem. Eventually, they get fixed, and the community appreciates their efforts.

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Good!

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@Calvin Lin , where did Skanda's account go?

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It got deleted by Skanda i.e. me

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click here for more information on LATEX

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Skanda Prasad please follow the guide lines below to write a link.

1) Firstly open the square brackets i.e, [ . . . ]

2) Next, write something in it. example [here]

3) Next go to the page where you want to connect this link. For example if you want to connect the link to you Sound problem, go to that page

4) Next, copy the URL written on the page. For example, in this page the URL can be found above(i.e, https://brilliant.org/discussions/thread/skanda-prasads-message-board/).

5) Now, open curved brackets i.e, ( . . . )(i.e, [here] and beside it, without space( . . . ))

6) Finally copy the URL in to the curved brackets and click post or preview to see it in

GREENS!For example, I've made a link to your Sound-2 problem, click hereLog in to reply

Hello every one!! Why don't you try this out??

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Oh Skanda , I happened to see this problem click here. The answer to this problem is wrong , infact Jon Hausmann has given proof to prove it false. I saw that you are one of its solvers.I am very curious to know its method. Can you explain to me how you arrived at this answer? That would be helpful .. thanks!

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With reference to your problem and problem, here is the proof that in

anycyclic hexagon, $\angle A + \angle C + \angle E = 360 ^ \circ$. It is independent of BDF being an equilateral triangle, which is not stated in the question, and cannot be used as an assumption in the problem.Solution:Say the (not necessarily regular) hexagon is ABCDEF. Then, $\angle A = 180 ^ \circ - \angle FDB$, $\angle C = 180 ^ \circ - \angle BFD$, $\angle E = 180 ^ \circ - \angle FDB$.Hence, $\angle A + \angle C + \angle E = 540 ^ \circ - ( \angle FDB - \angle BFD - \angle FDB ) = 540 ^ \circ - 180 ^ \circ = 360 ^ \circ$.

I do not abuse the power of being staff. Note that Jon Haussmann's dispute echoes mine. I have stated out the reasons for why your problem was deleted. Specifically, the 2 issues are:

In your question you did not state that $DEF$ is an equilateral triangle, as such that cannot be used in the solution. It is an unwarranted assumption, and hence a solution of "Let's assume this fact" is not a correct solution.

There are multiple ways to prove a fact. It is not reasonable to use MCQ to say "You can only prove this problem in this certain way". For example, I could have used a lot of parallel lines to show that the sum of those angles is 180, instead of using triangle $BFD$.

As such, the answer of "By construction proving triangle is equilateral and then by cyclic quadilateral property" is not correct because it is a false statement.

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If you do not know to solve in this way then please see the solution given below.

By construction, OB=OD=OF.

If you use a little of your brains, you might understand that BDF is now an equilateral triangle, by the isosceles triangle property of triangles OFD, OBD and OBF.

Hence all angles are equal in the triangle BDF as OB, OF, OD are the radii of the circle.

Now consider the quadrilateral DBAF, As angle B = 60 degrees, angle A = 120 degrees.

In the same way, it can be proved for the angles, C and E

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Are you claiming that if $OB = OD = OF$ then $BDF$ must be an equilateral triangle? This would imply that any 3 points on a circle always form an equilateral triangle.

Can you explain the following diagram them?

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Hence they are equal. Simple!! Isn't it Sir??

And also note that O is the centre of the circle. therefore radii are the same in the circle....!!!!!!

I have no pleasure in explaining the diagram because it is not the one related to my problem

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Yes, OB = OD = OF as length. In the diagram, I have given you points O, B, D, F such that OB = OD = OF. But BDF is not an equilateral triangle.

Note that even though OB= OD = OF, this does not imply that BD = DF = FB, which is necessary to show that it is an equilateral triangle.

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Consider traingle ODF OD=OF Hence angle ODF=OFD

But angle ODB=ODF and angle ODB+ODF= angle BDF

In the same way we can take the triangles OBF OB=OF Hence angle OBF=OFB

angle OFD+OFB=angle DFB

HENCE angles BDF = BFD = DBF (in the same way) AS ALL ANGLES IN A TRIANGLE ARE EQUAL, THEIR SIDES ARE EQUAL AND HENCEFORTH CONSIDERED TO BE AN EQUILATERAL TRIANGLE.

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Here's the proof...

Consider the triangle ODB in which angle ODB=OBD

Consider triangle ODF angle OFD=ODF

Come on!! you must know that the equal radius bring equal base angles.

Hence, angle ODB in the triangle OBD and angle ODF in triangle OFD have the same measures.

in the same way, if you add angles OFD and OFB you will get Angle DFB if you add angles OBF and OBD you will get angle DBF when all the angles in a triangle are equal, it is an equilateral triangle

IT IS TRUE FOR ALL SUCH ANGLES IN THE FIG. HENCE IT IS AN EQUILATERAL TRIANGLE.

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I m completely sure that my solution was correct and "meaningful" I donot wish continue this arguement, or whatever you might call it.....

THANK YOU FOR YOUR CO-OPERATION AND SUGGESTION, YOU ARE VERY BRAINY, INTELLIGENT, AND I AM AN ILLITERATE FELLOW WHO DOES'NT KNOWS TO READ............ HAPPY ?? PLEASE LEAVE ME ALONE AND I DO NOT EXPECT ANY MORE COMMENTS ON THE PROBLEMS I POST. PLEASE STOP DELETING PROBLEMS WHICH "YOU" DONOT LIKE, PEOPLE MIGHT HAVE POSTED WITH A MEANING ONLY AND NOT JUST FOR FUN.........

I dont give a damn about Jon Houseman or whoever he is... that is very unessential for me....

THANK YOU !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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this, which has now been deleted.

They were talking aboutLog in to reply

troll king contest by clicking here

Before talking about your previous comment, check out my question for theLog in to reply

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troll king contest by posting an April-fool-ish queston???

Well then, but why don't you join theLog in to reply

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thank you for your co-operation........

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Unless, the figure is trying to say that Angles A , C ,E are equal, in which case that should be clearly stated in the problem. And if so, then ABCDEF is a regular hexagon, and so there are 2 possible answers to your problem.

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Consider the quadrilateral DBFE, as angle DBF = 60 degrees, angle DEF = 120 degrees

Consider the quadrilateral DCBF as angle DFB = 60 degrees, angle DCB = 120 degrees

Hence Angles A + C + E = 120+120+120 = 360

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I am 100 % sure that it is accurate question and with a valid answer. As you would have noticed, I have posted my solution. If you prove that it is wrong, I will also stop posting this question again and again.

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