Let the side length of a regular octagon (an 8-sided polygon) be \(a\). Prove that the area of this regular octagon is \(2a^2 (1+\sqrt2) \).

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TopNewestLets try this problem https://brilliant.org/discussions/thread/rmo-2016/ – Ayush Rai · 4 months ago

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– Rohit Camfar · 3 weeks ago

@Ayush Rai I deactivated my account and forgot to ask a very important thing from you people. So, I broke my computer's firewall and opened Brilliant (hahaha). I had to ask you: Who is the regional coordinator of RMO? Is it someone like vice chancellor of the university or anyone else? I searched on internet but got nothing. I think it would be the person with the same post for every region. And one another thing is there any deadline to fill RMO form? I googled it and got that the filling started on 15th January. I was releaved thinking that the RMO is held in december so I have much time but no.Log in to reply

U can check this out – Ayush Rai · 3 weeks ago

the RMO coordinators are different for different regions.Log in to reply

– Rohit Camfar · 3 weeks ago

Hey you have given me the link to MP team.Log in to reply

– Ayush Rai · 3 weeks ago

oh..really .idkLog in to reply

– Rohit Camfar · 3 weeks ago

Are you kidding me?Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Does it ask to prove that BKCL are concyclic , if angle BAC = 90 degreess or vice versa .Log in to reply

– Ayush Rai · 3 months, 3 weeks ago

It asks to prove that BKCL are concyclic , if angle BAC = 90 degreesLog in to reply

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– Ayush Rai · 3 months, 3 weeks ago

Good one...+1Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Do you know , what is the reflection of a point about a line ???????Log in to reply

– Ayush Rai · 3 months, 3 weeks ago

yup.that is the easiest in co-geometryLog in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

Yeah ,I do know that . You mean the image of a point in the cartesian plane .Log in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

But I am asking you about the reflection of any point about a line on euclidean planeLog in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

I was solving a question " What is this line ? " On Brilliant when I encountered this term .Log in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

I took it easy and assumed that when a perpendicular drawn from the (actual) point to line is produced and a point at the same distance from the line (initial) as the actual point was from the line is cut off is the reflection of the point about that line .Log in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

I think that it was my misconceptionLog in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

So asked you .Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Have you applied for RMO (Sorry , I had earlier typed IMO by mistake )Log in to reply

– Ayush Rai · 3 months, 3 weeks ago

yeah.I wrote it this year.Could solve only 2.[both geometry]What about u?Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

I have came to know about this just now when I solved this queation I have had never heard about it earlier .Log in to reply

△ BDK ~ △ LDC ( By SAS similarity ) (Not used very often ) – Rohit Camfar · 3 months, 3 weeks ago

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– Rohit Camfar · 3 months, 3 weeks ago

Well , i Forgot to mention something . △ BDK ~ △ LDC By SAS similarity criteria ( Not used very often )Log in to reply

– Vishwash Kumar · 3 months, 3 weeks ago

Thanks !!!Log in to reply

– Ayush Rai · 3 months, 3 weeks ago

how do u know that KD=AD?Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Okay , then I have solvedLog in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Remember that in any right triangl the length of the line segment joining the midpoint of the hypo. to opposite vertex (containing right angle ) = 1/2 * hypotenuseLog in to reply

– Ayush Rai · 3 months, 3 weeks ago

Well even i solved it but iam lazy to write the solutionLog in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Remember that in any right triangl the length of the line segment joining the midpoint of the hypo. to opposite vertex (containing right angle ) = 1/2 * hypotenuseLog in to reply

– Rohit Camfar · 3 months, 3 weeks ago

How did you do it ??Log in to reply

– Ayush Rai · 3 months, 3 weeks ago

We note that the segments BC and KL cannot coincide, because the same would imply that AD is a perpendicular bisector of seg BC, and that ∆ABC is isosceles, contradicting the given information that the triangle is scalene. So without loss of generality, let A − B − K and A − L − C. We will separately prove the two parts of this problem:Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

I have a different solution .Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Hats OFF to my writing speed .Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Yeah , I was able to write my solution in 1 1/2 hours .Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

But I am gonna post it .Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Same here .Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Please have a look at RMO question.Log in to reply

– Rohit Camfar · 3 months, 3 weeks ago

Sorry , I saw your question yesterday . I am trying it .Log in to reply

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– Ayush Rai · 4 months ago

Even i wanted to show the same thing.Log in to reply

– Rohit Camfar · 4 months ago

niceLog in to reply

Now we use sin rule for the same triangle.So,\(Area(A)=\dfrac{1}{2}\times y^2\times cos45=\dfrac{1}{2}\times \dfrac{a^2\sqrt2}{2(\sqrt2-1)}\times \dfrac{1}{\sqrt2}=\dfrac{a^2}{4(\sqrt2-1)}.\)Now since all the triangles are congruent,the area of the Octagon\(=8\times \dfrac{a^2}{4(\sqrt2-1)}=\dfrac{2a^2}{\sqrt2-1}.\)Rationalizing the denominator we get it to be \(\boxed {2a^2(1+\sqrt2)}.\)Hence Proved. – Ayush Rai · 4 months ago

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– Rohit Camfar · 4 months ago

Can you do it using pure geometry . Without using trigonometry . It is under the head geometry.Log in to reply

– Ayush Rai · 4 months ago

Well i dont have any idea about that.Log in to reply

Quadrilateral in a triangle and I am not able to point out the mistake . can you help me ?

Well , I have a solution regarding this question . But I am messed up in an another geometry question right now . – Rohit Camfar · 4 months agoLog in to reply

– Ayush Rai · 4 months ago

can explain how u got ar.BPD,ar.BRE.... in the last statement.I am not able to figure the similar manner.Log in to reply

– Rohit Camfar · 4 months ago

Well , just in the similar manner as i have drawn ST || AC just like that drawing PM || AC and using the same technique , then RN ||\ AC and so onLog in to reply

– Ayush Rai · 4 months ago

I think so it is to the right of ELog in to reply

– Rohit Camfar · 4 months ago

I was able to point out my mistake ,wasted 7 hours . I was doing a silly mistake Ar. BQD sholud be 77/3 and I was taking it to be 154 / 3Log in to reply

– Ayush Rai · 4 months ago

Yes u are right!Log in to reply

– Rohit Camfar · 4 months ago

yeahLog in to reply

– Ayush Rai · 4 months ago

point M is to the left of E or right?Log in to reply

– Rohit Camfar · 4 months ago

It is to the right .Log in to reply

– Rohit Camfar · 4 months ago

But that does not matters , if you take it on left then DM = 8x/7 If you take it on right then EM = x/7 Both are same..............Log in to reply

– Ayush Rai · 4 months ago

can u give me only the question??all the details and informationsLog in to reply

– Rohit Camfar · 4 months ago

Well , I also need to rectify my solution . Please it is not too long and I am unable to point out the mistakeLog in to reply

– Rohit Camfar · 4 months ago

You can click on the link above Quadrilateral in a triangle above in my post . or you can search it as a level 5 problem .Log in to reply

– Ahmad Saad · 4 months ago

I can do that , but I want to know how post image file from my computer.Log in to reply

Quadrilateral in a triangle and I am not able to point out the mistake . can you help me ?

Well , I have a solution regarding this question . But I am messed up in an another geometry question right now . – Rohit Camfar · 4 months agoLog in to reply

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– Rohit Camfar · 4 months ago

Ar. BQD = 154/3 is not correct it should be 77/3Log in to reply

– Ayush Rai · 4 months ago

well thanks.now even i understood some tricks to solve problems.That was a nice problem and also construction.Any more geo problemsLog in to reply

– Rohit Camfar · 4 months ago

A similar question but somehow easier is A Classical geometry problem (level hard - geometry )Log in to reply

– Rohit Camfar · 4 months ago

Now i have arrived at my mistakeLog in to reply

– Rohit Camfar · 4 months ago

You are also telling the same thing . I have posted a solution by clearing how to proceed in similar manner . It will become too long so i haven't repeated it again and again .Log in to reply

– Ayush Rai · 4 months ago

you can read this one.https://brilliant.org/math-formatting-guide/#Log in to reply

– Ahmad Saad · 4 months ago

sorry, I read that but I still don't know how to use that. ThanksLog in to reply

– Rohit Camfar · 4 months ago

You must know how to post a solution to a general question as i have seen your solution . Just solve any easy question on brilliant you will have the right to post a solution and just insert a image there (the image of solution to this geometricc proof) you see that https...... something appears there copy that paste here in the comments and just post it ..................................................................Log in to reply

– Ahmad Saad · 4 months ago

Thanks.Log in to reply

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