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# Geometry Proof

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

Note by Rohit Camfar
12 months ago

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- 12 months 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.

- 8 months, 2 weeks ago

the RMO coordinators are different for different regions.U can check this out

- 8 months, 2 weeks ago

Hey you have given me the link to MP team.

- 8 months, 2 weeks ago

oh..really .idk

- 8 months, 2 weeks ago

Are you kidding me?

- 8 months, 2 weeks ago

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

- 11 months, 3 weeks ago

It asks to prove that BKCL are concyclic , if angle BAC = 90 degrees

- 11 months, 3 weeks ago

- 11 months, 3 weeks ago

Good one...+1

- 11 months, 3 weeks ago

Do you know , what is the reflection of a point about a line ???????

- 11 months, 3 weeks ago

yup.that is the easiest in co-geometry

- 11 months, 3 weeks ago

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

- 11 months, 3 weeks ago

But I am asking you about the reflection of any point about a line on euclidean plane

- 11 months, 3 weeks ago

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

- 11 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 .

- 11 months, 3 weeks ago

I think that it was my misconception

- 11 months, 3 weeks ago

- 11 months, 3 weeks ago

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

- 11 months, 3 weeks ago

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

- 11 months, 3 weeks ago

- 11 months, 3 weeks ago

Well I forgot to mention something

△ BDK ~ △ LDC ( By SAS similarity ) (Not used very often )

- 11 months, 3 weeks ago

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

- 11 months, 3 weeks ago

Thanks !!!

- 11 months, 3 weeks ago

how do u know that KD=AD?

- 11 months, 3 weeks ago

Okay , then I have solved

- 11 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 * hypotenuse

- 11 months, 3 weeks ago

Well even i solved it but iam lazy to write the solution

- 11 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 * hypotenuse

- 11 months, 3 weeks ago

How did you do it ??

- 11 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:

- 11 months, 3 weeks ago

I have a different solution .

- 11 months, 3 weeks ago

Hats OFF to my writing speed .

- 11 months, 3 weeks ago

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

- 11 months, 3 weeks ago

But I am gonna post it .

- 11 months, 3 weeks ago

Same here .

- 11 months, 3 weeks ago

Please have a look at RMO question.

- 11 months, 3 weeks ago

Sorry , I saw your question yesterday . I am trying it .

- 11 months, 3 weeks ago

- 12 months ago

Even i wanted to show the same thing.

- 12 months ago

nice

- 12 months ago

All the eight triangles are congruent isosceles triangles since it is a regular octagon.Since angle of each triangle(the angle opposite to the sides of the octagon).Now take one of those triangles and use cosine rule to find the two equal lengths of that triangle.let those sides be $$y.$$So,$$a^2=y^2+y^2-2\times y\times y\times cos 45\Rightarrow y^2=\dfrac{a^2\sqrt2}{2(\sqrt2-1)}.$$
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.

- 12 months ago

Can you do it using pure geometry . Without using trigonometry . It is under the head geometry.

- 12 months ago

Well i dont have any idea about that.

- 12 months ago

Well , I have a solution regarding this question . But I am messed up in an another geometry question right now . Quadrilateral in a triangle and I am not able to point out the mistake . can you help me ?

- 12 months ago

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

- 12 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 on

- 12 months ago

I think so it is to the right of E

- 12 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 / 3

- 12 months ago

Yes u are right!

- 12 months ago

yeah

- 12 months ago

point M is to the left of E or right?

- 12 months ago

It is to the right .

- 12 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..............

- 12 months ago

can u give me only the question??all the details and informations

- 12 months ago

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

- 12 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 .

- 12 months ago

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

- 12 months ago

Well , I have a solution regarding this question . But I am messed up in an another geometry question right now . Quadrilateral in a triangle and I am not able to point out the mistake . can you help me ?

- 12 months ago

- 12 months ago

Ar. BQD = 154/3 is not correct it should be 77/3

- 12 months ago

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

- 12 months ago

A similar question but somehow easier is A Classical geometry problem (level hard - geometry )

- 12 months ago

Now i have arrived at my mistake

- 12 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 .

- 12 months ago

- 12 months ago

sorry, I read that but I still don't know how to use that. Thanks

- 12 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 ..................................................................

- 12 months ago

Thanks.

- 12 months ago

Comment deleted 11 months ago