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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
1 month, 2 weeks ago

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Lets try this problem https://brilliant.org/discussions/thread/rmo-2016/ · 1 month, 1 week ago

Does it ask to prove that BKCL are concyclic , if angle BAC = 90 degreess or vice versa . · 1 month ago

It asks to prove that BKCL are concyclic , if angle BAC = 90 degrees · 1 month ago

· 1 month ago

Good one...+1 · 1 month ago

Do you know , what is the reflection of a point about a line ??????? · 1 month ago

yup.that is the easiest in co-geometry · 1 month ago

Yeah ,I do know that . You mean the image of a point in the cartesian plane . · 1 month ago

But I am asking you about the reflection of any point about a line on euclidean plane · 1 month ago

I was solving a question " What is this line ? " On Brilliant when I encountered this term . · 1 month 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 . · 1 month ago

I think that it was my misconception · 1 month ago

So asked you . · 1 month ago

Have you applied for RMO (Sorry , I had earlier typed IMO by mistake ) · 1 month ago

yeah.I wrote it this year.Could solve only 2.[both geometry]What about u? · 1 month ago

Well I forgot to mention something

△ BDK ~ △ LDC ( By SAS similarity ) (Not used very often ) · 1 month ago

Well , i Forgot to mention something . △ BDK ~ △ LDC By SAS similarity criteria ( Not used very often ) · 1 month ago

Thanks !!! · 1 month ago

how do u know that KD=AD? · 1 month ago

Okay , then I have solved · 1 month 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 · 1 month ago

Well even i solved it but iam lazy to write the solution · 1 month 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 · 1 month ago

How did you do it ?? · 1 month 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: · 1 month ago

I have a different solution . · 1 month ago

Hats OFF to my writing speed . · 1 month ago

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

But I am gonna post it . · 1 month ago

Same here . · 1 month ago

Please have a look at RMO question. · 1 month ago

Sorry , I saw your question yesterday . I am trying it . · 1 month ago

· 1 month, 1 week ago

Even i wanted to show the same thing. · 1 month, 1 week ago

nice · 1 month, 1 week 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. · 1 month, 2 weeks ago

Can you do it using pure geometry . Without using trigonometry . It is under the head geometry. · 1 month, 1 week ago

Well i dont have any idea about that. · 1 month, 1 week 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 ?

· 1 month, 1 week ago

can explain how u got ar.BPD,ar.BRE.... in the last statement.I am not able to figure the similar manner. · 1 month, 1 week 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 · 1 month, 1 week ago

I think so it is to the right of E · 1 month, 1 week 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 · 1 month, 1 week ago

Yes u are right! · 1 month, 1 week ago

yeah · 1 month, 1 week ago

point M is to the left of E or right? · 1 month, 1 week ago

It is to the right . · 1 month, 1 week 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.............. · 1 month, 1 week ago

can u give me only the question??all the details and informations · 1 month, 1 week ago

Well , I also need to rectify my solution . Please it is not too long and I am unable to point out the mistake · 1 month, 1 week 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 . · 1 month, 1 week ago

I can do that , but I want to know how post image file from my computer. · 1 month, 1 week 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 ?

· 1 month, 1 week ago

· 1 month, 1 week ago

Ar. BQD = 154/3 is not correct it should be 77/3 · 1 month, 1 week ago

well thanks.now even i understood some tricks to solve problems.That was a nice problem and also construction.Any more geo problems · 1 month, 1 week ago

A similar question but somehow easier is A Classical geometry problem (level hard - geometry ) · 1 month, 1 week ago

Now i have arrived at my mistake · 1 month, 1 week 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 . · 1 month, 1 week ago

you can read this one.https://brilliant.org/math-formatting-guide/# · 1 month, 1 week ago

sorry, I read that but I still don't know how to use that. Thanks · 1 month, 1 week 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 .................................................................. · 1 month, 1 week ago

Thanks. · 1 month, 1 week ago

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