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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
4 months ago

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

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@Ayush Rai @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. Rohit Camfar · 3 weeks ago

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@Rohit Camfar the RMO coordinators are different for different regions.U can check this out Ayush Rai · 3 weeks ago

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@Ayush Rai Hey you have given me the link to MP team. Rohit Camfar · 3 weeks ago

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@Rohit Camfar oh..really .idk Ayush Rai · 3 weeks ago

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@Ayush Rai Are you kidding me? Rohit Camfar · 3 weeks ago

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@Ayush Rai Does it ask to prove that BKCL are concyclic , if angle BAC = 90 degreess or vice versa . Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar It asks to prove that BKCL are concyclic , if angle BAC = 90 degrees Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai

Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar Good one...+1 Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai Do you know , what is the reflection of a point about a line ??????? Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar yup.that is the easiest in co-geometry Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai Yeah ,I do know that . You mean the image of a point in the cartesian plane . Vishwash Kumar · 3 months, 3 weeks ago

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@Vishwash Kumar But I am asking you about the reflection of any point about a line on euclidean plane Vishwash Kumar · 3 months, 3 weeks ago

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@Vishwash Kumar I was solving a question " What is this line ? " On Brilliant when I encountered this term . Vishwash Kumar · 3 months, 3 weeks ago

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@Vishwash Kumar 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 . Vishwash Kumar · 3 months, 3 weeks ago

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@Vishwash Kumar I think that it was my misconception Vishwash Kumar · 3 months, 3 weeks ago

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@Vishwash Kumar So asked you . Vishwash Kumar · 3 months, 3 weeks ago

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@Ayush Rai Have you applied for RMO (Sorry , I had earlier typed IMO by mistake ) Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar yeah.I wrote it this year.Could solve only 2.[both geometry]What about u? Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai I have came to know about this just now when I solved this queation I have had never heard about it earlier . Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Well I forgot to mention something

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

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@Ayush Rai Well , i Forgot to mention something . △ BDK ~ △ LDC By SAS similarity criteria ( Not used very often ) Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Thanks !!! Vishwash Kumar · 3 months, 3 weeks ago

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@Rohit Camfar how do u know that KD=AD? Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai Okay , then I have solved Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar 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 Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar Well even i solved it but iam lazy to write the solution Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai 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 Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai How did you do it ?? Rohit Camfar · 3 months, 3 weeks ago

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@Rohit Camfar 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: Ayush Rai · 3 months, 3 weeks ago

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@Ayush Rai I have a different solution . Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Hats OFF to my writing speed . Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Yeah , I was able to write my solution in 1 1/2 hours . Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai But I am gonna post it . Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Same here . Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Please have a look at RMO question. Rohit Camfar · 3 months, 3 weeks ago

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@Ayush Rai Sorry , I saw your question yesterday . I am trying it . Rohit Camfar · 3 months, 3 weeks ago

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Ahmad Saad · 4 months ago

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@Ahmad Saad Even i wanted to show the same thing. Ayush Rai · 4 months ago

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@Ayush Rai nice Rohit Camfar · 4 months ago

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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. Ayush Rai · 4 months ago

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@Ayush Rai Can you do it using pure geometry . Without using trigonometry . It is under the head geometry. Rohit Camfar · 4 months ago

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@Rohit Camfar Well i dont have any idea about that. Ayush Rai · 4 months ago

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@Ayush Rai 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 ?

Rohit Camfar · 4 months ago

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@Rohit Camfar can explain how u got ar.BPD,ar.BRE.... in the last statement.I am not able to figure the similar manner. Ayush Rai · 4 months ago

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@Ayush Rai 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 Rohit Camfar · 4 months ago

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@Rohit Camfar I think so it is to the right of E Ayush Rai · 4 months ago

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@Ayush Rai 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 Rohit Camfar · 4 months ago

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@Rohit Camfar Yes u are right! Ayush Rai · 4 months ago

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@Ayush Rai yeah Rohit Camfar · 4 months ago

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@Rohit Camfar point M is to the left of E or right? Ayush Rai · 4 months ago

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@Ayush Rai It is to the right . Rohit Camfar · 4 months ago

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@Rohit Camfar 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.............. Rohit Camfar · 4 months ago

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@Rohit Camfar can u give me only the question??all the details and informations Ayush Rai · 4 months ago

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@Ayush Rai Well , I also need to rectify my solution . Please it is not too long and I am unable to point out the mistake Rohit Camfar · 4 months ago

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@Ayush Rai 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 . Rohit Camfar · 4 months ago

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@Ayush Rai I can do that , but I want to know how post image file from my computer. Ahmad Saad · 4 months ago

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@Ahmad Saad 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 ?

Rohit Camfar · 4 months ago

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@Rohit Camfar

Ahmad Saad · 4 months ago

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@Ahmad Saad Ar. BQD = 154/3 is not correct it should be 77/3 Rohit Camfar · 4 months ago

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@Rohit Camfar well thanks.now even i understood some tricks to solve problems.That was a nice problem and also construction.Any more geo problems Ayush Rai · 4 months ago

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@Ayush Rai A similar question but somehow easier is A Classical geometry problem (level hard - geometry ) Rohit Camfar · 4 months ago

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@Rohit Camfar Now i have arrived at my mistake Rohit Camfar · 4 months ago

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@Rohit Camfar 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 . Rohit Camfar · 4 months ago

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@Ahmad Saad you can read this one.https://brilliant.org/math-formatting-guide/# Ayush Rai · 4 months ago

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@Ayush Rai sorry, I read that but I still don't know how to use that. Thanks Ahmad Saad · 4 months ago

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@Ahmad Saad 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 .................................................................. Rohit Camfar · 4 months ago

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@Rohit Camfar Thanks. Ahmad Saad · 4 months ago

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Comment deleted 3 months ago

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@Rohit Camfar

Ahmad Saad · 4 months ago

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