@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.

@Vishwash Kumar
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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 .

@Rohit Camfar
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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

@Ayush Rai
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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
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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:

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
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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 ?

@Ayush Rai
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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

@Ahmad Saad
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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
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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 .

@Ahmad Saad
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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 ..................................................................

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TopNewestLets try this problem https://brilliant.org/discussions/thread/rmo-2016/

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@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.

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

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

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

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

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Please have a look at RMO question.

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

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Even i wanted to show the same thing.

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nice

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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.

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

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Well i dont have any idea about that.

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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 .Log in to reply

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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 .Log in to reply

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Comment deleted 11 months ago

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