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Geometry Contest!

Rules of the contest:

1.Suppose problem \(i\) is posted. The one who solves the problem and posts the solution becomes able to publish the \(i+1\) problem.

2.This will continue until the problem posted is not answered within \(2\) hours. If the solution is not posted, the problem maker himself starts with the next problem, posting a solution of the previous one.

3.The deadline question will be declared later. Meaning, the question with which the contest ends.

4.Whosoever posts a new problem should post on slack in #general that new problem is up!

5.The new problem poster should know the solution of his posted problem.

6.If the new problem is not posted in 15 minutes of answering the previous question, ANYBODY can post a new question.

7.Marking scheme is none, This time the question setter will decide the marks, Out of 5 and the one who answers will be given that credit. The marks will be set acc. To the level of the question as in brilliant!

8.We are starting from Question 1.

Points table:

1.Prince Loomba:8 marks

2.Julian Poon:5 marks

3..Michael Fuller:4 marks

4.Ayush Rai:3 marks


EDIT: the contest has been moved

Note by Ayush Rai
4 months, 1 week ago

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Problem 9 [5 marks]

Let \(X\) and \(Y\) be points inside equilateral triangle \(ABC\). Let \(Y'\) be the reflection of \(Y\) in line \(BC\). Prove that

\[XY+XB+XC \geq Y'A\] Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa We will consider the worst case. Here YYprime is max. So Y is at A or YYprime=2hA (altitude from A to BC). Now to minimise LHS, we will take X as centroid otherwise XA+XB+XC will be greater. So we have to prove that 2/3 (Ma+Mb+Mc)>2hA

Or Ma+Mb+Mc>3hA Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Your first statement is unjustified, so technically, you are incorrect. You need not consider worst-case scenarios. In these questions, you must apply generally, not just in the worst-case scenario. Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa Why If we prove for worst case when LHS is minimum and RHS is max then also LHS is greater, inequality is proved. Thats a rule. Prince Loomba · 4 months, 1 week ago

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@Prince Loomba But why is the worst-case scenario at Y? These are the sorts of questions you must answer in a proof. Otherwise, your proof is considered null as it only considers one case. Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa I am ready to reply. Because Y'A is maximum when YY' is maximum. And maximum value of Y on perpendicular to BC can be A as after A it will move out of the triangle Prince Loomba · 4 months, 1 week ago

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@Prince Loomba I'm sorry, but this is not correct. This is still not justifying why this is true. Why is this the worst case scenario? What if Y=C (hypothetically)? This must all be answered. To define a worst-case scenario, you must describe why it is, why it can't occur anywhere else. Please, don't use wors-case scenarios in Geometry as it is very complicated. Sharky Kesa · 4 months, 1 week ago

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@Armain Labeeb Read carefully, Y is at A Prince Loomba · 4 months, 1 week ago

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@Sharky Kesa I have given you 10 cases Armain Labeeb · 4 months, 1 week ago

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@Sharky Kesa Ok I will try more cases.

See this,

IF X=Y=B or X=C=Y then the equation holds true that LHS=RHS.

IF X=Y is close to B or C the equation holds true that LHS>RHS

IF X=Y=A then the equation holds that LHS>RHS

IF X=Y is close to A then the equation holds that LHS>RHS

IF X=B or is close to B and Y=/=X and is far from B then the equation holds that LHS>RHS

IF X=B or is close to B and Y=/=X and is near B then the equation holds that LHS>RHS

IF X=C or is close to C and Y=/=X and is far from B then the equation holds that LHS>RHS

IF X=C or is close to C and Y=/=X and is near B then the equation holds that LHS>RHS.

IF X=centroid or is close to B and Y=/=X and is far from the centroid then the equation holds that LHS>RHS

IF X=centroid or is close to B and Y=/=X and is near the centroid then the equation holds that LHS>RHS

So LHS is greater than or equal to RHS

Done. Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb You will have to prove 'in geenral ' , not case by case coz there are infinite cases. Harsh Shrivastava · 4 months, 1 week ago

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@Harsh Shrivastava all types of cases are part of these cases though. Armain Labeeb · 4 months, 1 week ago

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@Sharky Kesa http://i.imgur.com/xmrurud.png Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb come to https://brilliant.org/discussions/thread/brilliant-geometry-contest-season-1/.This problem is shifted there. Ayush Rai · 4 months, 1 week ago

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@Sharky Kesa Ok I will still try to prove AGAIN using scenarios.

Ok so let's first minimise XY+XB+XC. Suppose X=B=Y then XB=XY=0 OR X=C=Y then XC=XY=0. We know that AB=AC=BC=B'C. Then X'=Y'=B' or X'=Y'=C. So A is the furthest point from X and Y. I will prove that the the worst scenario of LHS is the best scenario of RHS. Ok I will let X be a tpont B (works for C too). Since X=B=Y, XB=XY=0 and XY+XB+XC=XC=BC only. Now the furthest length from A that Y can be at is at point X which is either Point B or C. YA=Y'A=BC=AB=AC=XY+XB=XC. So it is possible that XY+XB+XC=Y'A. -eq(1)

Now to prove that LHS can be greater than RHS, let's take X=centroid of triangle than XA=XB=XC and Y be at the furthest point from A which is B and C. So XY+XB+XC=3XA=3XB=3XC. and AY'=AY. Clearly 3XA is greater than YA because Y is not close to being 3 times as far from X. SO, XY + XB+XC >Y'A. -eq(ii)

Combining two equations we have

\[XY+XB+XC \geq Y'A\] Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb You haven't shown why it is possible for X or Y to be placed elsewhere and satisfy. You've considered some cases, not the entire general thing. Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa I have proved that LHS can be greater than RHs and also that LHS can be equal to RHS and why LHs cannot be less than RHS. What more proof do you need? If I put X and Y in random points then will it be complete? Armain Labeeb · 4 months, 1 week ago

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@Sharky Kesa Let me prove this. You understood Y=A is worst case? Prince Loomba · 4 months, 1 week ago

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@Sharky Kesa So one more case will do it? Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb No you are on the wrong track. We have to prove this is the worst case i.e, LHS cant be less than this and RHS cant be greater that that Prince Loomba · 4 months, 1 week ago

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@Prince Loomba I did didnt I? Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb Thhats what I was saying the worst case is Y=A and X=centroid Prince Loomba · 4 months, 1 week ago

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@Sharky Kesa Am I right till now? Prince Loomba · 4 months, 1 week ago

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@Sharky Kesa I am only 13 and very bad at geometry. if there is a problem, pls ignore. also, my english is not good Armain Labeeb · 4 months, 1 week ago

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@Sharky Kesa I am very bad at geometry so this is my concept. For minimizing XY+XB+XC we need x to be infinitely close to either points B or C and Y to be infinitely close to X. so if X is infinitely close to point B then XB + XC+ XY= XC+0.000.....0001+0.000.....0001=XC+0+0=XC or if X is infinitely close to point C then XB + XC+ XY= XB+0.000.....0001+0.000.....0001=XB+0+0=XB. As y is infintely close to x, y prime = x prime. As x is infinitely close to B or C. y prime is equal to B prime or C prime in that case. Since the triangle is equilateral XA=XB=XC so we can say XY+XB+XC can be equal to Y'A. IF you take any other case you will find that XY+XB+XC is greater than Y'A. So the equation is true Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb Your reasoning is very flawed so I'm afraid you are incorrect. Sharky Kesa · 4 months, 1 week ago

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Problem 4 [2 points]

Suppose your house is at coordinate (1,1)

You want to visit your friend's house which is located at coordinate (-3, 5)

However, you want to first walk towards the sea to collect seashells, and then walk to your friend's house.

The coastline is described as the line \(y=0\), you can visit any point of the line to collect your seashells.

Now, what is the minimum distance you would have to walk? Julian Poon · 4 months, 1 week ago

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@Julian Poon My point is if you were to only go to your friends house you d have to cross the y axis. then you the minimum distance should be the just the distance from your house to your house which can be solved by pythagoras theorem. if the distance was a. then i have

a^2 = 2×4^2. because the difference between x and y coordinates of both x and y is the same which is 4.

Solving we have a=sqrt 32 Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb Yes I also think this Prince Loomba · 4 months, 1 week ago

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@Armain Labeeb The x- axis is the coastline and not y- axis. Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Right. Its sqrt 52 Prince Loomba · 4 months, 1 week ago

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@Julian Poon The answer is \(\sqrt{52}\) or \(2\sqrt{13}\) Julian Poon · 4 months, 1 week ago

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@Julian Poon Oh boy, that means I have to post another one Michael Fuller · 4 months, 1 week ago

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@Ayush Rai I dont know much but i think the answer is sqrt 32. I d use latex if i was on computer Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb oh damn sorry if y=0 then that means its x axis haha. srry for spamming Armain Labeeb · 4 months, 1 week ago

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@Julian Poon \(2\sqrt{13}\)

Reflect the point \((-3,5)\) in the \(x\) axis and walk from \((1,1)\) to \((-3,-5)\) in a straight line. Using Pythagoras this is \(\sqrt{6^2+4^2}=\sqrt{52}=\large \color{green}{\boxed{2\sqrt{13}}}\). Now reflect the part of the line below the \(x\) axis in the \(x\) axis, and this is the minimum distance Michael Fuller · 4 months, 1 week ago

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Problem 5 (\(x-1\) marks)

An equilateral triangle of side \(1\) has a circle inscribed in it, so that it touches all sides of the triangle at their medians. An equilateral triangle is inscribed in the circle.

The process of circle inscribing and triangle inscribing is repeated until infinity. Find the exact value of the sum of the areas of all triangles.

If the value is in the form \(\dfrac{\sqrt{x}}{y}\), find \(x+y\). Michael Fuller · 4 months, 1 week ago

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@Michael Fuller The answer is \(\frac{\sqrt{3}}{3}\).

Each smaller triangle's area can be seen to be 1/4 of the triangle bigger than it.

For instance, the first triangle's area is \(\sqrt{3}/4\). THe second triangle would be that area divided by 4. The third triangle would be the second divided by four again and so on.

Hence, summing all the areas, the total area gives \(\frac{\sqrt{3}}{3}\) Julian Poon · 4 months, 1 week ago

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@Julian Poon @Julian Poon the answer is 6 not sqrt3÷3 you have to x+y not the whole. But that doesnt matter because your solution is right. Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb Ah well... Julian Poon · 4 months, 1 week ago

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@Julian Poon Correct! \(x-1\) marks for you good sir. Michael Fuller · 4 months, 1 week ago

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@Michael Fuller Haha x-1 marks is x used up in the question, right? Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Indeed \(:)\) Michael Fuller · 4 months, 1 week ago

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Question 3 (3 marks)

The above figure shows a medal, comprised of a square of side \(1\) with a four-pointed star inscribed in it, such that the angle between the points is \(120^{\circ}\) as shown.

If the area of the circle inscribed in the star is \(\dfrac{a-\sqrt3}{b} \pi\) for integers \(a\) and \(b\), find \(ab\). Michael Fuller · 4 months, 1 week ago

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@Michael Fuller Area is \[\frac{2-\sqrt3}{6}\pi\] thus , answer is \(\color{green}{\boxed{12}}\) Aman Rajput · 4 months, 1 week ago

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@Aman Rajput Correct, but Julian answered first \(:)\) Michael Fuller · 4 months, 1 week ago

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@Aman Rajput Julian already got it. Ayush Rai · 4 months, 1 week ago

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@Michael Fuller The answer is 12.

To see why, the radius of the circle is half the length of the side minus the height of the triangle.

This gives

\[r=\frac{1}{2}-\frac{1}{\sqrt{3}}\]

THe area of the circle is \[\pi r^2\]

THis gives the area of the circle:

\[\frac{2-\sqrt{3}}{6}*\pi\] Julian Poon · 4 months, 1 week ago

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@Julian Poon Post next Prince Loomba · 4 months, 1 week ago

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@Julian Poon Winner! Michael Fuller · 4 months, 1 week ago

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@Michael Fuller Its 2D ?

And only of 3 marks? Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Yes, the figure is 2D. My solution is not too complicated, but also harder than your 2 marker, hence I will reward 3 points for it Michael Fuller · 4 months, 1 week ago

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@Michael Fuller Haha TIT FOR TAT

I posted 2 marks and you 3 marks comparing with me Prince Loomba · 4 months, 1 week ago

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Question 2 (2 marks)

Find perimeter of the figure. Prince Loomba · 4 months, 1 week ago

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@Prince Loomba The above figure is a square with an equilateral triangle \(AED\) "indented" into it, since all angles are \(60^{\circ}\). Therefore \(AE=ED=AD=12\), and the perimeter is \(5 \times 12 = \large \color{green}{\boxed{60}}\). Michael Fuller · 4 months, 1 week ago

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@Michael Fuller Nice solution. I was thinking of a solution when you posted this. You win haha :P Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb Hehe. Posted the next problem. Michael Fuller · 4 months, 1 week ago

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@Michael Fuller Right Post next problem Prince Loomba · 4 months, 1 week ago

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Question 1[3marks]

A semicircle of diameter \(1\) sits at the top of a semicircle of diameter \(2\),as shown.The shaded area inside the smaller semicircle and outside the larger semicircle is called lune.Determine the area of the lune. Ayush Rai · 4 months, 1 week ago

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@Ayush Rai

Prince Loomba · 4 months, 1 week ago

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@Prince Loomba It is the correct answer. Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Thanks, I will post new soon Prince Loomba · 4 months, 1 week ago

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https://brilliant.org/discussions/thread/brilliant-geometry-contest-season-1/#comment-a3f5119307e92 GO HERE Prince Loomba · 4 months, 1 week ago

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i want to take part in this where is the present question Abhishek Alva · 4 months, 1 week ago

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@Abhishek Alva goto the comments sorting option and click New, instead of Top. Anyways, Problem 9 is most recent. No one has solved properly. Sharky Kesa · 4 months, 1 week ago

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@Ayush Rai I've already posted Problem 9, since I was told to. Sharky Kesa · 4 months, 1 week ago

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Problem 8 [3 marks]

Prove that the locus of a centre of a circle touching both another circle and a straight line which in turn do not intersect is a parabola

eg.

Let \(C_1\) be the required circle and \(C_2\) given circle and \(L_1\) be the given line then:

  1. \(C_1\) touches \(L_1\) and \(C_2\).

  2. \(C_2\) does not touch \(L_1\).

Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Repost the problem 8. Harsh Shrivastava · 4 months, 1 week ago

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@Harsh Shrivastava Obviously yes Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Problem looked familiar. https://brilliant.org/problems/locus-of-circles-centre/ Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa You did a wrong thing by posting the link. You should have written your solution. Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Would've taken too long. Sorry. Also, I recognised the question (Eidetic/Photographic memory). Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa Try the proof without seeing. Its great really Prince Loomba · 4 months, 1 week ago

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@Sharky Kesa But the proof is great. Prince Loomba · 4 months, 1 week ago

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@Prince Loomba True, but I found this problem not too hard to show via co-ord geom. Still, my problem to post? Sharky Kesa · 4 months, 1 week ago

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@Prince Loomba

Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Its mine solution. So thats cheating. Prince Loomba · 4 months, 1 week ago

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@Prince Loomba well,didn't you do! Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Post the next. Harsh Shrivastava · 4 months, 1 week ago

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@Ayush Rai See the bedsheet behind the register, its also same! How can this be possible? Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Alternate universes.

Sharky Kesa · 4 months, 1 week ago

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@Ayush Rai You just copied his solution :P Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb i wrote it.How can you say that?our handwriting is same Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Well I took his solution and your solution and compared it pixel by pixel and compared colours. It is also mesmerizing that you took the photo at the same angle and from the smae distance and under the same light and in the same notebook of his. What is even more surprising? The solutions are the same.

_<

Armain Labeeb · 4 months, 1 week ago

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@Armain Labeeb One more thing see the bedsheet behind the register, Its also same Prince Loomba · 4 months, 1 week ago

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@Armain Labeeb Haha. Its mine. I can identify my writing Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Maybe you two are the same people from alternate universes. :P Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa maybe Ayush Rai · 4 months, 1 week ago

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@Ayush Rai So, do I get the points for it? Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa no Ayush Rai · 4 months, 1 week ago

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@Ayush Rai K, that's fine as long as problem remains. Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa He is a bad person. See he awarded marks to himself Prince Loomba · 4 months, 1 week ago

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Problem 7 [3 points] Suppose the medians of a triangle are 5,12,13 units.Find the sides of the triangle. Ayush Rai · 4 months, 1 week ago

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@Ayush Rai 26/3,2(sqrt244)/3,2(sqrt601)/3 Prince Loomba · 4 months, 1 week ago

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@Prince Loomba ya! sharky is right.@Prince Loomba post your solution. Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Its a direct formula. Side=\(\frac {2}{3} \times \sqrt {2 ((m_{a})^2+(m_b)^{2})-{m_{c}}^2}\) Prince Loomba · 4 months, 1 week ago

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@Prince Loomba oh..thats nice Ayush Rai · 4 months, 1 week ago

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@Ayush Rai Why am I awarded points even though I didn't posted the solution to previous problem? Harsh Shrivastava · 4 months, 1 week ago

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@Harsh Shrivastava you didn't.really? Ayush Rai · 4 months, 1 week ago

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@Ayush Rai BTW, my method is using formula \(AB^2 + AC^2 = 2(AD^2 + BD^2)\), then using simultaneous linear equations. Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa Even i used the same thing. Ayush Rai · 4 months, 1 week ago

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@Prince Loomba Beat me to it!!! Can we please post proofs of why it is true? Sharky Kesa · 4 months, 1 week ago

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@Prince Loomba Post next bro. Harsh Shrivastava · 4 months, 1 week ago

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@Prince Loomba correct! Ayush Rai · 4 months, 1 week ago

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Problem 6 [2 points]

What is the area of the largest circle possible with center at coordinate (0,0), but doesn't intersect the area bounded by \(y>\frac{1}{\left|x\right|}\)? Julian Poon · 4 months, 1 week ago

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@Julian Poon Simple, easy question. We wish to find the shortest distance between the Origin and this graph. But this can be easily done as follows: We have \(y = \frac{1}{|x|}\) and we wish to minimise \(x^2 + y^2\) (Using distance formula). Substituting \(y^2 = \frac {1}{x^2}\), we have \(x^2 + \frac {1}{x^2}\). Note that \(x^2\) is non-negative, so we can apply AM-GM to get that its minimum value is \(2\). Thus, the minimum distance between the curve and the origin is \(\sqrt{2}\). Thus, the largest circle satisfying has a radius of \(\sqrt{2}\), so it has an area of \(2\pi\). Sharky Kesa · 4 months, 1 week ago

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@Sharky Kesa I have a short trick. The nearest point is (1,1) and (-1,1). And if circle is bigger than that, then it will be excluded. So circle covering 1,1 and center origin area is 2pi Prince Loomba · 4 months, 1 week ago

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@Julian Poon Is the ander 2pi? Harsh Shrivastava · 4 months, 1 week ago

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@Harsh Shrivastava Post next bro Prince Loomba · 4 months, 1 week ago

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@Prince Loomba Sorry my net was down a bit. Harsh Shrivastava · 4 months, 1 week ago

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@Harsh Shrivastava Correct! Julian Poon · 4 months, 1 week ago

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@Harsh Shrivastava Same here, 2pi Prince Loomba · 4 months, 1 week ago

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