**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

## Comments

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TopNewestProblem 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 · 8 months ago

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Or Ma+Mb+Mc>3hA – Prince Loomba · 8 months ago

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– Sharky Kesa · 8 months ago

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.Log in to reply

– Prince Loomba · 8 months ago

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.Log in to reply

– Sharky Kesa · 8 months ago

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.Log in to reply

– Prince Loomba · 8 months ago

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 triangleLog in to reply

– Sharky Kesa · 8 months ago

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.Log in to reply

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– Prince Loomba · 8 months ago

Read carefully, Y is at ALog in to reply

– Armain Labeeb · 8 months ago

I have given you 10 casesLog in to reply

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 · 8 months ago

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– Harsh Shrivastava · 8 months ago

You will have to prove 'in geenral ' , not case by case coz there are infinite cases.Log in to reply

– Armain Labeeb · 8 months ago

all types of cases are part of these cases though.Log in to reply

– Armain Labeeb · 8 months ago

http://i.imgur.com/xmrurud.pngLog in to reply

– Ayush Rai · 8 months ago

come to https://brilliant.org/discussions/thread/brilliant-geometry-contest-season-1/.This problem is shifted there.Log in to reply

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 · 8 months ago

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– Sharky Kesa · 8 months ago

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.Log in to reply

– Armain Labeeb · 8 months ago

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?Log in to reply

– Prince Loomba · 8 months ago

Let me prove this. You understood Y=A is worst case?Log in to reply

– Armain Labeeb · 8 months ago

So one more case will do it?Log in to reply

– Prince Loomba · 8 months ago

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 thatLog in to reply

– Armain Labeeb · 8 months ago

I did didnt I?Log in to reply

– Prince Loomba · 8 months ago

Thhats what I was saying the worst case is Y=A and X=centroidLog in to reply

– Prince Loomba · 8 months ago

Am I right till now?Log in to reply

– Armain Labeeb · 8 months ago

I am only 13 and very bad at geometry. if there is a problem, pls ignore. also, my english is not goodLog in to reply

– Armain Labeeb · 8 months ago

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 trueLog in to reply

– Sharky Kesa · 8 months ago

Your reasoning is very flawed so I'm afraid you are incorrect.Log in to reply

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 · 8 months ago

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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 · 8 months ago

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– Prince Loomba · 8 months ago

Yes I also think thisLog in to reply

– Ayush Rai · 8 months ago

The x- axis is the coastline and not y- axis.Log in to reply

– Prince Loomba · 8 months ago

Right. Its sqrt 52Log in to reply

– Julian Poon · 8 months ago

The answer is \(\sqrt{52}\) or \(2\sqrt{13}\)Log in to reply

– Michael Fuller · 8 months ago

Oh boy, that means I have to post another oneLog in to reply

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– Armain Labeeb · 8 months ago

I dont know much but i think the answer is sqrt 32. I d use latex if i was on computerLog in to reply

– Armain Labeeb · 8 months ago

oh damn sorry if y=0 then that means its x axis haha. srry for spammingLog in to reply

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 · 8 months 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 · 8 months ago

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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 · 8 months ago

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@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 · 8 months ago

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– Julian Poon · 8 months ago

Ah well...Log in to reply

– Michael Fuller · 8 months ago

Correct! \(x-1\) marks for you good sir.Log in to reply

– Prince Loomba · 8 months ago

Haha x-1 marks is x used up in the question, right?Log in to reply

– Michael Fuller · 8 months ago

Indeed \(:)\)Log in to reply

Question 3 (3 marks)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 · 8 months ago

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– Aman Rajput · 8 months ago

Area is \[\frac{2-\sqrt3}{6}\pi\] thus , answer is \(\color{green}{\boxed{12}}\)Log in to reply

– Michael Fuller · 8 months ago

Correct, but Julian answered first \(:)\)Log in to reply

– Ayush Rai · 8 months ago

Julian already got it.Log in to reply

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 · 8 months ago

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– Prince Loomba · 8 months ago

Post nextLog in to reply

– Michael Fuller · 8 months ago

Winner!Log in to reply

And only of 3 marks? – Prince Loomba · 8 months ago

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– Michael Fuller · 8 months ago

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 itLog in to reply

I posted 2 marks and you 3 marks comparing with me – Prince Loomba · 8 months ago

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Question 2 (2 marks)Find perimeter of the figure. – Prince Loomba · 8 months ago

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– Michael Fuller · 8 months ago

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 \(5 \times 12 = \large \color{green}{\boxed{60}}\).Log in to reply

– Armain Labeeb · 8 months ago

Nice solution. I was thinking of a solution when you posted this. You win haha :PLog in to reply

– Michael Fuller · 8 months ago

Hehe. Posted the next problem.Log in to reply

– Prince Loomba · 8 months ago

Right Post next problemLog in to reply

Question 1[3marks]Log in to reply

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– Ayush Rai · 8 months ago

It is the correct answer.Log in to reply

– Prince Loomba · 8 months ago

Thanks, I will post new soonLog in to reply

https://brilliant.org/discussions/thread/brilliant-geometry-contest-season-1/#comment-a3f5119307e92 GO HERE – Prince Loomba · 8 months ago

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i want to take part in this where is the present question – Abhishek Alva · 8 months ago

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– Sharky Kesa · 8 months ago

goto the comments sorting option and click New, instead of Top. Anyways, Problem 9 is most recent. No one has solved properly.Log in to reply

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– Sharky Kesa · 8 months ago

I've already posted Problem 9, since I was told to.Log in to reply

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:

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

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

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– Harsh Shrivastava · 8 months ago

Repost the problem 8.Log in to reply

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– Prince Loomba · 8 months ago

Obviously yesLog in to reply

– Sharky Kesa · 8 months ago

Problem looked familiar. https://brilliant.org/problems/locus-of-circles-centre/Log in to reply

– Prince Loomba · 8 months ago

You did a wrong thing by posting the link. You should have written your solution.Log in to reply

– Sharky Kesa · 8 months ago

Would've taken too long. Sorry. Also, I recognised the question (Eidetic/Photographic memory).Log in to reply

– Prince Loomba · 8 months ago

Try the proof without seeing. Its great reallyLog in to reply

– Prince Loomba · 8 months ago

But the proof is great.Log in to reply

– Sharky Kesa · 8 months ago

True, but I found this problem not too hard to show via co-ord geom. Still, my problem to post?Log in to reply

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– Prince Loomba · 8 months ago

Its mine solution. So thats cheating.Log in to reply

– Ayush Rai · 8 months ago

well,didn't you do!Log in to reply

– Harsh Shrivastava · 8 months ago

Post the next.Log in to reply

– Prince Loomba · 8 months ago

See the bedsheet behind the register, its also same! How can this be possible?Log in to reply

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– Armain Labeeb · 8 months ago

You just copied his solution :PLog in to reply

– Ayush Rai · 8 months ago

i wrote it.How can you say that?our handwriting is sameLog in to reply

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– Prince Loomba · 8 months ago

One more thing see the bedsheet behind the register, Its also sameLog in to reply

– Prince Loomba · 8 months ago

Haha. Its mine. I can identify my writingLog in to reply

– Sharky Kesa · 8 months ago

Maybe you two are the same people from alternate universes. :PLog in to reply

– Ayush Rai · 8 months ago

maybeLog in to reply

– Sharky Kesa · 8 months ago

So, do I get the points for it?Log in to reply

– Ayush Rai · 8 months ago

noLog in to reply

– Sharky Kesa · 8 months ago

K, that's fine as long as problem remains.Log in to reply

– Prince Loomba · 8 months ago

He is a bad person. See he awarded marks to himselfLog in to reply

Problem 7 [3 points]Suppose the medians of a triangle are 5,12,13 units.Find the sides of the triangle. – Ayush Rai · 8 months agoLog in to reply

– Prince Loomba · 8 months ago

26/3,2(sqrt244)/3,2(sqrt601)/3Log in to reply

@Prince Loomba post your solution. – Ayush Rai · 8 months ago

ya! sharky is right.Log in to reply

– Prince Loomba · 8 months ago

Its a direct formula. Side=\(\frac {2}{3} \times \sqrt {2 ((m_{a})^2+(m_b)^{2})-{m_{c}}^2}\)Log in to reply

– Ayush Rai · 8 months ago

oh..thats niceLog in to reply

– Harsh Shrivastava · 8 months ago

Why am I awarded points even though I didn't posted the solution to previous problem?Log in to reply

– Ayush Rai · 8 months ago

you didn't.really?Log in to reply

– Sharky Kesa · 8 months ago

BTW, my method is using formula \(AB^2 + AC^2 = 2(AD^2 + BD^2)\), then using simultaneous linear equations.Log in to reply

– Ayush Rai · 8 months ago

Even i used the same thing.Log in to reply

– Sharky Kesa · 8 months ago

Beat me to it!!! Can we please post proofs of why it is true?Log in to reply

– Harsh Shrivastava · 8 months ago

Post next bro.Log in to reply

– Ayush Rai · 8 months ago

correct!Log in to reply

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 · 8 months ago

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– Sharky Kesa · 8 months ago

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 \(2\pi\).Log in to reply

– Prince Loomba · 8 months ago

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 2piLog in to reply

– Harsh Shrivastava · 8 months ago

Is the ander 2pi?Log in to reply

– Prince Loomba · 8 months ago

Post next broLog in to reply

– Harsh Shrivastava · 8 months ago

Sorry my net was down a bit.Log in to reply

– Julian Poon · 8 months ago

Correct!Log in to reply

– Prince Loomba · 8 months ago

Same here, 2piLog in to reply