**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

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## 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

\[XY+XB+XC \geq Y'A\]

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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

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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.

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Repost the problem 8.

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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

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Your reasoning is very flawed so I'm afraid you are incorrect.

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

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Am I right till now?

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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

\[XY+XB+XC \geq Y'A\]

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Thhats what I was saying the worst case is Y=A and X=centroid

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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.

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https://i.imgur.com/xmrurud.png

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

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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.

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

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I have given you 10 cases

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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?

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\(2\sqrt{13}\)

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

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The answer is \(\sqrt{52}\) or \(2\sqrt{13}\)

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Oh boy, that means I have to post another one

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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

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The x- axis is the coastline and not y- axis.

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Yes I also think this

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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.Log in to reply

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It is the correct answer.

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Question 2 (2 marks)Find perimeter of the figure.

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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 \(5 \times 12 = \large \color{green}{\boxed{60}}\).

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Right Post next problem

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Nice solution. I was thinking of a solution when you posted this. You win haha :P

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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 \(ab\).

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Its 2D ?

And only of 3 marks?

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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

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I posted 2 marks and you 3 marks comparing with me

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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:

\[\frac{2-\sqrt{3}}{6}*\pi\]

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Winner!

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Post next

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

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Julian already got it.

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Correct, but Julian answered first \(:)\)

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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 \(x+y\).

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

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Indeed \(:)\)

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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 \(\frac{\sqrt{3}}{3}\)

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Correct! \(x-1\) marks for you good sir.

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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.

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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 \(y>\frac{1}{\left|x\right|}\)?

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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 \(2\pi\).

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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

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Is the ander 2pi?

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Same here, 2pi

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Correct!

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Post next bro

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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.Log in to reply

26/3,2(sqrt244)/3,2(sqrt601)/3

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correct!

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Post next bro.

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

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ya! sharky is right.@Prince Loomba post your solution.

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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:

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

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

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Problem looked familiar. https://brilliant.org/problems/locus-of-circles-centre/

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But the proof is great.

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Try the proof without seeing. Its great really

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You did a wrong thing by posting the link. You should have written your solution.

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Its mine solution. So thats cheating.

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You just copied his solution :P

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

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i want to take part in this where is the present question

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

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https://brilliant.org/discussions/thread/brilliant-geometry-contest-season-1/#comment-a3f5119307e92 GO HERE

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