I am in.

Great, how to start this?

Yeah, I am there.

To start this thread. Someone post some problems...

Me too . Got 32 this year

Let's Begin with a question The Maths Boy has an equilateral triangle ABC of area 1. Put D on BC, E on CA, and F on AB, with BD = DC, CE = 2EA, and 2AF = FB. Note that AD, BE, and CF pass through a single point M. What is the area of triangle EMC?

Problem:You are given a n-sided convex polygon. Prove or disprove that you can construct a square with area same as that of the polygon.

TheMathsBoy wrote: Let's Begin with a question The Maths Boy has an equilateral triangle ABC of area 1. Put D on BC, E on CA, and F on AB, with BD = DC, CE = 2EA, and 2AF = FB. Note that AD, BE, and CF pass through a single point M. What is the area of triangle EMC? Is the answer $\boxed{\frac{1}{18}}$? Just use similarity of triangle, Menelaus, simple area manipulations. $\blacksquare$

Is it 1/6? I'm not sure.

@TheDarkPrince, I think the answer is yes, but I don't know how to solve for any n-sided non-regular polygon, please give some tips on how to approach such problems with non specified polygon?

Hello guys! I'm very fond of math. But unfortunately I'm weak at geometry which involves constructions. I need to upgrade my skills.Please I request someone to send some easy triangles and circles problems(those problems which have 'prove that').Please help me out. Thank you!

TheDarkPrince wrote: Problem:You are given a n-sided convex polygon. Prove or disprove that you can construct a square with area same as that of the polygon. Drunken_Master wrote: @TheDarkPrince, I think the answer is yes, but I don't know how to solve for any n-sided non-regular polygon, please give some tips on how to approach such problems with non specified polygon? Claim 1: We can construct a square with area same as that of a given right angled triangle. Proof: Let the legs of the right triangle be $a,b$. So, the area is $\frac{1}{2}ab$. Suppose $s$ is the side length of the square with area equal to the right triangle. Then, $s=\sqrt{\frac{1}{2}ab}$. Let $\frac{a}{2}=c$. \[s=\sqrt{bc}.\]Construction: Construct a semicircle $(AB)$ with diameter $b+c$. Let $C$ be on $AB$ such that $AC=b, BC=c$. Construct $CD\perp AB$, $D\in (AB)$. $CD=\sqrt{bc}$. It is not hard to show this by PoP. Claim 2: We can construct a square with area same as that of a given triangle. Proof: Let $ABC$ be any triangle. Let $l$ be a line parallel to $BC$ through $A$. Let $D$ be a point on $l$ such that $BD\perp BC$. Clearly $[ABC]=[DBC]$ and $\triangle DBC$ is a right triangle. So, by claim 1 we can construct a square with area equal to $[ABC]$. Claim 3: We are given two squares with side lengths $s_1,s_2$. Show that we can construct a square with area equal to $s_1^2+s_2^2$. Proof: Construction: Construct a right triangle with legs of side lengths $s_1,s_2$. Construct a square with one side as the hypotenuse. Clearly, the area of the square is $(\sqrt{s_1^2+s_2^2})^2=s_1^2+s_2^2$ as required. Main problem: Let $A_1A_2\dots A_n$ be the given convex $n$-gon. Join the line segments $A_1A_3,\dots A_1A_{n-1}$. We have divided the figure into $n-2$ triangles. From claim 2, we can construct $n-2$ squares with their sum of the areas equal to the area of $[A_1A_2\dots A_n].$ Now, from claim 3, we are done.

An easy one: Problem: In a $\Delta ABC$, $H$ be the orthocenter, denote $\odot(UVW)$ to be the circumcircle of $\Delta UVW$, $\odot(UV)$ to be circle with diameter $UV$. Let $AH\cap \odot(ABC)=X,BH\cap \odot(ABC)=Y,CH \cap \odot(ABC)=Z$. Suppose $YZ\cap \odot(AH)=A_1,A_2$. Define $B_1,B_2,C_1,C_2$ similarly. Prove that the points $A_1,A_2,B_1,B_2,C_1,C_2$ are concyclic.

Does the 'dot' Character represent a circle?

If you mean the geometric one ($\cong$) use \cong If you mean the modular one ($\equiv$) use \equiv Superguy wrote: I tried $mod23$ and $mod29$ and got that p=$10,7(mod23)$ and $15,2(mod29)$ But I couldn't solve further While I'm on the topic, using $\mod $ or $\pmod{n} $ instead of $mod$ looks better.

A and B are points on the same side of line l.Using reflection, construct the shortest route from A to B via a point P on the line l. Justify your answer.

Reflection of light bashes the problem.

Could you please send a full solution..

Reflect B in the line $l$ then joing $A$ and $B$'s reflection then use triangle inequality....

No.. The question says you have find the shortest distance from A to B VIA A POINT P IN LINE L....

Superguy wrote: Find the smallest prime $p$ such that there exist integer $a$ that satisfy $$ap \equiv 1 \pmod{667}$$and $$ a+p \equiv 17 \pmod{667}$$. Try this one and pls tell if it has solutions as I am not sure Could anyone solve it?

Sherlock_Holmes wrote: No.. The question says you have find the shortest distance from A to B VIA A POINT P IN LINE L.... Do you mean to say that P is fixed on line $L$? If that's the case then there is nothing else to find because the whole picture is determined now.

Superguy wrote: Superguy wrote: Find the smallest prime $p$ such that there exist integer $a$ that satisfy $$ap \equiv 1 \pmod{667}$$and $$ a+p \equiv 17 \pmod{667}$$. Try this one and pls tell if it has solutions as I am not sure Could anyone solve it? Hmm... In need of more time!!

Completely new to homothety and inversion so someone please show me an easy beginners solution using homothety and inversion.

For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$. Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and $N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number, prove that the value of the sum $ \frac{1}{P(2)N(2)} + \frac{1}{P(3)N(3)} + \cdot\cdot\cdot + \frac{1}{P(n)N(n)} = \frac{n-1}{2n+2} $

xXLordTenebrisXx wrote: For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$. Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and $N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number, prove that the value of the sum $ \frac{1}{P(2)N(2)} + \frac{1}{P(3)N(3)} + \cdot\cdot\cdot + \frac{1}{P(n)N(n)} = \frac{n-1}{2n+2} $ Can I prove this using induction, for two consecutive primes $k+1$ and $k+q+1$?

xXLordTenebrisXx wrote: xXLordTenebrisXx wrote: For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$. Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and $N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number, prove that the value of the sum $ \frac{1}{P(2)N(2)} + \frac{1}{P(3)N(3)} + \cdot\cdot\cdot + \frac{1}{P(n)N(n)} = \frac{n-1}{2n+2} $ Can I prove this using induction, for two consecutive primes $k+1$ and $k+q+1$? If so, then here is my proof: Since the result holds true for $n=2$ and $n=4$, suppose that it holds true for $n=k$ with $k+1$ being prime. Then, $$\sum_{2}^{k}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)}$$ Then suppose that $k+q+1$ is the next prime greater than $k+1$, then $\forall m$ such that $ k+1 \leq m \leq k+q+1$, $P(m) = k+1$ and $N(m) = k+q+1$. Therefore, $$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}} = \sum_{2}^{k}{\frac{1}{P(n)N(n)}} + \sum_{k+1}^{k+q}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)} +\frac{q}{(k+1)(k+q+1)}$$ Which, on manupulation yields $$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}}=\frac{k+q-1}{2(k+q+1)}$$Hence completes the proof. Is this correct or am I missing something?

xXLordTenebrisXx wrote: xXLordTenebrisXx wrote: xXLordTenebrisXx wrote: For any positive integer $ n > 1$, let $P(n)$ denote the largest prime not exceeding $n$. Let $N(n)$ denote the next prime larger than $P(n)$. (For example $P(10) = 7$ and $N(10) = 11$, while $P(11) = 11$ and $N(11) = 13$.) If $n + 1$ is a prime number, prove that the value of the sum $ \frac{1}{P(2)N(2)} + \frac{1}{P(3)N(3)} + \cdot\cdot\cdot + \frac{1}{P(n)N(n)} = \frac{n-1}{2n+2} $ Can I prove this using induction, for two consecutive primes $k+1$ and $k+q+1$? If so, then here is my proof: Since the result holds true for $n=2$ and $n=4$, suppose that it holds true for $n=k$ with $k+1$ being prime. Then, $$\sum_{2}^{k}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)}$$ Suppose that $k+q+1$ is the next prime greater than $k+1$, then $\forall m$ such that $ k+1 \leq m < k+q+1$ then, $P(m) = k+1$ and $N(m) = k+q+1$. Therefore, $$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}} = \sum_{2}^{k}{\frac{1}{P(n)N(n)}} + \sum_{k+1}^{k+q}{\frac{1}{P(n)N(n)}} = \frac{k-1}{2(k+1)} +\frac{q}{(k+1)(k+q+1)}$$ Which, on manupulation yields $$\sum_{2}^{k+q}{\frac{1}{P(n)N(n)}}=\frac{k+q-1}{2(k+q+1)}$$Hence completes the proof. Is this correct or am I missing something?

@above. Yup, it's very correct and elegant.

hansu wrote: @above. Yup, it's very correct and elegant. Thanks, I am somewhat new to problem solving.

Problem(an easy and well known f.e) find all functions $f$: $N$ to $N$ such that $f(f(m)+f(n))=m+n$.

One will surely be f(x)= x

It's a PEN Problem! Claim $f$ is injective Proof Suppose $f(m) = f(n)$. \begin{align*} m + n = f(f(m)+f(n)) &= f(f(m)+f(m)) = 2m \\ \implies m &= n \quad \square \end{align*} $f(f(n+1)+f(1))=n+2=f(f(n)+f(2))$, thus by injectivity: \begin{align*} \implies f(n+1) &= f(n) + f(2) - f(1) \\ \implies f(n) &= f(n-1)+f(2) - f(1) \\ &= f(n-2)+2(f(2)-f(1))\\ &= f(n-3) + 3(f(2)-f(1)) \\ &\ \vdots \\ &= f(1) + (f(2)-f(1))(n-1) \end{align*}Putting this in the original equation we get that $\boxed{f(n) = n}$

Anybody plz check my (partial)solution I have a doubt in this.. by injectivity, $f(n-k)+f(n+k)=f(n)+f(n)$ By transposition and putting $k=1$ , we see that either $f(1)>f(2)>$..... Or $f(1)<f(2)$.... In first case, put $m=n=1$ in problem statement to get $f(2f(1))=2$......(I) And because $f$ is strictly decreasing, therefore there is no candidate for $f(2f(1)+2)$. Contradiction. Hence the function is strictly increasing. And now using (I) we can easily arrive at the conclusion.

ubermensch wrote: Also p2 just required that isosceles trapeziums are cyclic right? I did so, took the cases of 5 red 0 blue, 4 red 1 blue, etc.

Math-wiz wrote: ubermensch wrote: Also p2 just required that isosceles trapeziums are cyclic right? I did so, took the cases of 5 red 0 blue, 4 red 1 blue, etc. Hey, it was just PHP, still its correct.....

Thank God, I got 1 correct

I almost solved the first question, but the time was up and I could not write it in the answer booklet

ubermensch wrote: And p3 was m=n+1,n+2 na When |m-n| = 2, n must be even.

@AhirGauss How many could you solve?

Math-wiz wrote: ubermensch wrote: Also p2 just required that isosceles trapeziums are cyclic right? I did so, took the cases of 5 red 0 blue, 4 red 1 blue, etc. Cases of 5 red 0 blue and 4 red 1 blue are not really cases- all regular pentagons are cyclic-that's trivial. The whole point of the problem is where 3 are red and 2 blue(or vice versa)

ubermensch wrote: Math-wiz wrote: ubermensch wrote: Also p2 just required that isosceles trapeziums are cyclic right? I did so, took the cases of 5 red 0 blue, 4 red 1 blue, etc. Cases of 5 red 0 blue and 4 red 1 blue are not really cases- all regular pentagons are cyclic-that's trivial. The whole point of the problem is where 3 are red and 2 blue(or vice versa) I think he is talking about the colour of A vertex of each $n$

The discussion is now on at the post INMO 2019 forum https://artofproblemsolving.com/community/q2h1770676p11618442

^^ I think ^^^ wants to say that the problem is not just to show that there would exist atleast 1 adjacent pair of same coloured vertices. Greetings!

When will the rmo 2019 and inmo 2020 group be created? I'm waiting for it...

seashvar wrote: When will the rmo 2019 and inmo 2020 group be created? I'm waiting for it... You can create it yourself, as INMO 2019 is over and the chances of re-exam are almost nil.

Wizard_32 wrote: Electron_Madnesss wrote: Problem. Let $\theta_1,\theta_2,\dots,\theta_n \in \mathbb{R}$. Prove that $(\sum_{i=1}^n \cos \theta_i)^2+(\sum_{j=1}^n \sin \theta_j)^2 \le n^2.$ P.S.I hope Wizard_32 does not post a solution to this problem. Seriously?

for others for a different solution. got it ! solved it!

\requestlock

No one was there to chat further here. It's just that you bumped it. Don't unnecessarily request lock for these threads.