Beginner's Marathon
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
They start, from Mirpur, one with the bike($ A$) and the other($ B$) on foot, and from Lalmatia($ C$) on foot of course. After one hour $ A$ will give the bike to $ C$, and $ B$ will stop, and wait for the bike. $ A$ will reach Lamatia after 2 hours. $ C$ rides the bike until he meets $ B$ , and gives it to him. After this they'll reach their targets in one hour, so $ 2$:$ 40$ in total.
Frankly, my dear, I don't give a damn.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
Problem $23$
Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.
Let $a,b,c$ be positive integers. Prove that it is impossible to have all of the three numbers $a^2+b+c,b^2+c+a,c^2+a+b$ to be perfect squares.
Frankly, my dear, I don't give a damn.
Re: Beginner's Marathon
$\text{Problem 23}$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
 Charles Caleb Colton
 Charles Caleb Colton
Re: Beginner's Marathon
$\text{Problem 24}$
Let $n$ be a positive integer and let $a_1, a_2,.....a_k$(here $k$ > 1) be distinct integers in the set {${1,2.....n}$} such that $n$ divides $a_i(a_{i+1}1)$ for $i = 1,2,.....k1$. Prove that $n$ does not divide $a_k(a_1  1)$
Let $n$ be a positive integer and let $a_1, a_2,.....a_k$(here $k$ > 1) be distinct integers in the set {${1,2.....n}$} such that $n$ divides $a_i(a_{i+1}1)$ for $i = 1,2,.....k1$. Prove that $n$ does not divide $a_k(a_1  1)$
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
 Charles Caleb Colton
 Charles Caleb Colton
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
Solution to problem $24$ (OVERKILL)
We prove by contradiction.
Assume that $n$ divides $a_k(a_11)$.
Assume that $(a_i, n)=1$. Then $n a_i(a_{i+1}1)$ implies $na_{i+1}1$ which implies $a_{i+1}=1$. Then $n1(a_{i+2}1)$ which is only possible if $a_{i+2}=1$ but this contradicts the distinction.
So there exist some prime $p$ such that $pn$ and $pa_i$.
Now we inductively prove that $p$ divides all $a_j$ such that $1\leq j \leq n$. We have already shown the base case. Now assume that $pa_m$. We have, $p n a_{m1}(a_m1)$. So, we must have $pa_{m1}$. Here, the indices are taken modulo $k$. So, $p$ divides all $a_j$ such that $1\leq j \leq n$. Note that the induction was done by starting from $a_i$. We get that if some prime divides at least one of the $a_j$'s, that prime divides all the $a_j$'s.
We will strengthen the statement more with dealing with prime powers.
Assume that $p^xn$. Then,as $p^xna_j(a_{j+1}1)$ for all $j$, $p^xa_j$ for all $j$.
So, we get that $(n,a_1)=(n,a_2)=.......=(n,a_k)= C$ where C is an integer greater than $1$.
Let, $b_j=\frac {a_j}{C}$ for all $j$. Also let, $n_1=\frac {n}{C}$. Then, we have $(n_1, C)=(n_1,b_j)=1$ for all $j$.
Now, as $n_1Cb_jC(b_{j+1}C1)$, we get $n_1b_{j+1}C1$ for all $j$. So, $n_1b_xC1b_yC+1=C(b_xb_y)$.
We get that $n_1b_xb_y$. Which is impossible since $b_x, b_y$ are different integers less than $n_1$. So we get a contradiction. So our first assumption is false. Now, this solution is really unclear. Someone tell me where does the logic breaks if our first assumption is false.
We prove by contradiction.
Assume that $n$ divides $a_k(a_11)$.
Assume that $(a_i, n)=1$. Then $n a_i(a_{i+1}1)$ implies $na_{i+1}1$ which implies $a_{i+1}=1$. Then $n1(a_{i+2}1)$ which is only possible if $a_{i+2}=1$ but this contradicts the distinction.
So there exist some prime $p$ such that $pn$ and $pa_i$.
Now we inductively prove that $p$ divides all $a_j$ such that $1\leq j \leq n$. We have already shown the base case. Now assume that $pa_m$. We have, $p n a_{m1}(a_m1)$. So, we must have $pa_{m1}$. Here, the indices are taken modulo $k$. So, $p$ divides all $a_j$ such that $1\leq j \leq n$. Note that the induction was done by starting from $a_i$. We get that if some prime divides at least one of the $a_j$'s, that prime divides all the $a_j$'s.
We will strengthen the statement more with dealing with prime powers.
Assume that $p^xn$. Then,as $p^xna_j(a_{j+1}1)$ for all $j$, $p^xa_j$ for all $j$.
So, we get that $(n,a_1)=(n,a_2)=.......=(n,a_k)= C$ where C is an integer greater than $1$.
Let, $b_j=\frac {a_j}{C}$ for all $j$. Also let, $n_1=\frac {n}{C}$. Then, we have $(n_1, C)=(n_1,b_j)=1$ for all $j$.
Now, as $n_1Cb_jC(b_{j+1}C1)$, we get $n_1b_{j+1}C1$ for all $j$. So, $n_1b_xC1b_yC+1=C(b_xb_y)$.
We get that $n_1b_xb_y$. Which is impossible since $b_x, b_y$ are different integers less than $n_1$. So we get a contradiction. So our first assumption is false. Now, this solution is really unclear. Someone tell me where does the logic breaks if our first assumption is false.
Last edited by ahmedittihad on Mon Jun 26, 2017 4:00 am, edited 1 time in total.
Frankly, my dear, I don't give a damn.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
Solution to Problem $24$ (NOT OVERKILL)
As the previous solution we solve by contradiction.
So, assume that $a_1a_k \equiv a_k (mod n)$.
$ a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k2}\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k$.
Therefore, $ 0 < a_1a_k < n$ is divisible by $ n$. Contradiction.
EID MUBARAK
As the previous solution we solve by contradiction.
So, assume that $a_1a_k \equiv a_k (mod n)$.
$ a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k2}\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k$.
Therefore, $ 0 < a_1a_k < n$ is divisible by $ n$. Contradiction.
EID MUBARAK
Frankly, my dear, I don't give a damn.
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: Beginner's Marathon
As this is the Beginner's Marathon, I request everyone to not give shortlist problems.
Problem $25$
Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
Problem $25$
Let $ABCD$ be a trapezoid with $AB\parallel CD$ and $ \Omega $ is a circle passing through $A,B,C,D$. Let $ \omega $ be the circle passing through $C,D$ and intersecting with $CA,CB$ at $A_1$, $B_1$ respectively. $A_2$ and $B_2$ are the points symmetric to $A_1$ and $B_1$ respectively, with respect to the midpoints of $CA$ and $CB$. Prove that the points $A,B,A_2,B_2$ are concyclic.
Frankly, my dear, I don't give a damn.
Re: Beginner's Marathon
Let $AD \cap \Omega = P$. Now, note that $AP = BB_1, AA_1 = CA_2$ and $BB_1 = CB_2$. Now, power of point implies that $AP.AD = AC.AA_1 \Rightarrow BB_1.BC = CA_2.AC \Rightarrow CA_2.AC=CB_2.BC$. Therefore, $A,B,A_2,B_2$ are cyclic.
Somebody post the next problem.
Somebody post the next problem.
The study of mathematics, like the Nile, begins in minuteness but ends in magnificence.
 Charles Caleb Colton
 Charles Caleb Colton

 Posts: 57
 Joined: Sun Dec 11, 2016 2:01 pm
Re: Beginner's Marathon
Problem $26$
Given an acute angled triangle $\bigtriangleup ABC$.Inscribe in it a triangle $UVW$ of least possible perimeter.
Given an acute angled triangle $\bigtriangleup ABC$.Inscribe in it a triangle $UVW$ of least possible perimeter.