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Re: 14x14 grid

Put $1$ and $0$ alternately in that grid. Then in any $T$ shape, $2$ configuration is possible: #1 three $1$'s and one $0$'s #2 three $0$'s and one $1$'s In both configuration, the sum of the numbers in that $T$ is odd. The grid is $14 x 14$. So, we need an odd number of $T$ - tiles...
by Atonu Roy Chowdhury
Thu Apr 27, 2017 12:12 am
 
Forum: Junior Level
Topic: 14x14 grid
Replies: 1
Views: 57

Re: JBMO:NT

It is obvious that $x \ge 1$ . Case 1: $y=0$ Then, $2^x - 1 = 5^z . 7^w$ . Subcase 1.1: $z > 0$ So, $2^x \equiv 1 (mod 5) $ . So, $ 4|x$. $2^{4k} - 1 = 16^k - 1$ which is divisible by $3$ . But $5^z . 7^w$ is not divisible by $3$. . Subcase 1.2: $z=0$ So, $2^x - 7^w = 1$ $x=1 \rightarrow w = 0$ $x=2...
by Atonu Roy Chowdhury
Wed Apr 26, 2017 12:13 am
 
Forum: Number Theory
Topic: JBMO:NT
Replies: 1
Views: 84

Re: USA(J)MO 2017 #3

WLOG $P$ lies on the shorter arc $BC$ . So, $[DEF]=[AEF]-[ABC]-[BDF]-[CDE]$ $\angle BAD = \alpha $ Use Sine Law to find $BD$,$DC$,$BF$,$CE$ in terms of $a$ and sine of $\alpha$ and $60-\alpha$, where $a$ is the length of the sides of $\triangle ABC$ . Then we'll use these lengths to find $[AEF]$,$[B...
by Atonu Roy Chowdhury
Mon Apr 24, 2017 11:26 am
 
Forum: Geometry
Topic: USA(J)MO 2017 #3
Replies: 6
Views: 162

Re: USA(J)MO 2017 #2

Substitute $a=x+y$ and $b=x-y$ and after some simplification, we get
$a^6 = b^6(4b+1)$
So, $4b+1=(2n+1)^6$
Here we'll find a value of $b$ in terms of $n$. Then $a=(2n+1)b$, here we'll input the value of $b$ and get a value of $a$ in terms of $n$. $x=\frac{a+b}{2}$ and $y=\frac{a-b}{2}$ .
by Atonu Roy Chowdhury
Sun Apr 23, 2017 8:44 am
 
Forum: Algebra
Topic: USA(J)MO 2017 #2
Replies: 1
Views: 74

Re: At last an ineq USAMO '17 #6

We'll show that $\sum_{cyc} \frac {a}{b^3 + 4} \ge \frac{2}{3} $ Subtracting $\frac{a}{4} + \frac{b}{4} + \frac{c}{4} + \frac{d}{4} = 1$ from both sides, we get $\sum_{cyc} (\frac{a}{b^3 + 4} - \frac{a}{4}) \ge \frac{-1}{3}$ After some simplification, we get $\sum_{cyc}\frac {3ab^3}{b^3 +4} \le 4$ B...
by Atonu Roy Chowdhury
Sat Apr 22, 2017 2:00 pm
 
Forum: Algebra
Topic: At last an ineq USAMO '17 #6
Replies: 3
Views: 97

Re: USAJMO/USAMO 2017 P1

Quite easy as USAMO #1 We'll show $(a,b)=(2n+1,2n-1)$ works. It is easy to see that $gcd(2n+1, 2n-1) =1 $. $(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv (2n+1) + (2n-1) \equiv 0 (mod 4n) $ because $(2n \pm 1)^2 \equiv 4n^2 \pm 4n +1...
by Atonu Roy Chowdhury
Sat Apr 22, 2017 1:33 pm
 
Forum: Number Theory
Topic: USAJMO/USAMO 2017 P1
Replies: 3
Views: 141

At last an ineq USAMO '17 #6

Find the minimum possible value of $\frac{a}{b^3+4}+\frac{b}{c^3+4}+\frac{c}{d^3+4}+\frac{d}{a^3+4}$ given that $a$, $b$, $c$, $d$ are nonnegative real numbers such that $a+b+c+d=4$.
by Atonu Roy Chowdhury
Sat Apr 22, 2017 9:32 am
 
Forum: Algebra
Topic: At last an ineq USAMO '17 #6
Replies: 3
Views: 97

Re: When everyone is busy solving USA(J)MO 2017,I am solving

Solution with angle chasing only. But seems ugly to me. Let $M$ be the midpoint of arc $BPC$. We will show that $M$ is our desired point. So, it suffices to show $\angle I_BMI_C = \angle I_BPI_C = \frac {180 - \angle A}{2} = \angle B $ or $ \angle I_BMI = \angle I_CMC $ where $I$ is the incenter of ...
by Atonu Roy Chowdhury
Fri Apr 21, 2017 12:15 pm
 
Forum: Geometry
Topic: When everyone is busy solving USA(J)MO 2017,I am solving2016
Replies: 1
Views: 81

Re: USA TST 2011/1

Similarity on $\triangle HQA$ and $\triangle HEA$ implies $\frac {HQ}{HF} = \frac {HE}{HA}$ Similarity on $\triangle HDR$ and $\triangle ADP$, $\triangle ABD$ and $\triangle AEP$, $\triangle BEA$ and $\triangle HEC$ implies $\frac {HD}{HR} = \frac {AD}{AP} = \frac {AB}{AE} = \frac {HC}{HE}$ . These ...
by Atonu Roy Chowdhury
Thu Apr 20, 2017 10:34 pm
 
Forum: Geometry
Topic: USA TST 2011/1
Replies: 3
Views: 112

Re: Beginner's Marathon

Problem $17$ : Let $\bigtriangleup$ $ABC$ be an acute triangle.$D$ is the foot of perpendicular drawn from $C$ on $AB$.Let the bisector of $\angle$ $ABC$ intersect $CD$ at $E$ and $\bigcirc$ $ADE$ at $F$.If $\angle$ $ADF$ =45˚,prove that $CF$ is tangent to $\bigcirc$ $ADE$. My solution: Just show $...
by Atonu Roy Chowdhury
Thu Apr 20, 2017 9:55 pm
 
Forum: Junior Level
Topic: Beginner's Marathon
Replies: 48
Views: 1449
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