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 $y=\frac{a-b}{2}$ .

## Search found 64 matches

- Sun Apr 23, 2017 8:44 am
- Forum: Algebra
- Topic: USA(J)MO 2017 #2
- Replies:
**1** - Views:
**3807**

- Sat Apr 22, 2017 2:00 pm
- Forum: Algebra
- Topic: At last an ineq USAMO '17 #6
- Replies:
**3** - Views:
**5148**

### 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...

- Sat Apr 22, 2017 1:33 pm
- Forum: Number Theory
- Topic: USAJMO/USAMO 2017 P1
- Replies:
**3** - Views:
**2228**

### Re: USAJMO/USAMO 2017 P1

Quite easy as USAMO #1

- Sat Apr 22, 2017 9:32 am
- Forum: Algebra
- Topic: At last an ineq USAMO '17 #6
- Replies:
**3** - Views:
**5148**

### 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 $a+b+c+d=4$.

- 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:
**1573**

### 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 ...

- Thu Apr 20, 2017 10:34 pm
- Forum: Geometry
- Topic: USA TST 2011/1
- Replies:
**3** - Views:
**2399**

### 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 ...

- Thu Apr 20, 2017 9:55 pm
- Forum: Junior Level
- Topic: Beginner's Marathon
- Replies:
**68** - Views:
**22694**

### 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 $...

- Fri Apr 14, 2017 3:05 pm
- Forum: Geometry
- Topic: Geometric Ineq
- Replies:
**1** - Views:
**1436**

### Re: Geometric Ineq

No one even tried? Should I say now "AJ GORIB BOLE" ?

- Thu Apr 13, 2017 8:43 am
- Forum: Geometry
- Topic: Comparing the inradii
- Replies:
**1** - Views:
**1337**

### Re: Comparing the inradii

My solution Trivial angle chasing yields that the angles of $\triangle A_1B_1C_1$ are $\frac{\angle A + \angle B}{2}$ , $\frac{\angle B + \angle C}{2}$ and $\frac {\angle C + \angle A}{2}$ . We know that $r = 4R \sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2}) $. So, it remains to show that $\sin...

- Thu Apr 13, 2017 8:24 am
- Forum: Geometry
- Topic: Comparing the inradii
- Replies:
**1** - Views:
**1337**

### Comparing the inradii

Let $ABC$ be an acute triangle with circumcircle $\Gamma$. Let $A_1,B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC,CBA$ and $ACB$ of $\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$.