I think it will be $R \equiv Q$

Anyway, Nice solution...

## Search found 155 matches

- Wed Feb 12, 2014 2:42 pm
- Forum: Geometry
- Topic: Don't touch my circles [externally;)]
- Replies:
**2** - Views:
**857**

- Tue Feb 11, 2014 4:05 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**4488**

### Re: warm-up problems for national BdMO'14

I think it will be $B$$K \cap \odot ABC=N$.Tahmid Hasan wrote: Let $A$$K \cap \odot ABC=N$.

- Mon Feb 10, 2014 9:27 pm
- Forum: Physics
- Topic: Falling Body, Thought Experiment
- Replies:
**2** - Views:
**1197**

### Re: Falling Body, Thought Experiment

I think so.... For any body of mass $m$ gravitational force acting on the body, $F=G\frac{M \times m}{R^2}$ (usual notations). So, Acceleration of the body, $a=\frac{F}{m}=\frac{GM}{R^2}$, which is a constant. $\therefore$ the acceleration of any freely falling body is constant. If two bodies are at...

- Mon Feb 10, 2014 8:52 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**4488**

### Re: warm-up problems for national BdMO'14

Could you explain a bit more?? How did you get this equation??Fatin Farhan wrote: $$\binom{n}{n-1}n=3n$$

So, $$n=3$$

And instead of using $*$ you can us the command "

**", if you want to write "$\times$".**

*\times*- Mon Feb 10, 2014 8:25 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**4488**

### Re: warm-up problems for national BdMO'14

There's a bug in the solution. $y$ and $x$ need not be integers. Only $N$ needs to be a positive integer. $\sqrt{1+8N}$ and $\sqrt{9+8N}$ may be rational, or, irrational.Fatin Farhan wrote: $$y^2-x^2=8$$

So, $$(y+x)=4,8$$ and $$(y-x)=2,1$$

- Sun Feb 09, 2014 6:22 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**4488**

### Re: warm-up problems for national BdMO'14

I've got some more: $[4]$( JBMO'07 ) Let $ABCD$ be a convex quadrilateral with $\angle DAC = \angle BDC = 36^\circ , \angle CBD = 18^\circ$ and $\angle BAC = 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle APD$ in degrees. $[5]$( Self-Made ) Show that, there is...

- Sun Feb 09, 2014 2:27 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**4488**

### Re: warm-up problems for national BdMO'14

A much easier solution to No. (2): $7^{2}\equiv (-1)\pmod{25} \Rightarrow 7^4\equiv 1\pmod{25}$. Again, $7^2\equiv 1 \pmod4 \Rightarrow 7^4\equiv 1\pmod4$ $\therefore 7^4\equiv 1\pmod{100}$ Now, $1996 \equiv 0 \pmod4$ $\displaystyle \therefore 7^{1996}\equiv 7^0\equiv 1\pmod{100}$ And, the result f...

- Sun Feb 09, 2014 2:11 pm
- Forum: National Math Olympiad (BdMO)
- Topic: warm-up problems for national BdMO'14
- Replies:
**25** - Views:
**4488**

### Re: warm-up problems for national BdMO'14

*Solution of No.(2): We need to calculate $\pmod {100}$. First of all, $100=2^2\times 5^2$. $\therefore \varphi(100)=100\times(1-\frac{1}{2})\times(1-\frac{1}{5})=40$. $\because gcd(7,100)=1$, $\therefore 7^{40}\equiv 1\pmod{100}$. Now, $1996\equiv36\pmod{40}$. $\displaystyle \therefore 7^{1996}\eq...

- Wed Feb 05, 2014 12:05 am
- Forum: Algebra
- Topic: Trouble in Trigonometry
- Replies:
**3** - Views:
**921**

### Re: Trouble in Trigonometry

In this solution, you did some pretty ugly calculations, which are very tough to do without a calculator.

For example, how did you solve the equations for $\sin{6^\circ}$ and $\sin{12^\circ}$ ??

Would you please elaborate...?

For example, how did you solve the equations for $\sin{6^\circ}$ and $\sin{12^\circ}$ ??

Would you please elaborate...?

- Fri Jan 17, 2014 7:01 pm
- Forum: Algebra
- Topic: Trouble in Trigonometry
- Replies:
**3** - Views:
**921**

### Trouble in Trigonometry

$\displaystyle \tan \theta = \sqrt{3}-4\sin 24^\circ$

where, $0<\theta < 90^\circ$.

Find the measure of $\theta$ in degrees.

where, $0<\theta < 90^\circ$.

Find the measure of $\theta$ in degrees.