Hope this will help.

http://bit.ly/XCQxfT

## Search found 188 matches

- Sat Oct 20, 2012 2:28 pm
- Forum: Chemistry
- Topic: More preferable Fuel
- Replies:
**7** - Views:
**5951**

- Thu Oct 18, 2012 2:36 pm
- Forum: Secondary Level
- Topic: a trigonometry problem
- Replies:
**2** - Views:
**1281**

### Re: a trigonometry problem

I am confused about the question .Why this $"\beta"$ angle is a variable when the equations tell,

$\displaystyle \sin \beta = \pm \sqrt {\frac{-1 \pm \sqrt {65}}{32}}$.But if the complex angles(!) are not allowed,then,

$\displaystyle \sin \beta = \pm \sqrt{\frac{-1+ \sqrt {65}}{32}}$.

$\displaystyle \sin \beta = \pm \sqrt {\frac{-1 \pm \sqrt {65}}{32}}$.But if the complex angles(!) are not allowed,then,

$\displaystyle \sin \beta = \pm \sqrt{\frac{-1+ \sqrt {65}}{32}}$.

- Wed Oct 17, 2012 11:04 pm
- Forum: Junior Level
- Topic: Find positive integers
- Replies:
**4** - Views:
**1996**

### Re: Find positive integers

Yes,the only solutions are $(a,b)=(1,998),(998,1)$. It is easy to notice one of the variables $(a,b)$ is odd and another is even.WLOG let us assume $b=2k$(even). Case1:When $a \leq 1$,the only solutions found by a little investigation is $(a,b)=(1,998),(998,1)$. Case2:When $a,b >1$,$a^{2k}+(2k)^{a}=...

### Re: Triangle

At first prove that,$PCQR$ is a rhombus. (Hint: Use congruence of triangles and the Angle Bisector Theorem) $\Delta PQR \cong \Delta PCQ$.[Because,$RQ=QC,PR=PC$ and $PQ$ is a common side.].Let,$O$ be the intersection point of $RC$ and $PQ$. $\therefore \angle RPQ=\angle CPQ,\angle RQP=\angle CQP$.In...

### Re: Triangle

I have found a generalized proof already.

### Re: Triangle

Is $\Delta ABC$ acute angled or it's not mentioned in the question? I asked this because the proof can be easily obtained then.

- Mon Oct 15, 2012 4:33 pm
- Forum: Combinatorics
- Topic: n x n board
- Replies:
**10** - Views:
**4861**

### Re: n x n board

Sorry,but how??? .Check if I am misunderstanding.

If you just paint the field of the left bottom corner, the others have no green field at all or having 0(even) greens.

If you just paint the field of the left bottom corner, the others have no green field at all or having 0(even) greens.

- Sun Oct 14, 2012 4:44 pm
- Forum: Combinatorics
- Topic: n x n board
- Replies:
**10** - Views:
**4861**

### Re: n x n board

Okay,Draw a $n \times n$ board where $n$ is even. Now choose all the fields of the leftmost column and the bottom-most row.So, $2n-1$ fields are chosen.Now however You paint them green, it satisfies neither $(a)$ nor $(b)$.If it really happens,then problem becomes a fallacy.

- Wed Oct 10, 2012 2:19 pm
- Forum: Junior Level
- Topic: Easy-quizy
- Replies:
**6** - Views:
**2892**

### Re: Easy-quizy

I also used Apollopnius for the reverse proof.But these proofs are smarter!sowmitra wrote:Well, there are many approaches. I tried Appolonius' Theorem. Similarity also gives a nice solution.

- Tue Oct 09, 2012 2:25 pm
- Forum: Junior Level
- Topic: Easy-quizy
- Replies:
**6** - Views:
**2892**

### Easy-quizy

That is easy but not a spoiler!,I think.

Problem: In $\Delta ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively.Prove that, $AB=AC$ if and only if $BE=CD$ .

Problem: In $\Delta ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively.Prove that, $AB=AC$ if and only if $BE=CD$ .