problem 16 :

In triangle $ABC$, the centroid is $G$ and $N$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $X$. Prove that $\angle AXC = \angle NGC$ if and only if angle $ACB$ is a right angle

## Search found 244 matches

- Mon Dec 17, 2012 5:51 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57337**

- Fri Dec 14, 2012 3:04 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62462**

### Re: IMO Marathon

মাসুম ভাই দোন্ত উই ডিসারভ আ লিটল হিন্ট ?

You've almost killed the marathon

You've almost killed the marathon

- Tue Dec 11, 2012 5:41 pm
- Forum: Secondary Level
- Topic: Edited version of a known problem
- Replies:
**4** - Views:
**1973**

### Re: Edited version of a known problem

Note that if we move the point $(0,0)$ to $(m,n)$ then it will take $m$ right operation and $n$ up operation . So the number of required ways is number of ways to choose $m$ numbers from $m+n$ distinct numbers . I mean $\displaystyle \binom {m+n}{m}$. Now we will count in how many ways one can move ...

- Mon Dec 03, 2012 5:35 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57337**

### Re: Secondary and Higher Secondary Marathon

P $13$ A number ($ \geq2$), is called product-perfect if it is equal to the product of all of its proper divisors. For example, $6=1×2×3$, hence $6$ is product-perfect. How many product-perfect numbers are there below $50$? Note: A proper divisor of a number $N$ is a positive integer less than $N$ t...

- Mon Dec 03, 2012 5:28 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57337**

### Re: Secondary and Higher Secondary Marathon

Just find the possible patterns for this six patterns of central $2 \times 2$ array .

এডিট : আমার ২০০ তম পোস্ট ।

Then consider their possible rotations and Count .এডিট : আমার ২০০ তম পোস্ট ।

- Sun Dec 02, 2012 11:04 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57337**

### Re: Secondary and Higher Secondary Marathon

If the Ans is $90$ . Then I'm ready to post my solution ..

joty ভাই , confirmation দেন ।

joty ভাই , confirmation দেন ।

- Sun Dec 02, 2012 10:16 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62462**

### Re: IMO Marathon

(Just trying .. It may have bug) Let $P(x,y): f(x^2-y^2)=f(x)f(y)$ Now $P(x,-x): f(0)=f(x)f(-x)$ . For distinct $x,y$; $P(x,y),P(x,-y)$ imply $f(x)f(y)=f(x)f(-y)$ … $(St1)$ Consider some cases . Case 1 : $f(x)=0 \forall x$ Case 2 : There is only one integer $i$ so that $f(i) \neq 0$ If $i$ is of for...

- Sun Nov 18, 2012 5:27 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**127** - Views:
**57337**

### Re: Secondary and Higher Secondary Marathon

Prove that if $n$ is a natural number,

$\displaystyle \frac {n^5}{5}+\frac {n^4}{2}+\frac {n^3}{3}−\frac {n}{30}$

is always an integer.

Source : An NT textbook .

$\displaystyle \frac {n^5}{5}+\frac {n^4}{2}+\frac {n^3}{3}−\frac {n}{30}$

is always an integer.

Source : An NT textbook .

- Sun Nov 18, 2012 3:55 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62462**

### Re: IMO Marathon

Problem 11 . Let quadrilateral $ABCD$ be inscribed in a circle. Suppose lines $AB$ and $DC$ intersect at $ P$ and lines $AD$ and $BC$ intersect at $Q$. From $Q$, construct the two tangents $QE$ and $QF$ to the circle where $E$ and $F$ are the points of tangency. Prove that the three points $P, E, F$...

- Fri Nov 16, 2012 7:23 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**62462**

### Re: IMO Marathon

Solution 10 : Let $AD \cap BE=F$ Using sine rule on $\triangle KFE ,\triangle KEA ,\triangle KFA$ We can write this equation . $\displaystyle \frac {sin{(45+\frac{A}{4})}}{sin {\frac{3A}{4}}}=\frac {cos{\frac{A}{2}}}{sin {45}}$ . $\displaystyle \rightarrow \frac{sin {\frac{A}{4}}+cos {\frac{A}{4}}}{...