Note that ,

$A,Q,P,D$ are harmonic .

Now pencil $C(A,Q,P,D)$ imply , $F,Q,E,G(FQ \cap BC)$ are harmonic . $AD \perp BC$ lead to desired result .

## Search found 244 matches

- Sat May 11, 2013 3:20 pm
- Forum: Geometry
- Topic: Angle Equity
- Replies:
**1** - Views:
**762**

- Wed May 01, 2013 1:39 pm
- Forum: Geometry
- Topic: Prove P,Q,A,O concyclic
- Replies:
**2** - Views:
**937**

### Re: Prove P,Q,A,O concyclic

Let $E,F$ be the midpoints of $BD,DC$ respectively . $AD' \perp BC$ ($D' \in BC$). $E',F',P',Q'$ are midpoints of $BD',D'C,AB,AC$ respectively . Note that , $EE'=FF'$ Now $EE'=PP'\cos {B}$ and $FF'=QQ' \cos {C}$ so $\displaystyle \frac {PP'}{QQ'}=\frac{\cos {C}}{\cos {B}}=\frac {OP'}{OQ'}$ and $\ang...

- Tue Feb 26, 2013 8:24 pm
- Forum: Geometry
- Topic: Prove concyclic
- Replies:
**4** - Views:
**1085**

### Re: Prove concyclic

Let $AB \cap DC =H$ Note that $D,G,C,H$ build a harmonic range . So $\displaystyle \frac{DG}{CG}=\frac{DH}{CH}$ $\Rightarrow MC^2=(DG-MG)(CG+MG)=MG.MH$ $\Rightarrow DG.CG=MG.GH$ Now $HB.HA=HC.HD=HG.HM$ $\Leftrightarrow HC.(HC+CD)=(HC+CG).(HC+CM)$ $\Leftrightarrow HC.CD=HC.CM+HC.CG+CG.CD$ $\Leftright...

- Sun Feb 03, 2013 5:45 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**31608**

### Re: IMO Marathon

(unless I'm not being silly)

What is the difference between "circumcircle of triangle $ABC$" and $w$ ? :S

What is the difference between "circumcircle of triangle $ABC$" and $w$ ? :S

- Sun Feb 03, 2013 4:37 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**31608**

### Re: IMO Marathon

Problem $\boxed {28}$: Trapezoid $ABCD$, with $AB$ parallel to $CD$, is inscribed in circle $w$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $w$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $AB$ intersect $BD$ and $BC$ at points $R$ and $S$, r...

- Fri Feb 01, 2013 5:17 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Let us help one another preparing for BdMO national 2013
- Replies:
**12** - Views:
**3179**

### Re: Let us help one another preparing for BdMO national 2013

How come ?Fahim Shahriar wrote:

And it doesn't have factor 5. Hence it's not divisible by 5.

- Fri Feb 01, 2013 12:25 pm
- Forum: National Math Olympiad (BdMO)
- Topic: Let us help one another preparing for BdMO national 2013
- Replies:
**12** - Views:
**3179**

### Re: Let us help one another preparing for BdMO national 2013

($x_1=a=3+\sqrt{8},y_1=b=3-\sqrt{8}$) Now Using vieta we have $(a+b)=6,ab=1$ $a^n+b^n=(3+\sqrt{8})^n+(3-\sqrt{8})^n$ We now that $r+1$th term of the expantion of $(3+\sqrt{8})^n$ is $\displaystyle \binom {n}{r}3^{n-r}.(\sqrt{8})^r$ , and for $(3-\sqrt{8})^n$ it is $\displaystyle \binom {n}{r}3^{n-r}...

- Thu Jan 31, 2013 7:09 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**126** - Views:
**23629**

### Re: Secondary and Higher Secondary Marathon

WoW . Waht a Co-incidence . I did post quite same solution there .

- Thu Jan 31, 2013 3:05 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**126** - Views:
**23629**

### Re: Secondary and Higher Secondary Marathon

Problem $\boxed {36}$ (A beautiful problem)

Suppose $ABCD$ is a rectangle and $P,Q,R,S$ are points on the sides $AB, BC, CD. DA$ respectively. Show that

$PQ+QR+RS+SP \geq 2AC$

source : AoPs

Suppose $ABCD$ is a rectangle and $P,Q,R,S$ are points on the sides $AB, BC, CD. DA$ respectively. Show that

$PQ+QR+RS+SP \geq 2AC$

source : AoPs

- Thu Jan 31, 2013 1:58 pm
- Forum: Higher Secondary Level
- Topic: Secondary and Higher Secondary Marathon
- Replies:
**126** - Views:
**23629**

### Re: Secondary and Higher Secondary Marathon

Well now the full solution : 35 $n^2-n=n(n-1) \equiv 0 (mod10^5)$ Note that $(n,n-1)=1$ So we have two cases Case 1 : $n=5^5a, n-1=2^5b $ subtract . You will get a diophentine equation $5^5a-2^5b=1$ It has a solution like $(a,b)=(-3+32t,-293+3125t) \forall t \in \mathbb {Z}$ Its easy to verify that ...