## Search found 18 matches

- Sat Oct 07, 2017 4:32 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2012/04
- Replies:
**3** - Views:
**1281**

### Re: APMO 2012/04

We'll have to show, $BF/CF=AB/AC$ which means $BF/CF=sin\angle{BCF}/sin\angle{CBF}=(BK*AC)/(AB*CK)$ We have to show now, $AB^2/AC^2=BK/CK$ or $AF$ is a symmedian. We get $A,O,P$ collinear Then $AEDM$ is cyclic Again, $\angle{AMC}=180-\angle{AMB}=\angle{AEF}=\angle{ABF}$ Also $\angle{AFB}=\angle{ACM}...

- Wed Oct 04, 2017 10:49 pm
- Forum: Geometry
- Topic: CGMO 2002/4
- Replies:
**2** - Views:
**246**

### Re: CGMO 2002/4

I found this in EGMO(Euclidean geometry in mathematical Olympiad). Then it's their mistake. Not mine.

- Sat Sep 30, 2017 12:02 pm
- Forum: Geometry
- Topic: CGMO 2002/4
- Replies:
**2** - Views:
**246**

### CGMO 2002/4

Circles $T_1$ and $T_2$ intersect at two points $B $ and $C$, and $BC$ is the diameter of $T_1$. Construct a tangent line to circle $T_1$ at $C$ intersecting $T_2$ at another point $A$. Line $AB$ meets $T_1$ again at $E $and line $CE $ meets $T_2$ again at $F $. Let $H $ be an arbitrary point on seg...

- Sat Jul 29, 2017 12:38 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2017 P4
- Replies:
**1** - Views:
**703**

### Re: IMO 2017 P4

We get $TA$ parallel to $KR$ because $\angle{ATS}=\angle{SJK}=\angle{SRK}$ We extend $KS$ to $P$ where $KS$ intersects $TA$ at $P$ Now, It's easy to prove that $TPRK$ is a rombus So, $\angle{TPK}=\angle{PKR}$ Again, $\angle{ARS}=\angle{SKR}$ So, $\angle{TPK}=\angle{ARS}$ So, $APSR$ is cyclic. $\angl...

- Sat Jul 29, 2017 12:29 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2017 P4
- Replies:
**1** - Views:
**703**

### IMO 2017 P4

Let R and S be dierent points on a circle Ω such that RS is not a diameter. Let be the tangent line to Ω at R. Point T is such that S is the midpoint of the line segment RT. Point J is chosen on the shorter arc RS of Ω so that the circumcircle Γ of triangle JST intersects at two distinct points. Let...

- Sat Jun 03, 2017 3:21 pm
- Forum: Geometry
- Topic: CGMO 2007/5
- Replies:
**1** - Views:
**233**

### CGMO 2007/5

Point $D$ lies inside triangle $ABC$ such that $$\angle{DAC}=\angle{DCA}=30^o$$ and $$\angle{DBA}=60^o$$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE$ I $EF$.

- Fri Jun 02, 2017 3:04 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2001 Problem1
- Replies:
**2** - Views:
**237**

### Re: IMO 2001 Problem1

$$\angle{A}+\angle{COP} <90° \Rightarrow \angle{COP} <90°-\angle{COD}$$ $$\Rightarrow \angle{COP} <\angle{OCP} \Rightarrow CP<PO$$ is enough to prove. $\angle C -\angle B \geq 30°$ as both $(\angle{C}-\angle{B})$ and 30° are less than $90°$, $$\begin{align*}&\sin(\angle{C}-\angle{B})\geq \sin30°\\ \...

- Fri Jun 02, 2017 3:01 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2001 Problem1
- Replies:
**2** - Views:
**237**

### IMO 2001 Problem1

Consider an acute angled triangle $ABC$. Let $P$ be the foot of the altitude of triangle $ABC$ issuing from the vertex $A$, and let $O$ be the circumvented of triangle $ABC$. Assume that $$\angle{C}\geq\angle{B}+30°$$. Prove that $$\angle{A}+\angle{COP}<90°$$

- Fri Jun 02, 2017 1:59 pm
- Forum: Geometry
- Topic: IMO 2007 Problem 4
- Replies:
**4** - Views:
**880**

### Re: IMO 2007 Problem 4

Let's draw $RX$ perpendicular on $PK$ which intersects thevcircumcirle of triangle $ABC$ at $M$ $RX$ is parallel to $BC$ So, $BRMC$ is a trapizoid. So, $BR$=$MC$ Again, $BR$=$AR$ is known as $CR$ in the angle bisector of angle $ACB$ So, $AR$=$MC$ and it makes $ARCM$ cyclic trapizoid Then $MR$=$b$ $X...

- Thu Apr 27, 2017 4:31 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BDMO 2017 National round Secondary 2
- Replies:
**1** - Views:
**586**

### Re: BDMO 2017 National round Secondary 2

Here, $AD$ is perpendicular on $BC$ Hence, $BE=CE$ We can show that $$\triangle$$$BDE$ and $$\triangle$$$CFE$ are congruent. So, $FE=ED=OF=\frac{1}{3}OD=\frac{1}{3}OB$ In the right angle triangle $OBE$, $OE^2+BE^2=OB^2$ or,$4OF^2+5=9OF^2$ or,$OF=1=DE$ in rigth angle triangle $CED$, $CE^2+DE^2=CD^2$ ...