When the altitudes intersect, 2 acute angles and 2 obtuse angles are formed. If my check is right, then $OH$ always divides the acute ones.

Do we need to prove this?

And also, there is an easy euclidean proof taking this fact into account.

## Search found 550 matches

- Sun Jan 26, 2014 6:12 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO National 2012: Higher Secondary, Secondary 07
- Replies:
**4** - Views:
**2467**

- Thu Jan 02, 2014 4:21 pm
- Forum: Number Theory
- Topic: function (USAMO 2012, Balkan 2012, IMO proposal 2010)
- Replies:
**3** - Views:
**2054**

### Re: function (USAMO 2012, Balkan 2012, IMO proposal 2010)

@asif e elahi

You have left out the solutions: $f(n)\equiv 1 \forall n\in \mathbb{N}, f(n)\equiv 2 \forall n\in \mathbb{N}$.

And I can't understand how you deduced $f(p-2)=p-2$. Could you please make it clear?

You have left out the solutions: $f(n)\equiv 1 \forall n\in \mathbb{N}, f(n)\equiv 2 \forall n\in \mathbb{N}$.

And I can't understand how you deduced $f(p-2)=p-2$. Could you please make it clear?

- Wed Aug 28, 2013 4:34 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 3
- Replies:
**6** - Views:
**4715**

### Re: [OGC1] Online Geometry Camp: Day 3

Let $C'A\cap QB'=X$. Now $180^{\circ}-\angle C'PQ=\angle C'PB=\angle C'AB=\angle C'XQ$ implies $C',X,Q,P$ are concyclic. Let $C'Q\cap PX=E, AC\cap PX=S$. Then $\angle PEQ=\angle C'ES=\angle CSE=\angle ASX$. Also, $\angle PQE=\angle PQC'=\angle PXC'=\angle SXA$. Thus $\triangle PEQ\sim \triangle ASX$...

- Wed Aug 28, 2013 3:08 pm
- Forum: Geometry
- Topic: two circles
- Replies:
**3** - Views:
**1247**

### Re: two circles

plz give me the solve of this problem..... $\omega 1$ and $ \omega 2$ are two concentric circles where $\omega 1> \omega 2$. A tangent of $\omega 2$ at point B touches $\omega 1$ at points A and C. another line through point A touches $\omega 2$ at points E and F such that the perpendicular bisecto...

- Tue Aug 27, 2013 5:31 pm
- Forum: National Math Camp
- Topic: [OGC1] Online Geometry Camp: Day 3
- Replies:
**6** - Views:
**4715**

### Re: [OGC1] Online Geometry Camp: Day 3

In $\triangle OMN$, $OM=ON$ and $\angle MOR=\angle NOR$, so $OR$ is the perpendicular bisector of $MN$. Now for $\triangle AMN, R$ is on the perpendicular bisector of $MN$, and at the same time, it is on the angle bisector of $\angle MAN$. So it is a well-known fact that $R$ is on $\bigodot AMN$. L...

- Tue Jul 30, 2013 4:26 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2013, Day 1-P3
- Replies:
**3** - Views:
**2969**

### Re: IMO 2013, Day 1-P3

I have used traditional notations here. Assume that $O\in \widehat{BAC}$. Then $\angle B_{1}A_{1}C_{1}$ is obtuse. Let us assume that $P$ is the midpoint of $\widehat{BAC}$. Then, $\angle C_{1}BP=\angle ABP=\angle ACP=\angle B_{1}CP$ and $PB=PC$. Also,$BC_{1}=s-a=B_{1}C$. Thus $\triangle PBC_{1}\con...

- Tue Jul 30, 2013 3:09 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2013, Day 2-P4
- Replies:
**2** - Views:
**2131**

### Re: IMO 2013, Day 2-P4

Let $V=\omega_1\cap\omega_2$. Now according to the definitions of $X,Y, \angle XVW=\angle YVW=90^{\circ}$ and thus $V\in XY$. Notice that $W,M,N$ are three points on the sides $BC,CA,AB$ of $\triangle ABC$, and $\bigodot BWN$ and $\bigodot CWM$ pass through $V$. So $V$ is the Miquel point. So, $M\in...

- Tue Jul 30, 2013 2:39 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2013, Day 1-P1
- Replies:
**1** - Views:
**1387**

### Re: IMO 2013, Day 1-P1

We will use induction on $k$. Note that if $k=1$, then $\forall n\in \mathbb{N}$, we have $\displaystyle 1+\frac{2^{1}-1}{n}=1+\frac{1}{n}$ and it satisfies the given condition for $k=1$. This our base case. Next,let us assume thatwe can find an expression for $\displaystyle 1+\frac{2^{k}-1}{n}$ for...

- Fri May 10, 2013 11:12 pm
- Forum: Geometry
- Topic: Angle Equity
- Replies:
**1** - Views:
**1219**

### Angle Equity

$AD$ is the altitude of an acute $\triangle ABC$. Let $P$ be an arbitrary point on $AD$. $BP,CP$ meet $AC,AB$ at $M,N$ respectively. $MN$ intersects $AD$ at $Q$. $F$ is an arbitrary point on side $AC$. $FQ$ intersects line $CN$ at $E$. Prove that $\angle FDA = \angle EDA$.

- Fri May 10, 2013 11:06 pm
- Forum: Asian Pacific Math Olympiad (APMO)
- Topic: APMO 2013 Problem 1
- Replies:
**1** - Views:
**1783**

### Re: APMO 2013 Problem 1

Note that, $\triangle OAF=\frac{1}{2}\cdot AF\cdot AO\cdot \sin (\angle OAF)=\frac{1}{2}\cdot \cos A\cdot AC\cdot R\cdot \sin (90^{\circ}-\angle C)=\frac{R}{2}\cdot b\cdot \cos C$ Derive similar equations for the other triangles to show that, $\triangle OAF=\triangle OCD$, $\triangle OBF=\triangle O...