surely it wants to be $MN||AO$ ??Thanic Nur Samin wrote: $MN||AH$

## Search found 6 matches

### Re: A Lemma?

### Re: A Lemma?

a proof not elegant

1.

$\angle ABH_a =\angle AH_bH_c $ and

$\angle BAH_a =\angle H_bAO $ as $O$ and $H$ are isogonal conjugates

2.

definitely not elegant

let the tangent meet $AB$ and $BC$ at $E$,$F$

$\angle H_bHF=\angle EHB =\angle HCB=\angle HH_bH_c $

and gives $EF\|H_bH_c$

1.

$\angle ABH_a =\angle AH_bH_c $ and

$\angle BAH_a =\angle H_bAO $ as $O$ and $H$ are isogonal conjugates

2.

definitely not elegant

let the tangent meet $AB$ and $BC$ at $E$,$F$

$\angle H_bHF=\angle EHB =\angle HCB=\angle HH_bH_c $

and gives $EF\|H_bH_c$

- Thu Apr 13, 2017 1:20 pm
- Forum: Number Theory
- Topic: Numbers expressible as $\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}}$
- Replies:
**1** - Views:
**766**

### Re: Numbers expressible as $\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{

$d_n$ the binary representation of a number that can't be expressed with a subtraction of two $n$ bit strings having a total of $n$ $1's$ together. we will prove that for a $d_n < 2^n $ there exists such $n$ bit strings for a number $x < 2^n$ is a $n$ bit string with atmost $n$ $1's$ now for $a_0$ a...

- Sat Feb 25, 2017 11:27 pm
- Forum: Algebra
- Topic: When the solution is easier than the question
- Replies:
**1** - Views:
**895**

### Re: When the solution is easier than the question

True for $ n=2, n=3 $... assume its true for $k \le n$ let $S_n$ denote the sum of the squares of the products of the subsets of $ \{1,2,....,n \} $ now, from the set $\{1,2,....,k+1\}$ the element $k+1$ has each non-neighbouring elements from non-neighbouring subsets of $\{1,2,....,k-1\}$ that, $S...

- Tue Feb 02, 2016 4:05 am
- Forum: Junior Level
- Topic: Easy Chess Tournament Problem
- Replies:
**1** - Views:
**1000**

### Easy Chess Tournament Problem

On a chess tournament 12 players took part, and any two of them played exactly one match.Any draw gives 0.5 point, win -1 point, loss -0.It turned out after the tournament that the first three together gained three times more points then the last five. How did the match between the seventh and eight...

- Sat Nov 21, 2015 8:18 pm
- Forum: Secondary Level
- Topic: geometry
- Replies:
**3** - Views:
**2423**

### Re: geometry

its enough to prove that $\angle BAO=\angle BAC$ when $BD$ is a tangent to the circle...

now,if $BD$ is a tangent $\angle OBD=\angle OBA+\angle ABD=90$.

again,$\angle D=90.\angle BAD+\angle ABD=90$.

then,$\angle OBA=\angle CAB$ and $\angle BAO=\angle BAC$ as,$OB=OA$.

now,if $BD$ is a tangent $\angle OBD=\angle OBA+\angle ABD=90$.

again,$\angle D=90.\angle BAD+\angle ABD=90$.

then,$\angle OBA=\angle CAB$ and $\angle BAO=\angle BAC$ as,$OB=OA$.