surely it wants to be $MN||AO$ ??Thanic Nur Samin wrote: $MN||AH$
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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: 2497
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: 2620
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: 2730
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: 13736
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,$OB=OA$.