## Search found 6 matches

Mon Apr 17, 2017 8:01 pm
Forum: Geometry
Topic: A Lemma?
Replies: 8
Views: 2365

### Re: A Lemma?

Thanic Nur Samin wrote: $MN||AH$
surely it wants to be $MN||AO$ ??
Mon Apr 17, 2017 7:52 pm
Forum: Geometry
Topic: A Lemma?
Replies: 8
Views: 2365

### 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 $EF\|H_bH_c$
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: 976

### 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$bit strings having a total of$n1'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$bit string with atmost$n1's$now for$a_0$a... Sat Feb 25, 2017 11:27 pm Forum: Algebra Topic: When the solution is easier than the question Replies: 1 Views: 1087 ### Re: When the solution is easier than the question True for$ 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: 1197

### 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: 7031

### 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,$OB=OA$.