Search found 110 matches
- Sun Jun 21, 2015 12:56 pm
- Forum: Geometry
- Topic: cyclic quad
- Replies: 3
- Views: 4483
cyclic quad
let $ABC$ be a acute angled triangle , and let $P$ and $Q$ be two points on side $BC$ . construct a point ${C}_{1}$ in such a way that the convex quad $APB{C}_{1}$ is cyclic , $Q{C}_{1}||CA$ and the point ${C}_{1}$ and $Q$ lie on the opposite side of line $AB$ .construct a point ${B}_{1}$ in such a ...
- Wed May 06, 2015 12:34 pm
- Forum: Geometry
- Topic: Inscribed-Quad in an Excribed-Quad
- Replies: 5
- Views: 5276
Re: Inscribed-Quad in an Excribed-Quad
i could not visit the forum last 2 days for internet troubles . so didn't notice the typo update .
was trying to solve the previous .....then hssss
whatever, now i got the solution which nirjhor posted
was trying to solve the previous .....then hssss
whatever, now i got the solution which nirjhor posted
- Sun May 03, 2015 12:22 pm
- Forum: Geometry
- Topic: triangular inequality [sides and area]
- Replies: 3
- Views: 3828
Re: triangular inequality [sides and area]
my solution is same as nirjhor
yeah sowmitra vaia this is IMO 1961/2
yeah sowmitra vaia this is IMO 1961/2
- Sat May 02, 2015 1:15 am
- Forum: Geometry
- Topic: triangular inequality [sides and area]
- Replies: 3
- Views: 3828
triangular inequality [sides and area]
Prove that for any triangle $ABC$ with sides $a,b,c$ and area $A$,
$a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A$
$a^{2}+b^{2}+c^{2}\geq 4\sqrt{3}A$
- Fri May 01, 2015 12:34 am
- Forum: Algebra
- Topic: Inequality (sin, r and s)
- Replies: 2
- Views: 3307
Re: Inequality (sin, r and s)
my solution is too much long :| $\frac{1}{\sqrt{2sinA}}+\frac{1}{\sqrt{2sinB}}+\frac{1}{\sqrt{2sinC}}\leq \sqrt{\frac{s}{r}}$ or,$\sqrt{\frac{R}{a}}+\sqrt{\frac{R}{b}}+\sqrt{\frac{R}{c}}\leq \sqrt{\frac{s}{r}}$ or,$\frac{R}{a}+\frac{R}{b}+\frac{R}{c}+\frac{2R}{\sqrt{ab}}+\frac{2R}{\sqrt{bc}}+\frac{2...
Re: cool geo
vaia your solution is very nice . but i did it just using angle chasing . :) let the internal angle bisector intersects the side $BC$ at $X$ $\angle NAP = 90-\angle AXD=90- \angle A/2 -\angle C$ $\angle NPA=\angle XPM=180-\angle PMX-\angle PXM=180-(180-\angle NFD)-\angle A/2-\angle B$ $=180-180+\ang...
- Mon Apr 27, 2015 10:59 am
- Forum: Introductions
- Topic: O hai all!
- Replies: 3
- Views: 8863
Re: O hai all!
zawad vai, nobody believes this kind of swear
Re: cool geo
here is a little typo . $\angle APH= \angle AFH$ should be $90$sowmitra wrote:$AEHF$ is cyclic. $\therefore P \in \odot AEF$ iff $\angle AHP= \angle AHF =90^{\circ}$.
- Fri Apr 24, 2015 12:21 am
- Forum: Combinatorics
- Topic: An interesting combinatorial identity
- Replies: 3
- Views: 3759
Re: An interesting combinatorial identity
argument is as same as prosenjit .
but this time , just think about a $a\cdot b\cdot c$ cube .
we want to place two rooks such that they are not in the same plane .
and in $R.H.S$ a little bit of inclusion and exclusion is needed
but this time , just think about a $a\cdot b\cdot c$ cube .
we want to place two rooks such that they are not in the same plane .
and in $R.H.S$ a little bit of inclusion and exclusion is needed
cool geo
In $\Delta ABC$, $AD,BE,CF$ are the feets of the perpendiculars . $M,N$ are the midpoints of $BC$ and $AH$ respectively [$H$ orthocentre] . $MN$ intersects the internal and the external angle bisector of $\angle BAC$ at $P$ and $Q$ respectively. Prove that $P,Q$ lies on the circumcircle of $\Delta A...