## Search found 110 matches

- Sun Jun 21, 2015 12:56 pm
- Forum: Geometry
- Topic: cyclic quad
- Replies:
**3** - Views:
**2641**

### 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:
**2974**

### 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:
**2174**

### 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:
**2174**

### 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:
**1912**

### 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:
**4312**

### 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:
**2125**

### 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...