## Search found 155 matches

- Tue Dec 30, 2014 9:40 pm
- Forum: Higher Secondary Level
- Topic: Looking for projective geometry books
- Replies:
**6** - Views:
**2481**

### Re: Looking for projective geometry books

I'm going to mention and give some links to Projective Geometry books and pdf's. (Since, the idea of pole-polars are used the most in olympiad problems, I'll give some links seperately about that as well.) I hope others will continue the thread. $[1]$ Projective Geometry , by, H.S.M. Coxeter $[2]$ A...

- Fri Dec 19, 2014 12:29 am
- Forum: Geometry
- Topic: Side, Angle, and, (Ex)-Circle
- Replies:
**4** - Views:
**1048**

### Side, Angle, and, (Ex)-Circle

In $\triangle ABC$, $\angle A=2\angle C$. $(O)$ is its $A-$excircle, and, $M$ is the mid-point of $AC$. $OM\cap BC=D$. Show that,

$[1]$ $AD$ bisects $\angle OAC$

$[2]$ $BA=BD$

$[1]$ $AD$ bisects $\angle OAC$

$[2]$ $BA=BD$

- Sat Nov 29, 2014 8:20 pm
- Forum: Geometry
- Topic: powers are equal
- Replies:
**3** - Views:
**860**

### Re: powers are equal

$A,B,C$ all lie on the circles they are meant to be respected to. Not true. First, prove that, $AB',$ etc. are tangent to $\odot B'PC,$ etc. Then, the powers of $A, B, $ and $C$ w.r.t. the given circles are equal to the squares of $AB', BC', CA'$. So, we'll be done if we can show $AB'=BC'=CA'$. Now...

- Tue Nov 11, 2014 7:53 pm
- Forum: Junior Level
- Topic: A confusion
- Replies:
**2** - Views:
**1021**

### Re: A confusion

No, this isn't true...

- Wed Oct 29, 2014 9:51 pm
- Forum: Secondary Level
- Topic: Floor sum
- Replies:
**1** - Views:
**758**

### Re: Floor sum

Let, $S=\sum^{n}_{k=1}\lfloor\sqrt k\rfloor$. For all $1\leq k\leq (a^2-1)$, $\lfloor\sqrt k\rfloor=i$, where, $1\leq i\leq (a-1)$, $i\in \mathbb{N}$, and, $i$ appears exactly $(i+1)^2-i^2=(2i+1)$ times in $S$. \[\therefore \sum^{a^2-1}_{k=1}\lfloor\sqrt k\rfloor=\sum_{i=1}^{a-1}i(2i+1)=2\sum i^2+\s...

- Sun Oct 26, 2014 8:48 pm
- Forum: Secondary Level
- Topic: A circle through incenter
- Replies:
**2** - Views:
**1015**

### Re: A circle through incenter

Clarification please... I think there's something wrong with the question

- Wed Oct 22, 2014 12:19 am
- Forum: Algebra
- Topic: (Angle)^(Side)[Inequality]
- Replies:
**7** - Views:
**1623**

### Re: (Angle)^(Side)[Inequality]

Oops... sorry I miscalculated by taking $\frac{1}{x}$ instead of $\frac{1}{\sin x}$ while differentiating...

- Tue Oct 21, 2014 12:04 am
- Forum: Algebra
- Topic: (Angle)^(Side)[Inequality]
- Replies:
**7** - Views:
**1623**

### Re: (Angle)^(Side)[Inequality]

\[f(x)=\frac{1}{\sin{x}}\cdot\ln\left(\frac{2x}{\pi}\right)\]

- Mon Oct 20, 2014 8:26 pm
- Forum: Algebra
- Topic: (Angle)^(Side)[Inequality]
- Replies:
**7** - Views:
**1623**

### Re: (Angle)^(Side)[Inequality]

I think one Jensen can suffice...

- Sun Oct 19, 2014 10:42 pm
- Forum: Algebra
- Topic: (Angle)^(Side)[Inequality]
- Replies:
**7** - Views:
**1623**

### (Angle)^(Side)[Inequality]

Let, $ABC$ be a triangle with angles $A,B,C$, and, sides $a,b,c$ (usual notations). Let, $R$ be the circum-radius. Prove that, \[\left(\frac{2A}{\pi}\right)^\frac{1}{a}\left(\frac{2B}{\pi}\right)^\frac{1}{b}\left(\frac{2C}{\pi}\right)^\frac{1}{c}\leq \left(\frac{2}{3}\right)^{\frac{\sqrt{3}}{R}}\]