Search found 110 matches
- Sun Feb 19, 2017 11:16 am
- Forum: Algebra
- Topic: $2009$ USA TST Inequality
- Replies: 1
- Views: 2314
$2009$ USA TST Inequality
Prove that for positive real numbers $x$, $y$, $z$, $$x^3(y^2+z^2)^2 + y^3(z^2+x^2)^2+z^3(x^2+y^2)^2 \geq xyz\left[xy(x+y)^2 + yz(y+z)^2 + zx(z+x)^2\right]$$
- Sun Feb 19, 2017 11:15 am
- Forum: Number Theory
- Topic: Numbers expressible as $\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}}$
- Replies: 1
- Views: 2380
Numbers expressible as $\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}}$
Let $n$ be a positive integer. Find, with proof, the least positive integer $d_{n}$ which cannot be expressed in the form \[\sum_{i=1}^{n}(-1)^{a_{i}}2^{b_{i}},\]
where $a_{i}$ and $b_{i}$ are nonnegative integers for each $i.$
where $a_{i}$ and $b_{i}$ are nonnegative integers for each $i.$
- Sun Feb 19, 2017 11:12 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 112397
Re: IMO Marathon
$\text{Problem } 53$ Let $n$ be a positive integer and let $a_1 \le a_2 \le \dots \le a_n$ and $b_1 \le b_2 \le \dots \le b_n$ be two nondecreasing sequences of real numbers such that $$ a_1 + \dots + a_i \le b_1 + \dots + b_i \text{ for every } i = 1, \dots, n$$ and $$ a_1 + \dots + a_n = b_1 + \do...
- Sun Feb 19, 2017 11:02 am
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies: 48
- Views: 43635
Combi Marathon
This thread is dedicated for a marathon on combinatorics problems. The rules are similar to the Geo Marathon . A poster must post both the solution of the previous problem and the next problem of the marathon. If no one solves the next problem within $2$ days, he or she must provide a solution or a ...
- Sun Feb 19, 2017 10:45 am
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 187663
Re: Geometry Marathon : Season 3
$\text{Problem 29}$ In acute triangle $ABC$ , segments $AD; BE$ , and $CF$ are its altitudes, and $H$ is its orthocenter. Circle $\omega$, centered at $O$, passes through $A$ and $H$ and intersects sides $AB$ and $AC$ again at $Q$ and $P$ (other than $A$), respectively. The circumcircle of triangle ...
- Wed Feb 15, 2017 10:48 pm
- Forum: Algebra
- Topic: F.E. (You run out of these names fast)
- Replies: 1
- Views: 2086
Re: F.E. (You run out of these names fast)
A brief sketch of the solution. Plugging in $x=1$ in the second equality gives $f(1)=1$. By induction, it follows that $f(x)=x$ for all positive integers $x$. Note that $f(x+k)=f(x)+k$ for positive integer $k$. So, $f(m+\frac{k}{m})^2 = (f(\frac{k}{m})+m)^2$. By the second F.E condition, $f(m+\frac{...
- Sat Feb 04, 2017 6:48 pm
- Forum: Geometry
- Topic: Circle is tangent to circumcircle and incircle
- Replies: 3
- Views: 3835
Re: Circle is tangent to circumcircle and incircle
Without loss of generality, let $Y$ be closer to $A$ than $X$. The crux move is to show that $AD$ bisects $\angle SAT$. Let $L$ be the midpoint of $XY$. Clearly $X,Y,S,T$ all lie on a circle centered at $L$. Denote this circle by $\ell$. Lemma : $CY\perp AX$ and $BX\perp AX$ Proof : Well known. Chas...
- Fri Feb 03, 2017 8:21 pm
- Forum: Number Theory
- Topic: IMO SL 2005 N5
- Replies: 0
- Views: 1828
IMO SL 2005 N5
Denote by $d(n)$ the number of divisors of the positive integer $n$. A positive integer $n$ is called highly divisible if $d(n) > d(m)$ for all positive integers $m < n$. Two highly divisible integers $m$ and $n$ with $m < n$ are called consecutive if there exists no highly divisible integer $s$ sat...
- Fri Feb 03, 2017 5:30 pm
- Forum: Algebra
- Topic: A beautiful FE
- Replies: 2
- Views: 2465
Re: A beautiful FE
We will show that $f(x)=x+1$. Check to see that this indeed verifies the original equation. $P(0,x)$ implies $f(f(x))=f(0)((f(x)-1)+2$ whilst $P(1,x)$ implies $f(f(x))=\dfrac{f(x)f(1)+2}{2}$. Combining the two gives us $f(x)=\dfrac{2f(0)-2}{2f(0)-f(1)}$ implying $f$ is a constant, which is readily f...
- Fri Feb 03, 2017 4:10 pm
- Forum: Geometry
- Topic: Geometry Marathon : Season 3
- Replies: 146
- Views: 187663
Re: Geometry Marathon : Season 3
$\text{Solution of Problem } 20$ This problem is readily bary-able. Line $CP$ has the equation $x\cdot (b^2+c^2-a^2)-y\cdot (a^2-b^2+c^2)=0$ and line $AP$ has the equation $b^2z+c^2y=0$. We therefore proceed to find $K$. Redefine $K$ to be the point such that $KH$ is perpendicular to $AD$. It suffic...