Search found 181 matches
Re: Nice geo
$CH$ meets $ AB$ at $E$, $BH$ meets $AC$ at $F$. Let $X$ be the midpoint of $AH$. It suffices to prove that $X,P,M$ are colinear. Now, $X$ is the center of the circle with diameter $AH$. $P$ is the midpoint of arc $EF$. It's also well known that $ME , MF$ are tangent to the circle $AEF$. The result ...
- Sun Jan 01, 2017 2:11 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2016 Problem 2
- Replies: 2
- Views: 3461
Re: IMO 2016 Problem 2
\[ \begin{array}{|ccc|ccc|ccc|} I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M...
- Sun Jan 01, 2017 4:00 am
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO 2016 Problem 1
- Replies: 3
- Views: 8964
Re: IMO 2016 Problem 1
We will show the result by the** radical axis theorem**. So, we will show that $BFDX, BMDE, FEMX$ is concyclic. $ \angle ADF = \angle ACB$ by **spiral similarity**. $\angle FDC = 180 - \angle ADE - \angle CDX = 180 - 2\angle ADE - \angle ADF = 180 - \angle BFC - \angle ACB = 90 $. So, $BCDF$ is a cy...
- Sun Jan 01, 2017 1:35 am
- Forum: Algebra
- Topic: Cool problem. IMO 2013 A1 I guess
- Replies: 1
- Views: 2283
Cool problem. IMO 2013 A1 I guess
Let $n$ be a positive integer and let $a_1, \ldots, a_{n-1} $ be arbitrary real numbers. Define the sequences $u_0, \ldots, u_n $ and $v_0, \ldots, v_n $ inductively by $u_0 = u_1 = v_0 = v_1 = 1$, and $u_{k+1} = u_k + a_k u_{k-1}$, $v_{k+1} = v_k + a_{n-k} v_{k-1}$ for $k=1, \ldots, n-1.$ Prove tha...
- Sun Jan 01, 2017 1:22 am
- Forum: Number Theory
- Topic: USAMO 2015 P5
- Replies: 2
- Views: 2952
Re: USAMO 2015 P5
Not my solution Clearly $ac+bd>2$, so supposed for sake of contradiction that $ac+bd=p$, some odd prime. Assume without loss of generality that $a>c,$ or otherwise we could switch $a$ with $c$ and $b$ with $d$. So $a^4>c^4=a^4+b^4-d^4$ implying $b^4<d^4$ and $b<d$. Now note $a^4-c^4=d^4-b^4$ so \[ (...
- Sun Jan 01, 2017 1:14 am
- Forum: Combinatorics
- Topic: Olympiad Comninatorics Chapter 2 problem 7
- Replies: 1
- Views: 2354
Re: Olympiad Comninatorics Chapter 2 problem 7
This is all I could progress with. WLoG, let's assume that every point is connectable by a line parallel to the axes. If every point aren't, we can just treat the non connectable set of points as the problem itself. We can also reduce the problem to the points which have other points on their x and ...
- Thu Dec 29, 2016 2:07 am
- Forum: Combinatorics
- Topic: Olympiad Comninatorics Chapter 2 problem 7
- Replies: 1
- Views: 2354
Olympiad Comninatorics Chapter 2 problem 7
Given a finite set of points in the plane, each with integer coordinates, is it always possible to color the points red or white so that for any straight line L parallel to one of the coordinate axes the difference (in absolute value) between the numbers of white and red points on L is not greater t...
- Sun Dec 11, 2016 4:38 pm
- Forum: Geometry
- Topic: APMO 2013 P5
- Replies: 3
- Views: 3706
Re: APMO 2013 P5
Take a projective transformation that fixes ω and sends the pointAC ∩ BD to the center of the circle. Thus ABCD is a rectangle. Because ABCD is harmonic, it must in fact be a square. Thus P is the point at infinity along AB CD and the problem is not very hard now. -Copied from Evan Chen's book
- Sun Aug 28, 2016 8:47 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114920
Re: IMO Marathon
$\boxed{\text{Problem 48}}$ Prove that, for any positive integer set $\{a_1,a_2,...,a_n\}$ there exists a positive integer $b$ such that the set $\{ba_1,ba_2,...,ba_n\}$ consists of perfect powers.
Source -Number Theory Structures examples and problems.
Source -Number Theory Structures examples and problems.
- Sun Aug 28, 2016 8:03 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies: 184
- Views: 114920
Re: IMO Marathon
We can't save any units on the outside of the grid. So, we better fill them with triangles. The problem reduces to a 98*98 grid for which we could ignore the outside edges. We notice that out of any 2*1 grid, we must have one unit with a triangle. So there are at least 98*98/2=4802 units filled with...