## Search found 176 matches

- Mon Feb 20, 2017 6:09 pm
- Forum: Geometry
- Topic: A Problem for Dadu
- Replies:
**2** - Views:
**1278**

### Re: A Problem for Dadu

Solution: Let the circumcircle of $ABCD$ be the unit circle. We will apply complex numbers. Now, let $a,b,c,d,h_a,h_b,h_c,h_d$ denote $A,B,C,D,H_A,H_B,H_C,H_D$ respectively. Now, $h_a=b+c+d$ So, the midpoint of $AH_A$ has complex coordinate $\dfrac{a+h_a}{2}=\dfrac{a+b+c+d}{2}$. Due to the symmetric...

- Mon Feb 20, 2017 2:16 pm
- Forum: Geometry
- Topic: Looking for non-trig solution
- Replies:
**3** - Views:
**1348**

### Re: Looking for non-trig solution

Here is a solution that doesn't use trig. Let $AB$ be the $x$-axis and the center $O$ the origin. So, $B$ is $(2,0)$ and $A$ is $(-2,0)$. Reflect $C$ through the $x$-axis. Since $\angle COG=30^{\circ}$, $\angle COC'=60^{\circ}$, and since $OC=OC'$, it $\triangle COC'$ is equilateral. If we let $G$ b...

- Sun Feb 19, 2017 11:43 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**47528**

### Re: IMO Marathon

$\text{Problem 54}$ The following operation is allowed on a finite graph: choose any cycle of length $4$ (if one exists), choose an arbitrary edge in that cycle, and delete this edge from the graph. For a fixed integer $n \ge 4$, find the least number of edges of a graph that can be obtained by repe...

- Sun Feb 19, 2017 11:25 pm
- Forum: International Mathematical Olympiad (IMO)
- Topic: IMO Marathon
- Replies:
**184** - Views:
**47528**

### Re: IMO Marathon

Solution to problem $53$: Take the sum of all differences for both sequences, since each difference in one sequence must have a corresponding equal difference in the other, the quantity would be same. Multiply both sides with $(-1)$. Add $(n-1)(a_1 + \dots + a_n) = (n-1)(b_1 + \dots + b_n)$ to both ...

- Sun Feb 19, 2017 7:04 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**22236**

### Re: Combi Marathon

Problem 2 Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_1,A_2, . . . ,A_k$ such that for all integers $n\ge 15$ and all $i \in \{1, 2, . . . , k\}$ there exist two distinct elements of $A_i$ who...

- Sun Feb 19, 2017 6:59 pm
- Forum: Combinatorics
- Topic: Combi Marathon
- Replies:
**48** - Views:
**22236**

### Re: Combi Marathon

We claim that $f(n)=\dfrac{1}{3}\dbinom{n}{2}\dbinom{n-2}{2}$. Note that the sum simply counts one third of the quadruplet of points $(P_i,P_j,P_k,P_t)$ so that $P_t$ is inside the circle. (Since the $P$'s can permute in the quadruplet). Now, for a non-reflex angle $\theta$, we intruduce a function ...

- Sat Feb 18, 2017 12:59 am
- Forum: Geometry
- Topic: IMO Shortlist 2010 G7
- Replies:
**1** - Views:
**1281**

### IMO Shortlist 2010 G7

Three circular arcs $\gamma_1, \gamma_2,$ and $\gamma_3$ connect the points $A$ and $C.$ These arcs lie in the same half-plane defined by line $AC$ in such a way that arc $\gamma_2$ lies between the arcs $\gamma_1$ and $\gamma_3.$ Point $B$ lies on the segment $AC.$ Let $h_1, h_2$, and $h_3$ be thre...

- Fri Feb 17, 2017 10:49 am
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 junior/10
- Replies:
**2** - Views:
**1126**

### Re: BdMO 2017 junior/10

Notice that the hocus pocus sum is essentially square of the sum of all numbers minus the sum of the squares of the numbers divided by two. So, if there are $x$ $(-1)$'s and $y$ $2$'s, then the sum is $\dfrac{(2y-x)^2-(4y+x)}{2}=\dfrac{(3y-100)^2-3y-100}{2}=\dfrac{(3y-100.5)^2-200.25}{2}$. It clearl...

- Wed Feb 15, 2017 9:25 pm
- Forum: Algebra
- Topic: There should be a list of generic titles for sourceless prob
- Replies:
**1** - Views:
**777**

### Re: There should be a list of generic titles for sourceless

Let $P(x,y)$ be the FE. $$P(0,x)\Rightarrow f(x)+f(-x)=2f(0)\cos x=2A\cos x$$ $$P\left (x+\dfrac{\pi}{2}\right )\Rightarrow f(x)+f(x+\pi)=0$$ $$P\left (\dfrac{\pi}{2},\dfrac{\pi}{2}+x\right )\Rightarrow f(-x)+f(x+\pi)=-2f\left (\dfrac{\pi}{2}\right )\sin x=-2B\sin x$$ So, $f(x)=A\cos x+B\sin x$. Sub...

- Tue Feb 14, 2017 7:15 pm
- Forum: National Math Olympiad (BdMO)
- Topic: BdMO 2017 National Round Secondary 8
- Replies:
**2** - Views:
**1527**

### Re: BdMO 2017 National Round Secondary 8

The sequence is actually $\dfrac{(1+i)^n+(1-i)^n}{2}$. It is easy to prove it by induction. Now, $(1+i)^8=(1-i)^8=2^4$ and $2016=8\times 252$, so $a_{2016}=2^{1008}$. Now, note that $\displaystyle \prod_{k=1}^{1008}2k=2^{1008}\times 1008!$ Again, $\displaystyle \prod_{k=1}^{1008}2i=\prod_{k=1}^{504}...