New answers tagged contest-math
2
votes
Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.
A simple long division method yields
$p(x)=ax^5+bx^4+1=(x^2-x-1)((ax^3)+(a+b)x^2+(2a+b)x+(3a+2b))+\ (5a+3b)x+(3a+2b+1)$
as we know $(x^2-x-1)$ is factor of $p(x)$, The remainder $(5a+3b)x+(3a+2b+1)$ ...
- 833
2
votes
Accepted
Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.
Your method is on-target. You just need to reduce the quintic until both the equations are of same degree to compare coefficients. I will show an example with highest degree of $2$ (reduction to ...
- 2,332
1
vote
Knowing $x^2-x-1$ is a factor of $p(x)=ax^5 + bx^4 + 1$ , find a,b.
Notice that $\frac{1\pm\sqrt{5}}{2}$ are the roots of $p(x)$ as they are the roots of $x^2-x-1$. Hence, we would have two equations:
$$p\left(\frac{1+\sqrt{5}}{2}\right) = a\left(\frac{1+\sqrt{5}}{2}\...
- 1,448
1
vote
Monochromatic $4$-Cycle in Bipartite Complete Graph
Let one part be $\{\,v_1, v_2, \ldots, v_n\,\}$ and another part be $\{\,u_1, u_2, \ldots, u_n\,\}$.
Let's show that for $n = 5$ the graph always has a monochromatic $4$-cycle. Suppose the opposite. ...
- 6,370
2
votes
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Disclaimer: Not a full solution
If one of $a, b, c=0$:
WLOG let $a=0$. Since $ab+bc+ca>0$ we know that $b, c\neq0$.
$$\begin{align}
& \frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ca}}+\frac{1}{\...
- 2,920
6
votes
counting sequences of elements of the set {1,2,3,4} with given property
Disclaimer
This answer skips some details, but hopefully it's sufficient to guide you and let you research the details yourself.
The Answer
$$
x_{4+2m}=
\left(\frac{5-3\sqrt{5}}{10}\right)
\left(\frac{...
- 10.6k
1
vote
counting sequences of elements of the set {1,2,3,4} with given property
Here's a recurrence I got.
Let $f_n$ be the number of ways to have such a sequence of length $n$ with $a_n = 2$.
Case 1: $a_{n-1}=1$
$a_{n-2}$ must be 2, this accounts for $f_{n-2}$ ways.
Case 2: $a_{...
3
votes
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Some thoughts.
By AM-GM, it suffices to prove that
$$\frac{2}{\frac{a^2 + 4bc}{2 + bc/2} + (2 + bc/2)} + \frac{2}{\frac{b^2 + 4ca}{2 + ca/2} + (2 + ca/2)} + \frac{2}{\frac{c^2 + 4ab}{2 + ab/2} + (2 + ...
- 36.5k
5
votes
Accepted
If $a+b+c+abc=4,$ prove $\frac{1}{\sqrt{a^2+4bc}}+\frac{1}{\sqrt{b^2+4ac}}+\frac{1}{\sqrt{c^2+4ba}}\ge \frac{5}{4}.$
Some thoughts.
By Holder inequality, we have
\begin{align*}
&\left(\sum_{\mathrm{cyc}} \frac{1}{\sqrt{a^2 + 4bc}} \right)^2\cdot \sum_{\mathrm{cyc}} (a^2 + 4bc)(4b + 4c - bc + 4ab + 4ac)^3 \\
\...
- 36.5k
3
votes
Accepted
IMO 1987/P1 - Combinatoric approach
Your claim that $p_n(k)=\binom nk(n-k-1)!$ is not correct. In the case where $k=0$, your formula would imply that there are $(n-1)!$ permutations with zero fixed points, so $(n-1)!$ derangements of $n$...
- 72.1k
1
vote
Accepted
German MO combinatorics problem 1995
Just to clarify user 10354138 answer beacause there might be a small mistake:
$$\frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}\ge 2\cdot \frac{n+1}{n+2}$$
$$\iff \frac{k}{2n-k+2}+\frac{2n-k+1}{k+1}-\frac{2(n+1)}{...
- 26
4
votes
Accepted
Proving the existence of integers $a, b, c$ such that $\left\lvert a + b\sqrt{2} + c\sqrt{3}\right\rvert < 10^{-5}$
DISCUSSION :
There are a couple of Issues with your nice approach. When those Issues are rectified , the Solution will work out.
(1) The range you have taken is $10^3$ , though the Correct range is $...
- 7,401
0
votes
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
Leaving the discontinuous case,The function might be of Two types,Either continuous Differentiable or continuous non-Differentiable
Now either the Function is Constant function or It is a non-...
- 833
2
votes
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
Suppose that $f(x) = 0$ for some $x \neq 0$. Then, substituting in any value of $y$ shows that $0 = x^2f(y)$, so $f(y)=0$ for all $y \in \mathbb R$. Meanwhile, letting $y=x=0$ gives $f(0)^3=0$, so $f(...
2
votes
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
Swapping $x$ and $y$, for any $x$ and $y$ we have:
$f(y)f(x)f(y-x)=y^2f(x)-x^2f(y)=-f(x)f(y)f(x-y) \Rightarrow f(x)=0~or~f(y-x)=-f(x-y) $.
Assuming a nontrivial solution, this means that $f(x)$ is an ...
- 556
3
votes
Accepted
Thailand MO $f(x)f(y)f(x-y) = x^2f(y) - y^2f(x)$
$$(1) \ \ \ \ \ f(0)=0$$
You can trivially show $f(x) = 0$ is a solution.
Otherwise,
$$(2) \ \ \ \ \ f(-t) = -f(t)$$
$$(3) \ \ \ \ f(2x)\cdot f(x) = 2x^2$$
$$(4) \ \ \ \ \ 2\cdot f(x)=f(2x)$$
Hence, $...
- 1,085
-4
votes
Proving the existence of integers $a, b, c$ such that $\left\lvert a + b\sqrt{2} + c\sqrt{3}\right\rvert < 10^{-5}$
The simplest one is to write a computer program that iterates through all b from -1000 to +1000, then through all c from -1000 to 1000 where $|b \sqrt 2 + c \sqrt 3| < 1000.5$, round the result to ...
- 9,599
0
votes
IGO 2017 P4 of Combinatorial Geometry
Let $ P_1, P_2, ..., P_{100} $ represent points on a circle. The initial count of clockwise triangles is $0$. As the points move, the orientation of each triangle $ P_iP_jP_k $ changes by one each ...
- 1
1
vote
Diagonals in an inscribed quadrilateral can simultaneously be angle bisectors of triangles involving their midpoints
May be this idea works: As can be seen in figure the circle passing the center of circumcircle of quadrilateral ABCD and the ends of it's diagonals passes through mid points of non corresponding ...
- 10.3k
4
votes
Accepted
BMO1 number theory question on fibonacci sequence and divisibility
Here's how I solved it.
The given problem can be formulated as: Prove that $f_n = anb^n$ mod $m$. I didn't want to think about whether various theorems hold over rings, so I'm going to initially hope ...
- 156
1
vote
Maximum value of $(1-a)(1-b)+(1-p)(1-q)$
By AM-GM and C-S we obtain:
$$(1-a)(1-b)+(1-p)(1-q)\leq\frac{(1-a)^2+(1-b)^2}{2}+\frac{(1-p)^2+(1-q)^2}{2}=$$
$$=\frac{4-2(a+b)}{2}+\frac{4-2(p+q)}{2}\leq\frac{4+2\sqrt{2(a^2+b^2)}}{2}+\frac{4+2\sqrt{...
- 197k
1
vote
Diagonals in an inscribed quadrilateral can simultaneously be angle bisectors of triangles involving their midpoints
Let $\angle CDE=\alpha, ~\angle CDM=\beta$ and $\angle MDB=\gamma$.
Simple angle chasing (using $MD=ME$ and $AC||DE$) leads to $\angle ADB=\beta$.
Hence, $DB$ is the $D$-symmedian of $\triangle ADC$. ...
- 3,354
0
votes
Maximum value of $(1-a)(1-b)+(1-p)(1-q)$
We can also use the method of Lagrange multipliers.
Maximize $f(x,y)=(1-x)(1-y)$ subject to $g(x,y)=x^2+y^2-2=0.$
$\frac{\partial f}{\partial x}=\lambda \frac{\partial g}{\partial x}$ and $\frac{\...
- 8,544
2
votes
Maximum value of $(1-a)(1-b)+(1-p)(1-q)$
As stated in the comments, you could check the method used by the AOPS community, but since you need help proceeding further in the trigonometric part, I will be posting that answer :
Since x, y are ...
- 503
3
votes
How long does it take to get to "Olympiad-Bronze-level" of math problem solving ability from no competition experience?
It's more about the time and effort that you put in. If you can set aside 2-4 hours a day to focus on (pretty much anything), you can make a lot of progress in a year.
After that it's having the ...
- 66.8k
5
votes
How long does it take to get to "Olympiad-Bronze-level" of math problem solving ability from no competition experience?
Maybe, but it took me 3 full years to make it from the moderate problem solving ability (I had practiced occasional problem solving for 7 years as a schoolboy before) to the IMO level with rather ...
- 17.8k
2
votes
Accepted
$a_{m+n}+a_{m-n}=\frac{1}{2}(a_{2m}+a_{2n}), a_1=1$, find $a_{1995}$ (craft)
Your conjecture that $a_{2k}=2^{2k}$ is false, in general. What actually holds though is $a_{2^k}=2^{2k}$, for every positive integer $k$.
Here is a proof: You have proven that $a_{2m}=4a_m$. Using ...
- 621
1
vote
Find the area of the quadrilateral $ABDC$
The “shoelace algorithm”
Place your points into two columns repeating the first point:
$\begin{array}{}
2&1\\
8&1\\
4&3\\
6&6\\
2&1\end{array}$
Multiply diagonally adding the ...
- 10k
1
vote
Accepted
Find the area of the quadrilateral $ABDC$
It's a known "triangle": the point $C$ is located within the triangle $ABD$, so if you want to calculate the surface of the quadrilateral, you need to subtract $ADC$ from $ABD$ instead of ...
- 1,603
0
votes
Prove that $a_n\ge 1$ for all $n\ge 1$ with equality iff $n=1,2,4,5$
The question is
Let $\alpha, \beta,\gamma \in \mathbb{C}$ be the three roots of $x^3 + x+1$. For any $n\in\mathbb{N}$, let $a_n = \dfrac{(\alpha^n-1)(\beta^n-1)(\gamma^n-1)}{(\alpha-1)(\beta-1)(\...
- 34.3k
0
votes
Show that for all $n$ there exist some $n$-digit number with no $0$ in it whose digit sum divides it.
It's a classical induction problem to show that
Lemma. Given $m\ge 1$, there is an $m-$digit multiple of $2^m$ whose digits are all $1$ or $2$.
Let $2^{m-1}\le n<2^m$ and $x$ be the $(m+1)-$digit ...
- 2,187
2
votes
Accepted
how to solve $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{x^2}{\sin(x)} \ln\left(2^{\sin^3(x)}+ 5^{\cos^3(x)} \right)dx$?
I am a bit skeptical about a possible closed form for
$$I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\frac{x^2}{\sin(x)} \log\left(2^{\sin^3(x)}+ 5^{\cos^3(x)} \right)\,dx$$
It can write (I suppose that ...
- 255k
6
votes
Accepted
Closed form of $\prod_{n=0}^{\infty}\frac1{1+x^{2^n}}$
This is a telescoping product.
$$(1-x^{2^n})(1+x^{2^n}) = 1 - x^{2^{n+1}}.$$
Therefore,
$$(1-x) \prod_{n=0}^N (1+x^{2^n}) = 1 - x^{2^{N+1}}.$$
The rest of the details I leave to you as an exercise.
- 129k
1
vote
Bases of three $n$-dimensional subspaces of $\mathbb R^{2n}$ which have pairwise trivial intersection.
Pushing through OP's approach
OP starts with a (any) basis $\{ c_i \}$ of $C$.
OP shows that there is a unique $a_i \in A, b_i \in B$ such that $a_i + b_i = c_i$.
As pointed out in the comments, we ...
- 66.8k
2
votes
Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$
Elaborating on my hint, show that
$(a+b+c)^2 \geq 3(ab+bc+ca)$
$a+b+c \geq 3$
let $ t = a+b+c$, then $LHS \geq t + \frac{ 9}{t^2}$.
Finally, show that for $ t \geq 3$, $ t + \frac{9}{t^2 } \geq 4$.
...
- 66.8k
0
votes
Why does the following condition holds in the geometric optimisation problem?
$n_1+n_2>9$ would mean that the circles around $A_1,A_2$ contain at least 3 common points -- they cannot possibly host 10 distinct points because the total number of points is 7.
But if 3 distinct ...
- 1,102
1
vote
Accepted
BMO2 2011/12 question on cyclic quadrilateral and showing two circumcircles have same radius
According to the diagram below, you can follow these steps:
Step $1$: $\triangle EBC$ and $\triangle EAD$ are similar. So, $\angle QEC= \angle SED.$
Step $2$: Similarly, you should be able to show ...
- 4,423
3
votes
Accepted
Given that $a,b,c>0$ and $abc=1$, prove that $a+b+c+\frac{3}{ab+bc+ca} \geq 4$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$3u+\frac{w^3}{v^2}\geq 4w,$$ which is true by AM-GM:
$$3u+\frac{w^3}{v^2}\geq4\sqrt[4]{\frac{u^3w^3}{v^2}}\geq4\sqrt[4]{\...
- 197k
4
votes
Partition a stable (Middle School Math)
"My daughter had this question .... she is a tad confused on this."
It is not her fault : I too am confused , because that Question is almost meaningless.
(1) When the Stable has concrete ...
- 7,401
3
votes
Minimizing $\sum_{cyc}\frac{\sqrt{5a+8bc}}{8a+5bc}$ with $\sum_{cyc}ab=1$
Not an solution, just some thoughts
Let $f(a,b,c)$ be the expression we want to minimize. Suppose one of the variables is $0$, and the other two are inverses of each other. WLOG, assume $a=0, b=\frac{...
- 198
2
votes
Erdos asserts about Sylvester sequence
I just find this short note which is really helpful for this problem. The crucial trick here is to wisely use Muirhead inequality combining an induction.
For the proof, see https://arxiv.org/pdf/math/...
- 323
1
vote
How to find $X$ with these given values?
An alternate solution with minimal trigonometry: let $B = (0, 0)$; C = $(0, 4)$. Then $(BA)$ would be represented by an equation $x - \sqrt{3}y=0$. The set of points from which $[DC]$ is seen at $30^\...
- 2,434
2
votes
Accepted
Values of $x_0$ for which the sequence defined by $x_{n+1}=\frac{(n^2+1)x_n^2}{x_n^3+n^2}$ is bounded.
You're correct that the sequences are monotone decreasing (and I believe they also converge to $0$, but I haven't confirmed this) for all $x_0 = k$ where $k \gt 1$ is an integer. However, your ...
- 46.5k
1
vote
Math olympiad : combinatorics problem about inequality
Hints towards a solution. If you're stuck, show what you've tried and explain why you couldn't push through.
Suppose we set $a_i$ in increasing order, and require that $ a_{i+1} - a_i \geq 1 + 3 \...
- 66.8k
2
votes
How to find $X$ with these given values?
Guess and make special triangles.
$\because \triangle ABC \sim \triangle DAC$
$\therefore AC^2 = BC*DC = 2DC^2$
$AC = \sqrt{2}DC$
Draw $CH\perp AD$ and the intersection is H.
$CH=\frac{1}{2}AC = \...
- 21
8
votes
Accepted
How to find $X$ with these given values?
Applying the law of sines:
$ \dfrac{a}{\sin(30^\circ) } = \dfrac{b}{\sin(120^\circ - x)} $
and
$ \dfrac{a}{\sin(x)} = \dfrac{b}{\sin(30^\circ)} $
Dividing these two out,
$ \dfrac{\sin(x)}{\sin(30^\...
- 19k
0
votes
BMO1 2003/04 Question 2 - Geometry Prolem
I did:
set $\measuredangle{ADP} = \beta$
$\measuredangle{APD} = 90 - \beta$ (because angles in triangle sum to $180$)
$\measuredangle{QDC} = 90 - \beta$
$\measuredangle{QCD} = \beta$
$\measuredangle{...
- 169
2
votes
How to find $X$ with these given values?
Let:
$\overline{AD} = R.$
$\overline{BD} = S = \overline{CD}.$
$\angle BAD = \alpha.$
Applying the Law of Sines, you have that
$$\frac{1/2}{S} = \frac{\sin(x)}{R} ~~~~\text{and}
~~~~\frac{1/2}{R} =...
- 33.6k
8
votes
Accepted
Minimizing $\sum_{cyc}\frac{\sqrt{5a+8bc}}{8a+5bc}$ with $\sum_{cyc}ab=1$
Some thoughts.
Let
$$x := 5a + 8bc, \quad y := 5b + 8ca, \quad z := 5c + 8ab,$$
$$u := 8a + 5bc, \quad v := 8b + 5ca, \quad w := 8c + 5ab.$$
Let
$$A := \frac{2\sqrt{2} - \sqrt{5}}{3}, \quad B := \frac{...
- 36.5k
1
vote
Accepted
Geometry question on with two circles from BMO1 1994
Good luck for the BMO this year! This should be a matter of using the alternate segment theorem, the theorem that opposite angles in a cyclic quadrilateral are supplementary (twice), and that the ...
- 143
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