Basic, Part B

Let $a,b>0, a^4+ab+b^4=60$ . Prove that
$2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}-1)}{4}$ $\frac{\sqrt{481}-1}{4}\leq a^2-ab+b^2 \leq 2\sqrt{15}$ Let $a,b>0, a^4-ab+b^4=60$ . Prove that
$2\sqrt{15} \leq a^2+ab+b^2 \leq \frac{3(\sqrt{481}+1)}{4}$ $5<a^2-ab+b^2 \leq 2\sqrt{15}$

1 reply

sqing

an hour ago

sqing

28 minutes ago

Geometry

Lukariman 4

N an hour ago by lbh_qys

Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.

4 replies

Lukariman

Yesterday at 12:43 PM

lbh_qys

an hour ago

Need help with barycentric

Sadigly 0

an hour ago

Hi,is there a good handout/book that explains barycentric,other than EGMO?

0 replies

Sadigly

an hour ago

0 replies

Combinatorics

P162008 3

N 2 hours ago by P162008

Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$

3 replies

P162008

5 hours ago

P162008

2 hours ago

Find min and max

lgx57 0

2 hours ago

Source: Own

$x_1,x_2, \cdots ,x_n\ge 0$ , $\displaystyle\sum_{i=1}^n x_i=m$ . $k_1,k_2,\cdots,k_n >0$ . Find min and max of
$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$

0 replies

lgx57

2 hours ago

0 replies

Find min

lgx57 0

2 hours ago

Source: Own

$a,b>0$ , $a^4+ab+b^4=60$ . Find min of
$4a^2-ab+4b^2$
$a,b>0$ , $a^4-ab+b^4=60$ . Find min of
$4a^2-ab+4b^2$

0 replies

lgx57

2 hours ago

0 replies

III Lusophon Mathematical Olympiad 2013 - Problem 5

DavidAndrade 2

N 3 hours ago by KTYC

Find all the numbers of $5$ non-zero digits such that deleting consecutively the digit of the left, in each step, we obtain a divisor of the previous number.

2 replies

DavidAndrade

Aug 12, 2013

KTYC

3 hours ago

Maximum number of terms in the sequence

orl 11

N 4 hours ago by navier3072

Source: IMO LongList, Vietnam 1, IMO 1977, Day 1, Problem 2

In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.

11 replies

orl

Nov 12, 2005

navier3072

4 hours ago

USAMO 2003 Problem 1

MithsApprentice 68

N 4 hours ago by Mamadi

Prove that for every positive integer $n$ there exists an $n$ -digit number divisible by $5^n$ all of whose digits are odd.

68 replies

MithsApprentice

Sep 27, 2005

Mamadi

4 hours ago

Centroid, altitudes and medians, and concyclic points

BR1F1SZ 3

N 4 hours ago by EeEeRUT

Source: Austria National MO Part 1 Problem 2

Let $\triangle{ABC}$ be an acute triangle with $BC > AC$ . Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$ . The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$ . Let $M$ be the point where $CS$ intersects $AB$ . The line $SF$ intersects the circle $\gamma$ at a point $Q$ , such that $F$ lies between $S$ and $Q$ . Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)

3 replies

BR1F1SZ

Monday at 9:45 PM

EeEeRUT

4 hours ago

Basic, Part B

CarSa 1

N Apr 26, 2025 by Math-lover1

For each four--digit number $\overline {abcd}$ , that is, with $a$ nonzero, let $P(\overline {abcd})$ be the product $(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)$ .
For example, $P(2022) = (2+0)(2+2)(2+2)(0+2)(0+2)(2+2) = 512$ and $P(1234) = (1+2)(1+3)(1+4)(2+3)(2+4)(3+4)$ .
How many numbers $\overline {abcd}$ with at least one $0$ amoung their digits satisfy that $P(\overline {abcd})$ is a power of 2?

1 reply

CarSa

Apr 24, 2025

Math-lover1

Apr 26, 2025

Basic, Part B

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CarSa

3 posts

CarSa

#1 h Apr 24, 2025, 8:45 PM • 1 Y

Y by PikaPika999

This post has been edited 2 times. Last edited by CarSa, Apr 24, 2025, 8:48 PM
Reason: yes

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Math-lover1

199 posts

Math-lover1

#2 h Apr 26, 2025, 10:41 PM • 2 Y

Y by PikaPika999, CarSa

My solution

First, we observe that $a, b, c, d$ must be either powers of $2$ or $0$ .

Notice that the cases $(2, 2, 2, 0), (2, 2, 0, 0), (2, 0, 0, 0), (4, 4, 4, 0).... (8, 0, 0, 0)$ work, where each element is either a fixed $a = 2^m$ or $0$ . By counting the cases and keeping in mind that $a \neq 0$ , we get $(3+3+1)\cdot3 = \boxed{21}$

Hopefully I'm not too late

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