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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Easy geo
oVlad   3
N 2 minutes ago by Primeniyazidayi
Source: Romania EGMO TST 2019 Day 1 P1
A line through the vertex $A{}$ of the triangle $ABC{}$ which doesn't coincide with $AB{}$ or $AC{}$ intersectes the altitudes from $B{}$ and $C{}$ at $D{}$ and $E{}$ respectively. Let $F{}$ be the reflection of $D{}$ in $AB{}$ and $G{}$ be the reflection of $E{}$ in $AC{}.$ Prove that the circles $ABF{}$ and $ACG{}$ are tangent.
3 replies
oVlad
2 hours ago
Primeniyazidayi
2 minutes ago
abc(a+b+c)=3, show that prod(a+b)>=8 [Indian RMO 2012(b) Q4]
Potla   28
N 5 minutes ago by mihaig
Let $a,b,c$ be positive real numbers such that $abc(a+b+c)=3.$ Prove that we have
\[(a+b)(b+c)(c+a)\geq 8.\]
Also determine the case of equality.
28 replies
Potla
Dec 2, 2012
mihaig
5 minutes ago
NT with repeating decimal digits
oVlad   1
N 7 minutes ago by kokcio
Source: Romania EGMO TST 2019 Day 1 P2
Determine the digits $0\leqslant c\leqslant 9$ such that for any positive integer $k{}$ there exists a positive integer $n$ such that the last $k{}$ digits of $n^9$ are equal to $c{}.$
1 reply
oVlad
2 hours ago
kokcio
7 minutes ago
Inequalities make a comeback
MS_Kekas   2
N 8 minutes ago by ZeroHero
Source: Kyiv City MO 2025 Round 1, Problem 11.5
Determine the largest possible constant \( C \) such that for any positive real numbers \( x, y, z \), which are the sides of a triangle, the following inequality holds:
\[
\frac{xy}{x^2 + y^2 + xz} + \frac{yz}{y^2 + z^2 + yx} + \frac{zx}{z^2 + x^2 + zy} \geq C.
\]
Proposed by Vadym Solomka
2 replies
1 viewing
MS_Kekas
Jan 20, 2025
ZeroHero
8 minutes ago
Inequalities
sqing   4
N 2 hours ago by sqing
Let $x,y\ge 0$ such that $ 13(x^3+y^3) \leq 125(1+xy)$. Prove that
$$  k(x+y)-xy\leq  5(2k-5)$$Where $k\geq 5.6797. $
$$  6(x+y)-xy\leq 35$$
4 replies
sqing
Yesterday at 1:04 PM
sqing
2 hours ago
Inscribed Semi-Circle!!!
ehz2701   2
N 5 hours ago by mathafou
A right triangle $ABC$ with legs $AB = a$ and $BC = b$ is drawn with a semicircle inscribed into the triangle. What is the smallest possible radius of the semi-circle and the largest possible radius?

2 replies
ehz2701
Sep 11, 2022
mathafou
5 hours ago
geometry
carvaan   1
N 5 hours ago by vanstraelen
OABC is a trapezium with OC // AB and ∠AOB = 37°. Furthermore, A, B, C all lie on the circumference of a circle centred at O. The perpendicular bisector of OC meets AC at D. If ∠ABD = x°, find last 2 digit of 100x.
1 reply
carvaan
Yesterday at 5:48 PM
vanstraelen
5 hours ago
Inequalities
nhathhuyyp5c   1
N Today at 9:09 AM by Mathzeus1024
Let $a, b, c$ be non-negative real numbers such that $a^2 + b^2 + c^2 = 3$. Find the maximum and minimum values of the expression
\[
P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}.
\]
1 reply
nhathhuyyp5c
Yesterday at 6:35 AM
Mathzeus1024
Today at 9:09 AM
weird permutation problem
Sedro   2
N Today at 8:56 AM by alexheinis
Let $\sigma$ be a permutation of $1,2,3,4,5,6,7$ such that there are exactly $7$ ordered pairs of integers $(a,b)$ satisfying $1\le a < b \le 7$ and $\sigma(a) < \sigma(b)$. How many possible $\sigma$ exist?
2 replies
Sedro
Yesterday at 2:09 AM
alexheinis
Today at 8:56 AM
In a school of $800$ students, $224$ students play cricket, $240$ students play
Vulch   2
N Today at 8:12 AM by MathBot101101
Hello everyone,
In a school of $800$ students, $224$ students play cricket, $240$ students play hockey and $336$ students play basketball. $64$ students play both basketball and hockey, $80$ students play both cricket and basketball, $40$ students play both cricket and hockey, and $24$ students play all three: basketball, hockey, and cricket. Find the number of students who do not play any game.

Edit:
In the above problem,I just want to know that why the number of students who don't play any game shouldn't be 0, because,if we add 224,240 and 336 it comes out to be 800.I have solution,but I just want to know how to explain it without theoretically.Thank you!
2 replies
Vulch
Yesterday at 11:41 PM
MathBot101101
Today at 8:12 AM
Combinatorial proof
MathBot101101   8
N Today at 5:55 AM by MathBot101101
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}=1-\frac{1}{{n+1)!}
without induction and using only combinatorial arguments?

Induction proof wasn't quite as pleasing for me.
8 replies
MathBot101101
Yesterday at 7:37 AM
MathBot101101
Today at 5:55 AM
Inequalities
sqing   25
N Today at 3:58 AM by sqing
Let $   a,b    $ be reals such that $  a^2+ab+b^2 =3$ . Prove that
$$ \frac{4}{ 3}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{11}{4 }$$$$ \frac{13}{ 4}\geq \frac{1}{ a^2+5 }+ \frac{1}{ b^2+5 }+ab \geq -\frac{2}{3 }$$$$ \frac{3}{ 2}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq -\frac{17}{6 }$$$$ \frac{19}{ 6}\geq  \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2}$$Let $   a,b    $ be reals such that $  a^2-ab+b^2 =1 $ . Prove that
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
25 replies
sqing
Apr 16, 2025
sqing
Today at 3:58 AM
Three variables inequality
Headhunter   4
N Today at 3:18 AM by lbh_qys
$\forall a\in R$ ,$~\forall b\in R$ ,$~\forall c \in R$
Prove that at least one of $(a-b)^{2}$, $(b-c)^{2}$, $(c-a)^{2}$ is not greater than $\frac{a^{2}+b^{2}+c^{2}}{2}$.

I assume that all are greater than it, but can't go more.
4 replies
Headhunter
Yesterday at 6:58 AM
lbh_qys
Today at 3:18 AM
Indonesia Regional MO 2019 Part A
parmenides51   23
N Today at 2:08 AM by chinawgp
Indonesia Regional MO
Year 2019 Part A

Time: 90 minutes Rules


p1. In the bag there are $7$ red balls and $8$ white balls. Audi took two balls at once from inside the bag. The chance of taking two balls of the same color is ...


p2. Given a regular hexagon with a side length of $1$ unit. The area of the hexagon is ...


p3. It is known that $r, s$ and $1$ are the roots of the cubic equation $x^3 - 2x + c = 0$. The value of $(r-s)^2$ is ...


p4. The number of pairs of natural numbers $(m, n)$ so that $GCD(n,m) = 2$ and $LCM(m,n) = 1000$ is ...


p5. A data with four real numbers $2n-4$, $2n-6$, $n^2-8$, $3n^2-6$ has an average of $0$ and a median of $9/2$. The largest number of such data is ...


p6. Suppose $a, b, c, d$ are integers greater than $2019$ which are four consecutive quarters of an arithmetic row with $a <b <c <d$. If $a$ and $d$ are squares of two consecutive natural numbers, then the smallest value of $c-b$ is ...


p7. Given a triangle $ABC$, with $AB = 6$, $AC = 8$ and $BC = 10$. The points $D$ and $E$ lies on the line segment $BC$. with $BD = 2$ and $CE = 4$. The measure of the angle $\angle DAE$ is ...


p8. Sequqnce of real numbers $a_1,a_2,a_3,...$ meet $\frac{na_1+(n-1)a_2+...+2a_{n-1}+a_n}{n^2}=1$ for each natural number $n$. The value of $a_1a_2a_3...a_{2019}$ is ....


p9. The number of ways to select four numbers from $\{1,2,3, ..., 15\}$ provided that the difference of any two numbers at least $3$ is ...


p10. Pairs of natural numbers $(m , n)$ which satisfies $$m^2n+mn^2 +m^2+2mn = 2018m + 2019n + 2019$$are as many as ...


p11. Given a triangle $ABC$ with $\angle ABC =135^o$ and $BC> AB$. Point $D$ lies on the side $BC$ so that $AB=CD$. Suppose $F$ is a point on the side extension $AB$ so that $DF$ is perpendicular to $AB$. The point $E$ lies on the ray $DF$ such that $DE> DF$ and $\angle ACE = 45^o$. The large angle $\angle AEC$ is ...


p12. The set of $S$ consists of $n$ integers with the following properties: For every three different members of $S$ there are two of them whose sum is a member of $S$. The largest value of $n$ is ....


p13. The minimum value of $\frac{a^2+2b^2+\sqrt2}{\sqrt{ab}}$ with $a, b$ positive reals is ....


p14. The polynomial P satisfies the equation $P (x^2) = x^{2019} (x+ 1) P (x)$ with $P (1/2)= -1$ is ....


p15. Look at a chessboard measuring $19 \times 19$ square units. Two plots are said to be neighbors if they both have one side in common. Initially, there are a total of $k$ coins on the chessboard where each coin is only loaded exactly on one square and each square can contain coins or blanks. At each turn. You must select exactly one plot that holds the minimum number of coins in the number of neighbors of the plot and then you must give exactly one coin to each neighbor of the selected plot. The game ends if you are no longer able to select squares with the intended conditions. The smallest number of $k$ so that the game never ends for any initial square selection is ....
23 replies
parmenides51
Nov 11, 2021
chinawgp
Today at 2:08 AM
Concurrent lines
BR1F1SZ   4
N Apr 13, 2025 by NicoN9
Source: 2025 CJMO P2
Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$, where $BC\neq DA$. A circle passing through $C$ and $D$ intersects $AC, AD, BC, BD$ again at $W, X, Y, Z$ respectively. Prove that $WZ, XY, AB$ are concurrent.
4 replies
BR1F1SZ
Mar 7, 2025
NicoN9
Apr 13, 2025
Concurrent lines
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G H BBookmark kLocked kLocked NReply
Source: 2025 CJMO P2
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BR1F1SZ
555 posts
#1
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Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$, where $BC\neq DA$. A circle passing through $C$ and $D$ intersects $AC, AD, BC, BD$ again at $W, X, Y, Z$ respectively. Prove that $WZ, XY, AB$ are concurrent.
This post has been edited 1 time. Last edited by BR1F1SZ, Mar 7, 2025, 8:35 PM
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Burmf
27 posts
#2
Y by
Pascal on hexagon $XYCWZD$ gives that $XY \cap WZ \in AB$
(i probably missed something cus i didn't use the trapezoid condition
This post has been edited 1 time. Last edited by Burmf, Mar 7, 2025, 8:32 PM
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Maximilian113
549 posts
#3
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Lol, I showed $AWZB$ and $AXYB$ are cyclic, so radax
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khina
994 posts
#4
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My proposal, though I doubt it's truly original. My solution is the same as @above's.
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NicoN9
121 posts
#5
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Same as @above's:

It is suffice to show that $A, B, W, Z$, and $A, B, Y, X$ are concyclic, respectively. This is proved by\[
\measuredangle AWZ
=\measuredangle CWZ
=\measuredangle CDZ
=\measuredangle CDB
=\measuredangle ABD
\]and same for $A, B, Y, X$.
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