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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
J
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Cool one
MTA_2024   7
N 43 minutes ago by invisibleman
Prove that for all real numbers $a$ and $b$ verifying $a>b>0$ . $$(n+1) \cdot b^n \leq \frac{a^{n+1}-b^{n+1}}{a-b} \leq (n+1) \cdot a^n $$
7 replies
MTA_2024
Mar 15, 2025
invisibleman
43 minutes ago
J
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a/b + b/a never integer ?
MTA_2024   4
N an hour ago by invisibleman
Let $a$ and $b$ be 2 distinct positive integers.
Can $\frac a b +\frac b a $ be in an integer. Prove why ?
4 replies
MTA_2024
Mar 16, 2025
invisibleman
an hour ago
J
H
Does there exist a function $f:\mathbb N^* \rightarrow \mathbb Z$ other than a p
kyotaro   3
N 5 hours ago by quantam13
Does there exist a function $f:\mathbb N^* \rightarrow \mathbb Z$ other than a polynomial satisfying
$$f(a)-f(b) \mid a-b$$
3 replies
kyotaro
Yesterday at 10:51 AM
quantam13
5 hours ago
J
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solve the system of equations
Havu   3
N Today at 2:27 AM by lbh_qys
Solve the system of equations:
\[\begin{cases} 3x^2-2xy+3y^2+\dfrac{2}{x^2-2xy+y^2}=8\\ 2x+\dfrac{1}{x-y}=4 \end{cases}\]
3 replies
Havu
Yesterday at 7:52 AM
lbh_qys
Today at 2:27 AM
J
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Inequalities
sqing   2
N Today at 2:19 AM by sqing
Let $ a,b,c\geq 2 . $ Prove that
$$(a-1)(b^2-2)(c-1) - \frac{1}{2}abc\geq -2$$$$(a^2-2)(b-1)(c^2-2) - \frac{3}{2}abc\geq -6$$
2 replies
sqing
Yesterday at 1:42 PM
sqing
Today at 2:19 AM
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2019 Chile Classification / Qualifying NMO Juniors XXXI
parmenides51   5
N Today at 1:14 AM by liyufish
p1. Consider the sequence of positive integers $2, 3, 5, 6, 7, 8, 10, 11 ...$. which are not perfect squares. Calculate the $2019$-th term of the sequence.


p2. In a triangle $ABC$, let $D$ be the midpoint of side $BC$ and $E$ be the midpoint of segment $AD$. Lines $AC$ and $BE$ intersect at $F$. Show that $3AF = AC$.


p3. Find all positive integers $n$ such that $n! + 2019$ is a square perfect.


p4. In a party, there is a certain group of people, none of whom has more than $3$ friends in this. However, if two people are not friends at least they have a friend in this party. What is the largest possible number of people in the party?
5 replies
parmenides51
Oct 11, 2021
liyufish
Today at 1:14 AM
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2^n = p+3^p
reeh_haan   8
N Today at 12:20 AM by MajesticCheese
Find all pairs $(p, n)$ of positive integers which satisfy the equation $$2^n = p+3^p$$
8 replies
reeh_haan
Dec 28, 2021
MajesticCheese
Today at 12:20 AM
J
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2017 Mock ARML Team Round #7 Revenge of the incenters
parmenides51   1
N Yesterday at 10:16 PM by Giant_PT
Let $ABC$ be a triangle with side lengths $AB = 20$, $BC = 17$, $AC = 13$, incenter $I$ and circumcircle $\Gamma$, and let $\omega_A$ be the circle that is tangent to $AB$, $AC$, $\Gamma$ at $D$ ,$E$, $F$, respectively. If $I_F$ ,$ I_I$, $I_C$ denote the incenters of $CFI$, $FIE$, $ECF$, respectively, find the measure, in degrees, of the largest angle in $\vartriangle I_F I_II_C$.
1 reply
parmenides51
Jan 12, 2024
Giant_PT
Yesterday at 10:16 PM
J
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Don Sirloin Bowel's Algorithm
BadAtCompetitionMath21420   1
N Yesterday at 10:07 PM by ohiorizzler1434
Can you sum
\begin{align*} \sum_{k=1}^n F^m_k? \end{align*}
I coined a strategy Don Sirloin Bowel's algorithm because I found it and didn't see it anywhere else. Please tell me if this looks familiar because it can have absolutely scrumptious applications. (I hope that this stuff can help you sum that.) Starting with the "golden quadratic" or whatever the sigma it's called $x^2=x+1$. Multiplying by $x$ on both sides and resubstituting $x^2$, we find $x^3=x^2+x=(x+1)+x=2x+1$. Continuing this process, we find $x^4=3x+2$, $x^5=5x+3$, and so forth. We claim that $x^n=F_nx+F_{n-1}.$ We already have our base case. $x^n=F_nx+F_{n-1}$. Multiplying by $x$, $x^{n+1}=F_nx^2+F_{n-1}x=F_n(x+1)+F_{n-1}x=F_{n+1}x+F_{n}$. Hence, we have proved our claim by induction. Now, raising both sides to the $m$th power,
\begin{align*} x^{mn}&=\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^kx^m\\ &=\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^k(F_mx+F_{m-1})\\ &=x\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^kF_m+\sum_{k=0}^mF_{n}^{m-k}F_{n-1}^kF_{m-1}\\ &=F_{mn}x+F_{mn-1} \end{align*}So,
\begin{align*} F_{mn}=\sum_{k=0}^m \binom{m}{k}F_{n}^{m-k}F_{n-1}^kF_m \end{align*}and
\begin{align*} F_{mn-1}=\sum_{k=0}^mF_{n}^{m-k}F_{n-1}^kF_{m-1} \end{align*}Are these identities useful? And can you use them to compute the sum?
1 reply
BadAtCompetitionMath21420
Yesterday at 9:50 PM
ohiorizzler1434
Yesterday at 10:07 PM
J
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Chessboard
Ecrin_eren   5
N Yesterday at 8:44 PM by TrendCrusher
On an 8×8 checkerboard, what is the minimum number of squares that must be marked (including the marked ones) so that every square has exactly one marked neighbor? (We define neighbors as squares that share a common edge, and a square is not considered a neighbor of itself.)

5 replies
Ecrin_eren
Tuesday at 8:55 PM
TrendCrusher
Yesterday at 8:44 PM
J
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Why does the combined equation have two negative solutions?
Luking   1
N Yesterday at 7:30 PM by vanstraelen
It is known that the moving point $G(x,y)$ is on the curve $C_1: y^2 - x^2 = 1$. There is a parabola $C_2: x^2 = 4y$ with focus $F$. Two tangent lines to$C_2$ are drawn through a point $P$ on $C1$, and the tangent points are $A$ and $B$ respectively. The line $l$ parallel to the line $AB$ is tangent to $C_2$ at point $Q$. Question: When the line $l$ and $C_1$ have two intersection points, find the range of $|QF|$.\
This may make some of the information in the question useless, because I deleted the first two questions of this big question in order to avoid making the question too long and get straight to the point.\
According to the calculation, I get the analytical expression of the line $l$ as $y=\frac{x_0}{2} x - \frac{x_0^2}{4}$.\
At first, I thought it only needed to be not parallel to the parabola asymptotes.\
That is, the slope of the straight line $k \neq \frac{a}{b}$ , then $\frac{x_0}{2} \neq \frac{a}{b}$ , so $x_0 \neq \pm \frac{2a}{b} = \pm2$ ,and $x_0^2 \neq 4$.\
$|QF| = y_0 +1 \neq 5$.\
But when I checked the answer, it was wrong.\
It combines the straight line $l$ with the curve $C_1$ to get an equation and then uses Vieta's theorem.
\begin{cases}
y=\frac{x_0}{2} x - \frac{x_0^2}{4}
x^2=4y
\end{cases}
$$( 4 x_{0}^{2}-1 6 ) y^{2}-8 x_{0}^{2} y-x_{0}^{4}-4 x_{0}^{2}=0 $$The answer is that according to the question, the equation has two negative roots. I can't understand this, and this is exactly where my problem lies.\
Then it gets the following system of equations:
\begin{cases}
4x_0-16 \neq 0
\Delta > 0
x_1+x_2 <0
x_1\cdot x_2>0
\end{cases}
Solve, $2\sqrt{5}-2<x_0<4$.
So we get $|QF|=\frac {x_0^2}{4} + 1 \in (\frac {\sqrt{5}+1}{2},2)$.
I hope you can help me figure out why both roots of that equation are negative.
IMAGE
IMAGE
1 reply
Luking
Yesterday at 4:02 PM
vanstraelen
Yesterday at 7:30 PM
J
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Rolling a die
Ecrin_eren   2
N Yesterday at 6:16 PM by tofubear
John has a standard die with values from 1 to 6. He plays a game where, after each roll, the value on the top face permanently increases by 1. After rolling the die three times, the probability that at least one face has a value of at least 7 is given as a/b , where a and b are coprime. What is the value of a/b ?
2 replies
Ecrin_eren
Yesterday at 5:05 PM
tofubear
Yesterday at 6:16 PM
J
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Travel group
Ecrin_eren   0
Yesterday at 5:12 PM
A group of 5 students will form a travel group of 3 students. Each student is asked to select two other students they would like to travel with. Given that the probability of forming a 3-person group where each member has selected the other two is a/b , where a and b are coprime positive integers, what is the value of a/b ?

0 replies
Ecrin_eren
Yesterday at 5:12 PM
0 replies
J
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Area problem
MTA_2024   1
N Yesterday at 4:46 PM by vanstraelen
Let $\omega$ be a circle inscribed inside a rhombus $ABCD$. Let $P$ and $Q$ be variable points on $AB$ and $AD$ respectively, such as $PQ$ is always the tangent line to $\omega$.
Prove that for any position of $P$ and $Q$ the area of triangle $\triangle CPQ$ is the same.
1 reply
MTA_2024
Mar 16, 2025
vanstraelen
Yesterday at 4:46 PM
J
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J
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Complex + Radical Evaluation
Saucepan_man02   1
N Mar 17, 2025 by MathRook7817
Evaluate: (without calculators)
$$ (\sqrt{6 - 2 \sqrt{5}} + i \sqrt{2 \sqrt{5} + 10})^5 + (\sqrt{6 - 2 \sqrt{5}} - i \sqrt{2 \sqrt{5} + 10})^5$$
1 reply
Saucepan_man02
Mar 17, 2025
MathRook7817
Mar 17, 2025
J
Complex + Radical Evaluation
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complex numbers
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Saucepan_man02
1299 posts
Saucepan_man02
#1 h  Mar 17, 2025, 1:08 PM
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Evaluate: (without calculators)
$$ (\sqrt{6 - 2 \sqrt{5}} + i \sqrt{2 \sqrt{5} + 10})^5 + (\sqrt{6 - 2 \sqrt{5}} - i \sqrt{2 \sqrt{5} + 10})^5$$
This post has been edited 1 time. Last edited by Saucepan_man02, Mar 17, 2025, 1:08 PM
Reason: EDIT
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MathRook7817
612 posts
MathRook7817
#2 h  Mar 17, 2025, 1:30 PM
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I think you just find the magnitude of each one, then evaluate the angle, and then double it by 5 then evaluate the trig of it to find it

too lazy to evaluate it
This post has been edited 1 time. Last edited by MathRook7817, Mar 17, 2025, 1:31 PM
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