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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
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Twin Prime Diophantine
awesomeming327.   23
N 12 minutes ago by HDavisWashu
Source: CMO 2025
Determine all positive integers $a$, $b$, $c$, $p$, where $p$ and $p+2$ are odd primes and
\[2^ap^b=(p+2)^c-1.\]
23 replies
awesomeming327.
Mar 7, 2025
HDavisWashu
12 minutes ago
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Troublesome median in a difficult inequality
JG666   2
N 20 minutes ago by navid
Source: 2022 Spring NSMO Day 2 Problem 3
Determine the minimum value of $\lambda\in\mathbb{R}$, such that for any positive integer $n$ and non-negative reals $x_1, x_2, \cdots, x_n$, the following inequality always holds:
$$\sum_{i=1}^n(m_i-a_i)^2\leqslant \lambda\cdot\sum_{i=1}^nx_i^2,$$Here $m_i$ and $a_i$ denote the median and arithmetic mean of $x_1, x_2, \cdots, x_i$, respectively.

Duanyang ZHANG, High School Affiliated to Renmin University of China
2 replies
JG666
May 22, 2022
navid
20 minutes ago
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Iran TST Starter
M11100111001Y1R   1
N 27 minutes ago by DeathIsAwe
Source: Iran TST 2025 Test 1 Problem 1
Let \( a_n \) be a sequence of positive real numbers such that for every \( n > 2025 \), we have:
\[ a_n = \max_{1 \leq i \leq 2025} a_{n-i} - \min_{1 \leq i \leq 2025} a_{n-i} \]Prove that there exists a natural number \( M \) such that for all \( n > M \), the following holds:
\[ a_n < \frac{1}{1404} \]
1 reply
M11100111001Y1R
Yesterday at 7:36 AM
DeathIsAwe
27 minutes ago
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Concurrent lines, angle bisectors
legogubbe   0
35 minutes ago
Source: ???
Hi AoPS!

Let $ABC$ be an isosceles triangle with $AB=AC$, and $M$ an arbitrary point on side $BC$. The internal angle bisector of $\angle MAB$ meets the circumcircle of $\triangle ABC$ again at $P \neq A$, and the internal angle bisector of $\angle CAM$ meets it again at $Q \neq A$. Show that lines $AM$, $BQ$ and $CP$ are concurrent.
0 replies
legogubbe
35 minutes ago
0 replies
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help me ..
exoticc   2
N Today at 5:00 PM by sodiumaka
Find all pairs of functions (f,g) : R->R that satisfy:
f(1)=2025;
g(f(x+y))+2x+y-1=f(x)+(2x+y)g(y) , ∀x,y∈R
2 replies
exoticc
Today at 9:26 AM
sodiumaka
Today at 5:00 PM
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Inequalities
sqing   20
N Today at 3:32 PM by ytChen
Let $ a,b,c\geq 0 ,a+b+c\leq 3. $ Prove that
$$a^2+b^2+c^2+ab +2ca+2bc + abc \leq \frac{251}{27}$$$$ a^2+b^2+c^2+ab+2ca+2bc + \frac{2}{5}abc \leq \frac{4861}{540}$$$$ a^2+b^2+c^2+ab+2ca+2bc + \frac{7}{20}abc \leq \frac{2381411}{26460}$$
20 replies
sqing
May 21, 2025
ytChen
Today at 3:32 PM
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21st PMO National Orals #9
yes45   0
Today at 3:22 PM
In square $ABCD$, $P$ and $Q$ are points on sides $CD$ and $BC$, respectively, such that $\angle{APQ} = 90^\circ$. If $AP = 4$ and $PQ = 3$, find the area of $ABCD$.

Answer Confirmation
Solution
0 replies
yes45
Today at 3:22 PM
0 replies
J
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Inequalities
sqing   0
Today at 2:31 PM
Let $ a,b> 0 $ and $2a+2b+ab=5. $ Prove that
$$ \frac{a^4}{b^4}+\frac{1}{a^4}+42ab-a^4\geq 43$$$$ \frac{a^5}{b^5}+\frac{1}{a^5}+64ab-a^5\geq 65$$$$ \frac{a^6}{b^6}+\frac{1}{a^6}+90ab-a^6\geq 91$$$$ \frac{a^7}{b^7}+\frac{1}{a^7}+121ab-a^7\geq 122$$
0 replies
sqing
Today at 2:31 PM
0 replies
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Original Problem (qrxz17)
qrxz17   0
Today at 1:32 PM
Problem. Suppose that
\begin{align*} (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 &= 138 \\ (a^2+b^2+c^2)^2 &=100. \end{align*}Find \(a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1)\).
Answer: Click to reveal hidden text
Solution. Subtracting the two given equations, we get
\begin{align*} a^4+b^4+c^4=38. \end{align*}Taking the square root of the second equation, we get

\begin{align*} a^2+b^2+c^2 = 10. \end{align*}
Then,
\begin{align*} a^4+b^4+c^4-(a^2+b^2+c^2) &= a^4-a^2+b^4-b^2+c^4-c^2 \\ &= a^2(a^2-1)+b^2(b^2-1)+c^2(c^2-1) = \boxed{28}. \end{align*}
0 replies
qrxz17
Today at 1:32 PM
0 replies
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17th PMO Area Stage #5
yes45   0
Today at 1:31 PM
Triangle \(ABC\) has a right angle at \(B\), with \(AB = 3\) and \(BC = 4\). If \(D\) and \(E\) are points
on \(AC\) and \(BC\), respectively, such that \(CD = DE = \frac{5}{3}\), find the perimeter of quadrilateral
\(ABED\).
Answer Confirmation
Solution
0 replies
yes45
Today at 1:31 PM
0 replies
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Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Today at 1:30 PM
Problem. Suppose that \(a\), \(b\), \(c\) are positive integers such that \((a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 63\). Find all possible values of \(a+b+c\).
Answer. Click to reveal hidden text
Solution. Expanding and cancelling "like" terms, we get

\begin{align*} (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) &= (a^4 +b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2) - 2(a^4+b^4+c^4) \\ &= 2a^2b^2+2a^2c^2+2b^2c^2 - a^4 -b^4-c^4 \end{align*}
We multiply the entire equation by \(-1\) to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.

\begin{align*} a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 = -63. \end{align*}
Simplifying the left-hand expression, we get

\begin{align*} a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\ &= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\ &= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\ &= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\ &=[a^2-(b-c)^2][a^2-(b+c)^2] \\ &= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63. \end{align*}
If \(a=b=c\), we get
\begin{align*} -3a^4 &=-63 \\ a^4 &= 21 \end{align*}

in which there are no real solutions for \(a\).

If we either have \(a=b\), \(a=c\), or \(b=c\), we get something of the form

\[ a^2(a+2b)(a-2b)=-63. \]
Since \(1\) and \(9\) are the only square factors of \(63\), we have \(a = \{1, 3\}\).

In this case, the solutions for \((a, b, c)\) are \((1, 4, 4)\) and \((3, 2, 2)\), including all their permutations.

If \(a, b, c\) are all distinct positive integers, there are no solutions.

Therefore, all possible values of \(a+b+c\) are \(\boxed{7 \text{ and }9}\).
0 replies
qrxz17
Today at 1:30 PM
0 replies
J
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MATHirang MATHibay 2011 Orals Survival Round Average \#1
qrxz17   0
Today at 1:27 PM
Problem. Solve for all possible values of y: $\sqrt{y^2 + 4y + 8} + \sqrt{y^2+4y+4}=\sqrt{2(y^2+4y+6)}$
Answer. Click to reveal hidden text
Solution. Let \( N = y^2+4y+6\). We have
\begin{align*} \sqrt{N+2} + \sqrt{N-2} = \sqrt{2N} \end{align*}
Taking the square of the equation, we get

\begin{align*} (N+2)+(N-2)+2\sqrt{N^2-4}&=2N \\ \sqrt{N^2-4} &=0 \end{align*}
Taking the square of the equation again, we get

\begin{align*} N^2-4&=0 \\ N^2&=4 \end{align*}
If N is -2, then \(\sqrt{2N}\) is an imaginary number. Thus, N must be equal to +2. We then have

\begin{align*} y^2 +4y+6&=2 \\ y^2+4y+4&=0 \\ (y+2)^2&=0 \\ y+2&=0 \\ y &= \boxed{-2}. \end{align*}
0 replies
qrxz17
Today at 1:27 PM
0 replies
J
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Original Problem (qrxz17)
qrxz17   0
Today at 1:12 PM
Problem. Given that \( 2^{\log_{16} 71} = a \) and \( x^{\log_{256} 71} = 71a \), find the value of \(x\).
Answer. Click to reveal hidden text
Solution. Let \(k=\log_{16}71\). We have
\begin{align*} 2^k &= a \\ 16^k &= 71 \end{align*}
So,
\begin{align*} 2^k\cdot16^k&=a\cdot71 \\ 32^k &= 71a = x^{\log_{256} 71} \end{align*}
Substituting the value of k, we get

\begin{align*} 32^{\log_{16}71} &= x^{\log_{256} 71} \end{align*}
By using the change-of-base formula, we get

\begin{align*} 32^{\frac{\log_{256}71}{\log_{256}16}}=32^{\frac{\log_{256}71}{\frac{1}{2}}} = 32^{2 \cdot \log_{256}71} = 1024^{\log_{256}71} \end{align*}
Therefore, \(x=\boxed{1024}\).
0 replies
qrxz17
Today at 1:12 PM
0 replies
J
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Sipnayan 2016 SHS Semifinals Average \#4
qrxz17   0
Today at 1:11 PM
Problem. Find the maximum value of \( 8 \cdot 27^{\log_6{x}}+27\cdot 8^{\log_6x} - x^3\) as x varies over the positive real numbers.
Answer. Click to reveal hidden text
Solution. Simplifying the equation, we get
\begin{align*} 8 \cdot 27^{\log_6{x}} + 27 \cdot 8^{\log_6{x}} - x^3 \\ = 2^3 \cdot 3^{3\log_6{x}} + 3^3\cdot 2^{3\log_6{x}} - x^3 \\ = 2^3 \cdot 3^{\log_6{x^3}} + 3^3\cdot 2^{\log_6{x^3}} - x^3 \end{align*}
Let \(y=log_6{x}\). We have
\begin{align*} y &= \log_6{x^3} \\ 6^y &= x^3 \end{align*}
We can rewrite the equation into
\begin{align*} 2^3 \cdot 3^{y} &+ 3^3\cdot 2^{y} - 6^y \\ =2^3\cdot 3^y &+ 3^3 \cdot 2^y - 2^y \cdot 3^y \end{align*}
Then, by grouping terms, we get
\begin{align*} 3^y(2^3 - 2^y)&+ 3^2\cdot 2^y \text{ or} \\ 2^y(3^2-3^y)&+ 2^3\cdot 3^y \end{align*}
Either way works. For the first equation, we can get the maximum at y=3. For the second equation, we can get the maximum at y=2. This happens because each expression reaches a turning point (where the function changes from increasing to decreasing) when the terms in the parentheses equal zero. Both expressions produce the maximum value of \(\boxed{216}\).
0 replies
qrxz17
Today at 1:11 PM
0 replies
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A locus contained in an equilateral triangle
CatalinBordea   1
N Oct 1, 2018 by vanstraelen
Source: Romanian TST 1978, Day 3, P5
Find locus of points $ M $ inside an equilateral triangle $ ABC $ such that
$$ \angle MBC+\angle MCA +\angle MAB={\pi}/{2}. $$
1 reply
CatalinBordea
Sep 30, 2018
vanstraelen
Oct 1, 2018
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A locus contained in an equilateral triangle
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Reveal topic tags
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Source: Romanian TST 1978, Day 3, P5
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CatalinBordea
2142 posts
CatalinBordea
#1 h  Sep 30, 2018, 2:53 PM • 2 Y
Y by Adventure10, Mango247
Find locus of points $ M $ inside an equilateral triangle $ ABC $ such that
$$ \angle MBC+\angle MCA +\angle MAB={\pi}/{2}. $$
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vanstraelen
9062 posts
vanstraelen
#2 h  Oct 1, 2018, 7:07 PM • 2 Y
Y by Adventure10, Mango247
Locus of the points $M$ are the line segments $CD\ ,\ CD \bot AB\ ,\ D \in AB$, ...
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