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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
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Calculus, sets
wl8418   2
N 27 minutes ago by wl8418
Is empty set a proper set of an non empty set? Why or why not? Any clarification or insight is appreciated. Thanks in advance!
2 replies
wl8418
Today at 6:11 AM
wl8418
27 minutes ago
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Geometry
AlexCenteno2007   1
N 31 minutes ago by ohiorizzler1434
Given triangle ABC, it is true that BD = CF where D and F are points in the same half-plane with respect to line BC and it is also known that BD is parallel to AC and CF is parallel to AB. Show that BF, CD and the interior bisector of A are concurrent.
1 reply
AlexCenteno2007
2 hours ago
ohiorizzler1434
31 minutes ago
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How can i prove this?
marxs01   0
an hour ago
If $\frac{\cos{x}}{\cos{y}} + \frac{\sin{x}}{\sin{y}} = -1$ prove that $\frac{\cos^3{y}}{\cos{x}} + \frac{\sin^3{y}}{\sin{x}} = 1$
0 replies
marxs01
an hour ago
0 replies
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How do I prove this? - Sets and symmetric difference
smadadi1000   1
N 3 hours ago by KSH31415
For sets A, B and C, prove that (A Δ B) Δ C = (A Δ C) Δ (A \ B).

The textbook - Proofs by Jay Cummings - had this definition for symmetric difference:
A Δ B = ( A U B)/(A ∩ B)

this is exercise 3.37 (e).
1 reply
smadadi1000
Today at 3:23 PM
KSH31415
3 hours ago
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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
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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
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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
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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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Σ to ∞
phiReKaLk6781   3
N Apr 22, 2025 by Maxklark
Evaluate: $ \sum\limits_{k=1}^\infty \frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}}$
3 replies
phiReKaLk6781
Mar 20, 2010
Maxklark
Apr 22, 2025
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Σ to ∞
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phiReKaLk6781
2074 posts
phiReKaLk6781
#1 h  Mar 20, 2010, 7:29 PM • 4 Y
Y by Adventure10, Mango247, At777, and 1 other user
Evaluate: $ \sum\limits_{k=1}^\infty \frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}}$
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BanziC
37 posts
BanziC
#2 h  Mar 21, 2010, 11:34 AM • 4 Y
Y by AlastorMoody, Adventure10, Mango247, and 1 other user
$ \sum\limits_{k=1}^\infty\frac{1}{k\sqrt{k+2}+(k+2)\sqrt{k}} =\\ \sum\limits_{k=1}^\infty\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{(k\sqrt{k+2}+(k+2)\sqrt{k})((k+2)\sqrt{k}-k\sqrt{k+2})} = \\ \sum\limits_{k=1}^\infty\frac{(k+2)\sqrt{k}-k\sqrt{k+2}}{2k(k+2)} =\\ \frac{1}{2}\sum\limits_{k=1}^\infty\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+2}} =\\ \frac{1}{2}(1 + \frac{1}{\sqrt{2}})$
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P162008
226 posts
P162008
#3 h  Apr 22, 2025, 12:18 AM
Y by
Let $T_k$ denote the general term of the given sum $S$.

Then, $$T_k = \frac{1}{k\sqrt{k + 2} + (k + 2)\sqrt{k}} = \frac{1}{\sqrt{k(k + 2)} (\sqrt{k} + \sqrt{k + 2})} = \frac{\sqrt{k + 2} - \sqrt{k}}{2\sqrt{k(k + 2)}} = \frac{1}{2}\left(\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k + 2}}\right)$$
Now, $$S = \sum_{k=1}^\infty T_k = \frac{1}{2}\left(1 + \frac{1}{\sqrt{2}}\right)$$
Therefore, $$S = \boxed{\frac{\sqrt2 + 1}{2\sqrt2}}$$
This post has been edited 11 times. Last edited by P162008, Apr 22, 2025, 6:46 PM
Reason: Typo
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Maxklark
6 posts
Maxklark
#5 h  Apr 22, 2025, 6:12 PM
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