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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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Polar Coordinates
pingpongmerrily   4
N 8 minutes ago by K124659
Convert the equation $r=\tan(\theta)$ into rectangular form.
4 replies
1 viewing
pingpongmerrily
18 minutes ago
K124659
8 minutes ago
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Calculus, sets
wl8418   2
N an hour 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
Yesterday at 6:11 AM
wl8418
an hour ago
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Geometry
AlexCenteno2007   1
N an hour 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
1 viewing
AlexCenteno2007
2 hours ago
ohiorizzler1434
an hour 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 4 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
Yesterday at 3:23 PM
KSH31415
4 hours ago
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help me ..
exoticc   2
N Yesterday 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
Yesterday at 9:26 AM
sodiumaka
Yesterday at 5:00 PM
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Inequalities
sqing   20
N Yesterday 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
Yesterday at 3:32 PM
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21st PMO National Orals #9
yes45   0
Yesterday 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
Yesterday at 3:22 PM
0 replies
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Inequalities
sqing   0
Yesterday 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
Yesterday at 2:31 PM
0 replies
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Original Problem (qrxz17)
qrxz17   0
Yesterday 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
Yesterday at 1:32 PM
0 replies
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17th PMO Area Stage #5
yes45   0
Yesterday 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
Yesterday at 1:31 PM
0 replies
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Sipnayan 2019 SHS Orals Semifinal Round B Difficult \#3
qrxz17   0
Yesterday 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
Yesterday at 1:30 PM
0 replies
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MATHirang MATHibay 2011 Orals Survival Round Average \#1
qrxz17   0
Yesterday 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
Yesterday at 1:27 PM
0 replies
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Original Problem (qrxz17)
qrxz17   0
Yesterday 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
Yesterday at 1:12 PM
0 replies
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J
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BMT 2018 Algebra Round Problem 7
IsabeltheCat   5
N Apr 22, 2025 by P162008
Let $$h_n := \sum_{k=0}^n \binom{n}{k} \frac{2^{k+1}}{(k+1)}.$$Find $$\sum_{n=0}^\infty \frac{h_n}{n!}.$$
5 replies
IsabeltheCat
Dec 3, 2018
P162008
Apr 22, 2025
J
BMT 2018 Algebra Round Problem 7
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IsabeltheCat
4242 posts
IsabeltheCat
#1 h  Dec 3, 2018, 3:36 AM • 2 Y
Y by Adventure10, Mango247
Let $$h_n := \sum_{k=0}^n \binom{n}{k} \frac{2^{k+1}}{(k+1)}.$$Find $$\sum_{n=0}^\infty \frac{h_n}{n!}.$$
This post has been edited 1 time. Last edited by IsabeltheCat, Dec 3, 2018, 11:11 PM
Reason: wrong dummy variable was used previously
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rzlng
99 posts
rzlng
#2 h  Dec 3, 2018, 4:51 AM • 3 Y
Y by Srofller, Adventure10, Mango247
sketch
@below, thanks. fixed
This post has been edited 1 time. Last edited by rzlng, Dec 3, 2018, 5:11 AM
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TomCalc
1635 posts
TomCalc
#3 h  Dec 3, 2018, 5:05 AM • 1 Y
Y by Adventure10
IsabeltheCat wrote:
Find $$\sum_{k=0}^\infty \frac{h_n}{n!}.$$
The second sum has a wrong dummy variable and @above the answer is correct.
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NikoIsLife
9657 posts
NikoIsLife
#4 h  Dec 3, 2018, 10:59 AM • 3 Y
Y by Adventure10, Mango247, soryn
Solution
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IsabeltheCat
4242 posts
IsabeltheCat
#5 h  Dec 3, 2018, 11:11 PM • 2 Y
Y by Adventure10, Mango247
TomCalc wrote:
IsabeltheCat wrote:
Find $$\sum_{k=0}^\infty \frac{h_n}{n!}.$$
The second sum has a wrong dummy variable and @above the answer is correct.

Thanks for pointing that out. I fixed the error in the original post.
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P162008
226 posts
P162008
#6 h  Apr 22, 2025, 6:56 AM
Y by
Storage
This post has been edited 3 times. Last edited by P162008, Apr 28, 2025, 9:40 PM
Reason: Typo
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