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Combinatorial proof
MathBot101101   0
4 hours ago
Is there a way to prove
\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}
without induction and using only combinatorial arguments?

Induction proof wasn't quite as pleasing for me.
0 replies
MathBot101101
4 hours ago
0 replies
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What are vectors?
SomeonecoolLovesMaths   3
N 4 hours ago by programjames1
How does one define vectors with mathematical rigority? Most of the time it is stated as an object with a magnitude and a direction, but it in itself is very vague in my opinion. Like how do you define direction and magnitude itself?
3 replies
SomeonecoolLovesMaths
Today at 4:19 AM
programjames1
4 hours ago
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Inequalities
nhathhuyyp5c   0
5 hours ago
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
\[ P = \frac{a}{a^2 + 2} + \frac{b}{b^2 + 2} + \frac{c}{c^2 + 2}. \]
0 replies
nhathhuyyp5c
5 hours ago
0 replies
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Recursion
Sid-darth-vater   1
N Today at 5:14 AM by aidan0626
Help, I can't characterize ts and I dunno what to do
1 reply
Sid-darth-vater
Today at 3:02 AM
aidan0626
Today at 5:14 AM
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Inequalities
sqing   18
N Today at 2:38 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
$$ \frac{3}{ 2}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }+ab \geq \frac{4}{15 }$$$$ \frac{14}{ 15}\geq \frac{1}{ a^2+3 }+ \frac{1}{ b^2+3 }-ab \geq -\frac{1}{2 }$$$$ \frac{3}{ 2}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }+ab \geq \frac{13}{42 }$$$$ \frac{41}{ 42}\geq \frac{1}{ a^4+3 }+ \frac{1}{ b^4+3 }-ab \geq -\frac{1}{2 }$$
18 replies
sqing
Apr 16, 2025
sqing
Today at 2:38 AM
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weird permutation problem
Sedro   0
Today at 2:09 AM
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?
0 replies
Sedro
Today at 2:09 AM
0 replies
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Calculus BC help
needcalculusasap45   7
N Today at 1:42 AM by ehz2701
So basically, I have the AP Calculus BC exam in less than a month, and I have only covered until Unit 6 or 7 of the cirriculum. I am self studying this course (no teacher) and have not had much time to study bc of 6 other APs. I need to finish 8, 9, and 10 in less than 2 weeks. What can I do ? I would appreciate any help or resources anyone could provide. Could I just learn everything from barrons and princeton? Also, I have not taken AP Calculus AB before.

7 replies
needcalculusasap45
Yesterday at 1:51 PM
ehz2701
Today at 1:42 AM
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Some problems
hashbrown2009   2
N Today at 12:12 AM by UberPiggy
1. Real numbers a,b,c are satisfy a+1/b = b+1/c = c+1/a =x. If a,b,c are distinct, what is the value of x?
2. If x^2+y^2=1, then what is the value of : root(x^2-2x+1) + root(xy-2x+y-2) ?
3. Find the value of the sequence 2^2 + (3^2+1) + (4^2+2) + … + (97^2+95) + (98^2+96).
4. If x^2+x-1=0, then evaluate (1-x^2-x^3-x^4-…-x^2022-x^2023)/x^2022 .
5. If triangle XYZ has 3 sides that are all whole numbers, and the perimeter of XYZ is 24, what is the probability XYZ is a right triangle?

Note: If someone can latex-ify this it would help.
2 replies
hashbrown2009
Yesterday at 11:01 PM
UberPiggy
Today at 12:12 AM
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ez problem....
Cobedangiu   4
N Yesterday at 11:22 PM by iniffur
Let $x,y \in Z$ and $xy \cancel \vdots7$
Find $n \in Z^+$.
$x^2+y^2+xy=7^n$
4 replies
Cobedangiu
Apr 18, 2025
iniffur
Yesterday at 11:22 PM
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Twin Primes Digital Root
FerMath10   1
N Yesterday at 8:57 PM by Sedro
Hi,

I noticed something interesting while playing around with twin primes (pairs of primes that differ by 2). Here is what I noticed:
Conjecture: The product of twin primes—excluding the pair (3, 5)—always has a digital root of 8.

Just to clarify, the digital root of a number is the single-digit value you get by repeatedly summing its digits until only one digit remains. For example, the digital root of 77 is 7 + 7 = 14, and then 1 + 4 = 5.

I tested this on several examples, and it seems to hold, but I’m not sure if it’s a well-known result or something that breaks down for larger primes.

Is this an obvious consequence of some known number theory property? Would love to hear your thoughts!
1 reply
FerMath10
Yesterday at 8:51 PM
Sedro
Yesterday at 8:57 PM
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Geometry problem
kjhgyuio   1
N Apr 4, 2025 by Mathzeus1024
Source: smo
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
1 reply
kjhgyuio
Apr 1, 2025
Mathzeus1024
Apr 4, 2025
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Geometry problem
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kjhgyuio
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kjhgyuio
#1 h  Apr 1, 2025, 1:12 PM
Y by
In trapezium ABCD,AD is parallel to BC and points E and F are midpoints of AB and DC respectively. If
Area of AEFD/Area of EBCF =√3 + 1/3-√3 and the area of triangle ABD is √3 .find the area of trapezium ABCD
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Mathzeus1024
819 posts
Mathzeus1024
#2 h  Apr 4, 2025, 9:06 AM
Y by
In the first quadrant of the $xy-$plane, let:

$A(0,0); B(h\cot(A),h); C(k-h\cot(D),h); D(k,0); E\left(\frac{h\cot(A)}{2},\frac{h}{2}\right); F\left(k - \frac{h\cot(D)}{2}, \frac{h}{2}\right)$

define trapezoid $ABCD$ that is divided into trapezoids $AEFD$ and $EBCF$ by segment $EF$ (for $AD \parallel EF \parallel BC$).

We have $[ABD] = \frac{kh}{2}=\sqrt{3}$. If $[AEFD] = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right)[EBCF]$, then we obtain:

$\frac{h/2}{2}\left[k + \left(k-\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2}\right)\right] = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot \frac{h/2}{2}\left[\left(k-\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2}\right) + (k-h\cot(A)-h\cot(D))\right]$;

or $2k + -\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2} = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot \left[2k-\frac{3h\cot(A)}{2}-\frac{3h\cot(D)}{2}\right]$;

or $\frac{4\sqrt{3}}{h} -\frac{h\cot(A)}{2}-\frac{h\cot(D)}{2} = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot \left[\frac{4\sqrt{3}}{h}-\frac{3h\cot(A)}{2}-\frac{3h\cot(D)}{2}\right]$;

or $8\sqrt{3} - h^{2}[\cot(A)+\cot(D)] = \left(\sqrt{3}+\frac{1}{3-\sqrt{3}}\right) \cdot [8\sqrt{3} - 3h^{2}[\cot(A)+\cot(D)]$;

or $[\cot(A)+\cot(D]h^2 = \frac{56-8\sqrt{3}}{1+7\sqrt{3}}$ (i).

Now, $[ABCD] = \frac{h}{2}[2k-h\cot(A)-h\cot(D)] = hk-\frac{[\cot(A)+\cot(D]h^2}{2}$;

or $2\sqrt{3} - \frac{1}{2} \cdot \frac{56-8\sqrt{3}}{1+7\sqrt{3}} = 2\sqrt{3} - \frac{28-4\sqrt{3}}{1+7\sqrt{3}} = \textcolor{red}{\frac{56+46\sqrt{3}}{73}}$.
This post has been edited 4 times. Last edited by Mathzeus1024, Apr 4, 2025, 9:22 AM
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