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k a April Highlights and 2025 AoPS Online Class Information
jlacosta   0
Apr 2, 2025
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0 replies
jlacosta
Apr 2, 2025
0 replies
Transformation of a cross product when multiplied by matrix A
Math-lover1   2
N an hour ago by Math-lover1
I was working through AoPS Volume 2 and this statement from Chapter 11: Cross Products/Determinants confused me.
[quote=AoPS Volume 2]A quick comparison of $|\underline{A}|$ to the cross product $(\underline{A}\vec{i}) \times (\underline{A}\vec{j})$ reveals that a negative determinant [of $\underline{A}$] corresponds to a matrix which reverses the direction of the cross product of two vectors.[/quote]
I understand that this is true for the unit vectors $\vec{i} = (1 \ 0)$ and $\vec{j} = (0 \ 1)$, but am confused on how to prove this statement for general vectors $\vec{v}$ and $\vec{w}$ although its supposed to be a quick comparison.

How do I prove this statement easily with any two 2D vectors?
2 replies
Math-lover1
Yesterday at 10:29 PM
Math-lover1
an hour ago
unfair coin, points winning 2024 TMC AIME Mock #9
parmenides51   5
N 2 hours ago by Math-lover1
Krithik has an unfair coin with a $\frac13$ chance of landing heads when flipped. Krithik is playing a game where he starts with $1$ point. Every turn, he flips the coin, and if it lands heads, he gains $1$ point, and if it lands tails, he loses $1$ point. However, after the turn, if he has a negative number of points, his point counter resets to $1$. Krithik wins when he earns $8$ points. Find the expected number of turns until Krithik wins.
5 replies
parmenides51
Apr 26, 2025
Math-lover1
2 hours ago
BrUMO 2025 Team Round Problem 15
lpieleanu   1
N 3 hours ago by vanstraelen
Let $\triangle{ABC}$ be an isosceles triangle with $AB=AC.$ Let $D$ be a point on the circumcircle of $\triangle{ABC}$ on minor arc $AB.$ Let $\overline{AD}$ intersect the extension of $\overline{BC}$ at $E.$ Let $F$ be the midpoint of segment $AC,$ and let $G$ be the intersection of $\overline{EF}$ and $\overline{AB}.$ Let the extension of $\overline{DG}$ intersect $\overline{AC}$ and the circumcircle of $\triangle{ABC}$ at $H$ and $I,$ respectively. Given that $DG=3, GH=5,$ and $HI=1,$ compute the length of $\overline{AE}.$
1 reply
lpieleanu
Apr 27, 2025
vanstraelen
3 hours ago
trigonometric functions
VivaanKam   9
N 4 hours ago by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
9 replies
VivaanKam
Yesterday at 8:29 PM
aok
4 hours ago
No more topics!
[MAT 2022 #8] Farey Sequence?
I-_-I   17
N Jul 25, 2022 by I-_-I
Farey's calculator always displays exactly $10$ digits after the decimal point, and has one operation: $\sqrt{x}.$ Pressing the button will output the square root of the input, with everything beyond the $10$ digits truncated and forgotten by the calculator. Farey enters
\[ 1.000000\underline{abcd}\]where $a$, $b$, $c$, and $d$ are digits, and continues pressing the $\sqrt{x}$ operator until the calculator displays $1.0000000000$. Find the unique four-digit number $\underline{abcd}$ such that the number of times the rightmost digit displayed is even is maximized.

Aaron Guo
17 replies
I-_-I
Jul 23, 2022
I-_-I
Jul 25, 2022
[MAT 2022 #8] Farey Sequence?
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I-_-I
2047 posts
#1 • 2 Y
Y by Mango247, Mango247
Farey's calculator always displays exactly $10$ digits after the decimal point, and has one operation: $\sqrt{x}.$ Pressing the button will output the square root of the input, with everything beyond the $10$ digits truncated and forgotten by the calculator. Farey enters
\[ 1.000000\underline{abcd}\]where $a$, $b$, $c$, and $d$ are digits, and continues pressing the $\sqrt{x}$ operator until the calculator displays $1.0000000000$. Find the unique four-digit number $\underline{abcd}$ such that the number of times the rightmost digit displayed is even is maximized.

Aaron Guo
This post has been edited 1 time. Last edited by I-_-I, Jul 23, 2022, 11:59 PM
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brainfertilzer
1831 posts
#2
Y by
Is it $8193$? I just used the binomial expansion $(1 + x)^{1/2}$. Unless I'm trolling, this was pretty misplaced imo.
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Awesome3.14
1733 posts
#3
Y by
brainfertilzer wrote:
Is it $8193$? I just used the binomial expansion $(1 + x)^{1/2}$. Unless I'm trolling, this was pretty misplaced imo.

no I think the answer was 8190
however I put 8193 on the test
This post has been edited 1 time. Last edited by Awesome3.14, Jul 23, 2022, 11:45 PM
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miguel00
593 posts
#4
Y by
wait why, isn't it just 2^13??

Ohh oops. Yeah, I thought it was too easy for #8.
This post has been edited 1 time. Last edited by miguel00, Jul 23, 2022, 11:49 PM
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ihatemath123
3446 posts
#5
Y by
I put that too @above but put sqrt(1.0000008192) into wolfram
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brainfertilzer
1831 posts
#6
Y by
Awesome3.14 wrote:
brainfertilzer wrote:
Is it $8193$? I just used the binomial expansion $(1 + x)^{1/2}$. Unless I'm trolling, this was pretty misplaced imo.

no I think the answer was 8190
however I put 8193 on the test

huh
unless I'm using my calculator wrong, that gives an odd digit after the second sqrt?? I'm pretty sure 8193 is optimal since it gives even digits all the way until $1.0000000000$
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Arrowhead575
2281 posts
#7
Y by
i confirm 8190
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brainfertilzer
1831 posts
#8
Y by
Ok I probably made a mistake. Can someone please post a solution/sketch for this? Thanks.
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IAmTheHazard
5001 posts
#9 • 1 Y
Y by brainfertilzer
If the number formed by the last 4 digits is k, then pressing the sqrt operator changes it to floor((k-1)/2): separate into even and odd, since (1+(k/2)/10^-whatever)^2 is just a tiny bit greater than 1+k/10^-whatever
Then we can check that if f(n) is the least starting number that gives n even values for k (not including 0), so f(1)=2, that f(n+1)=2f(n)+1 or +2, but since we want f(n+1) to be even as well we have f(n+1)=2f(n)+2. inspection yields f(n)=2^(n+1)-2 or something (the exact exponent doesn't matter, just that it's 2 less than a power of 2), so ans is 8192-2
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brainfertilzer
1831 posts
#10
Y by
oh @above that makes sense. Thanks.
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megarnie
5603 posts
#11
Y by
I thought the formula was k/2

oops
This post has been edited 1 time. Last edited by megarnie, Jul 24, 2022, 12:16 AM
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miguel00
593 posts
#12 • 1 Y
Y by Mango247
Wait why does the formula turn out to be floor of k-1/2?

Oh thanks @below
This post has been edited 1 time. Last edited by miguel00, Jul 24, 2022, 12:58 AM
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brainfertilzer
1831 posts
#13
Y by
I think it's because $(1 + \epsilon)^{1/2} = 1 + \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 + O(\epsilon^3)$, and the quadratic term takes off a little bit, so you have to do $\lfloor ( \epsilon - 1)/2\rfloor$ instead of just $\epsilon/2$
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Mathelete101
252 posts
#14
Y by
How is the formula floor((k-1)/2)? I dont get @above's explanation. Why is $(1 + \epsilon)^{1/2} = 1 + \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 + O(\epsilon^3)$?
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brainfertilzer
1831 posts
#15
Y by
Mathelete101 wrote:
Why is $(1 + \epsilon)^{1/2} = 1 + \frac{1}{2}\epsilon - \frac{1}{8}\epsilon^2 + O(\epsilon^3)$?

Because binomial expansion: $(1 + x)^{1/2} = \sum_{j = 0}^{\infty}\binom{1/2}{j}x^j$
This post has been edited 1 time. Last edited by brainfertilzer, Jul 24, 2022, 3:27 PM
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superagh
1865 posts
#16 • 1 Y
Y by Arrowhead575
I don't think worrying too much about the forwards direction is necessary, the second half of IAmTheHazard's solution (backwards direction) is sufficient.

If we work backwards, we start from 1.0000000000 Then think about $(1+x)^2 = 1+2x+x^2$, where $x$ is really small and like a x2 pattern, but we must not neglect the $x^2$. The next number in the sequence can be 1.0000000002 because it's square root could be not be 1.0000000001 because of the $x^2$ tiny difference, the true square root would be something like 1.0000000000999..., which the calculator would output as 1.0000000000.

Next, by the same token 1.0000000004 yields 1.0000000001 because of the tiny difference. Therefore because of the x2 pattern we have possible 1.0000000005 or 1.0000000006, for which we choose the latter for the even number rule. We can keep using this idea/rule, continuously choosing the latter (out of 13 and 14 next, choose 14), to get a sequence of last four digits that goes like $0, 2, 6, 14, 30, ....$ which we can write an explicit formula for, $2^n - 2$, which yields the answer of $8192 - 2 = 8190$.
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MathPirate101
1603 posts
#17
Y by
IAmTheHazard wrote:
If the number formed by the last 4 digits is k, then pressing the sqrt operator changes it to floor((k-1)/2): separate into even and odd, since (1+(k/2)/10^-whatever)^2 is just a tiny bit greater than 1+k/10^-whatever
Then we can check that if f(n) is the least starting number that gives n even values for k (not including 0), so f(1)=2, that f(n+1)=2f(n)+1 or +2, but since we want f(n+1) to be even as well we have f(n+1)=2f(n)+2. inspection yields f(n)=2^(n+1)-2 or something (the exact exponent doesn't matter, just that it's 2 less than a power of 2), so ans is 8192-2

why do you need floor((k-1)/2) and not just k/2?
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I-_-I
2047 posts
#18 • 3 Y
Y by vvluo, MathPirate101, Mango247
MathPirate101 wrote:
IAmTheHazard wrote:
If the number formed by the last 4 digits is k, then pressing the sqrt operator changes it to floor((k-1)/2): separate into even and odd, since (1+(k/2)/10^-whatever)^2 is just a tiny bit greater than 1+k/10^-whatever
Then we can check that if f(n) is the least starting number that gives n even values for k (not including 0), so f(1)=2, that f(n+1)=2f(n)+1 or +2, but since we want f(n+1) to be even as well we have f(n+1)=2f(n)+2. inspection yields f(n)=2^(n+1)-2 or something (the exact exponent doesn't matter, just that it's 2 less than a power of 2), so ans is 8192-2

why do you need floor((k-1)/2) and not just k/2?

Basically we know if $x$ is $\frac{\text{last four digits}}{10^{10}}$ or whatever then $(1+x)^2 = 1 + 2x + x^2$; the $x^2$ term is small enough that it never really shows up on the calculator but its existence means that $\sqrt{1 + 2x}$ would be less than $1+x$.
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