1st USB S.0 2016: The Ultimate Step by Step Problem Olympiad

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374 posts

#1 h Mar 20, 2016, 12:47 AM • 17 Y

Y by jowanar, FlyingCucumber, Wave-Particle, nikolapavlovic, eshan, tarzanjunior, Magikarp1, acegikmoqsuwy2000, TheOneYouWant, aopser123, math101010, Kwow, SohamSchwarz119, Eray, Stupke, Adventure10, Mango247

1st Ultimate Step by Step (Problem) Olympiad-USBs(p)O
8. April, 08:00am GMT-09. April, 08:00am GMT, 2016
___________________________________________________________________________________

Winners of 1st USBs(p)O are:

InfinityDots: (30 points)
- ArseneLupin
- TacH
- talkon
- ThE-dArK-lOrD
Garfield (30 points)
Sarsil: (30 points)
- silouan
- Mikesar

2nd place is taken by:

TeamFatih: (27 points)
- Eray
- armann

And 3rd place is taken by:

Tibi Valoare Clan: (22 points)
- abnunc
- tbn456834678
SCAM: (20 points)
- SohamSchwarz119
- calculus_riju
- Abhinandan18
- Mathzoo

Here are the final result table

Congratulations to all of you, hope you had some fun, see you!

Previous instructions

Attachments:

USBspO 2016 problems.pdf (268kb)

USBspO 2016 Solutions.pdf (311kb)

This post has been edited 22 times. Last edited by mihajlon, Apr 9, 2016, 2:18 PM

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FlyingCucumber

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FlyingCucumber

#2 h Mar 20, 2016, 1:10 AM • 2 Y

Y by Adventure10, Mango247

What does the s and p stand for?

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Wave-Particle

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Wave-Particle

#3 h Mar 20, 2016, 1:55 AM • 2 Y

Y by Adventure10, Mango247

Sign-ups

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mihajlon

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mihajlon

#4 h Mar 20, 2016, 2:12 AM • 2 Y

Y by Adventure10, Mango247

FlyingCucumber wrote:

What does the s and p stand for?

If you mean in name then for "step" and "problem"(it is written above...)

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weilunsun28

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weilunsun28

#6 h Mar 20, 2016, 3:15 AM • 2 Y

Y by Adventure10, Mango247

Maximum number of people per team?

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Tega

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Tega

#7 h Mar 20, 2016, 3:17 AM • 2 Y

Y by Adventure10, Mango247

anandiyer12 wrote:

Sign-ups

You want to join my team for this contest?

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eshan

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eshan

#8 h Mar 20, 2016, 3:25 AM • 2 Y

Y by Adventure10, Mango247

Signups

This post has been edited 1 time. Last edited by eshan, Mar 20, 2016, 3:56 AM

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weilunsun28

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weilunsun28

#9 h Mar 20, 2016, 3:27 AM • 4 Y

Y by Wave-Particle, dragonmaster3000, Adventure10, Mango247

eshan wrote:

Signups

I'm pretty sure when he put "SOLO" next to his name, he meant that he wanted to compete "SOLO".

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m1234567

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m1234567

#10 h Mar 20, 2016, 3:41 AM • 2 Y

Y by Adventure10, Mango247

Please sign me up anyone (I can't latex) I'll be on someone's team

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mihajlon

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mihajlon

#11 h Mar 20, 2016, 11:36 AM • 2 Y

Y by Adventure10, Mango247

weilunsun28 wrote:

Maximum number of people per team?

As much as you want, but I think it would be more interesting if there is $<5$ people in each team.

m1234567 wrote:

I can't latex

You don't need to know how to write in LaTeX, you can just write it on paper and scan it or take the photo of it and send(but...)

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Eray

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Eray

#12 h Mar 20, 2016, 4:32 PM • 2 Y

Y by Adventure10, Mango247

When is the deadline for sign ups?

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Zoom

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Zoom

#13 h Mar 20, 2016, 4:36 PM • 2 Y

Y by Adventure10, Mango247

Do teams get points for each contestant, and then add it up, or what?

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mihajlon

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mihajlon

#14 h Mar 20, 2016, 4:40 PM • 2 Y

Y by Adventure10, Mango247

Eray wrote:

When is the deadline for sign ups?

You can sign-up to the contest until the begining of the contest i.e. $8.$ April...

Zoom wrote:

Do teams get points for each contestant, and then add it up, or what?

No, it's up to you if you want to compete with more team members(so you could(maybe...) solve it more easily, and feel team spirit

...), or want to check your skills and compete SOLO...
Max points per team is $30$ ...

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Zoom

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#15 h Mar 20, 2016, 5:35 PM • 3 Y

Y by lucasxia01, KenV, Adventure10

Then the $+2$ marks for solving all problems make no sense, because no one will have $29$ nor $28$ points?

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FlyingCucumber

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FlyingCucumber

#16 h Mar 20, 2016, 5:46 PM • 1 Y

Y by Adventure10

Can we make a fantasy step problem team math olympiad? I want canton on my roster. Betting 1 bitcoin.

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Garfield

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Garfield

#118 h Apr 9, 2016, 10:24 AM • 2 Y

Y by Adventure10, Mango247

Problem 2 is very nice,my solutin:
Clearly from $a,b,c<3$ we see that $a,b,c$ are sides of triangle. so $3-a=\frac{b+c-a}{2}$ which remind us on Heron's formula.We wil prove
$\prod_{cyc} \frac{a+b-c}{2}\cdot \frac{a^2+b^2+c^2}{(abc)^2}\leq \frac{3}{16}$ which is equivalent to $\prod_{cyc}\frac{a+b-c}{2}\cdot \frac{a+b+c}{2}\cdot \frac{a^2+b^2+c^2}{(abc)^2}\leq \frac{9}{16}$ which is equivalent (by Heron's formula to: $S^2\cdot \frac{a^2+b^2+c^2}{(abc)^2}\leq \frac{9}{16}$ now $S=2R\cdot \prod \sin(A)$ so applying in formula we need to prove $\sum_{cyc} \sin(A)^2\leq\frac{9}{4}$ which is equivalent to $\sum_{cyc} cos(A)^2\geq \frac{3}{4}$ which is well-known ,equality for $a=b=c=2$ so $M=\frac{3}{\frac{3}{16}}=16$

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Eray

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Eray

#119 h Apr 9, 2016, 10:25 AM • 3 Y

Y by armann, Adventure10, Mango247

Was anyone able to solve the case $m=q=1$ in Problem 1? I.e, $11.5^n-3^p.2=1$

And I'd like to note that Problem 3 is Serbia MO 2010 P5

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anantmudgal09

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anantmudgal09

#120 h Apr 9, 2016, 10:29 AM • 2 Y

Y by Adventure10, Mango247

oh, I couldn't submit

What are the answer by the way? I had like to try 4. Did 1,2,3

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mihajlon

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mihajlon

#121 h Apr 9, 2016, 10:32 AM • 2 Y

Y by Adventure10, Mango247

Here are the official solutions:(they are attached to the first post as well...)

Attachments:

USBspO 2016 Solutions.pdf (311kb)

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Garfield

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Garfield

#122 h Apr 9, 2016, 10:34 AM • 2 Y

Y by Adventure10, Mango247

That's not too hard $11\cdot 5^n=3\cdot 2+1$ ,if $n=1$ obviously we have solutions so let $n\geq 2$ ,now check at $mod 25$ ,trivialy we see that order for $3$ $mod 25$ is $20$ and $3^7\equiv 12 (mod 25)$ so $p\equiv 2 (mod 5)$ but taking $(mod 11)$ we see that $3^3\equiv 2 (mod 11)$ son $p\equiv 3 (mod5)$ so it's contradiction.

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Garfield

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Garfield

#123 h Apr 9, 2016, 10:55 AM • 2 Y

Y by Adventure10, Mango247

Mine solution of problem 4:
By simple angle-chasing $AIKF$ is cyclic so $\angle AFI=\angle AKB=90^{\circ}$ so we need to prove that $AE$ is perpendicular $OH$ .We have $C-X_{B}-H_{B}$ and $B-X_{C}-H_{B}$ ,let $H_{B}H_{A}$ , $X_{A}X_{B}$ meet in $R$ .By Papuse theorem we have $R-H-G$ ,where G is centroid of $\triangle ABC$ .Now let $S$ be center of nine point cyrcle (midpoint of OH),so we have $R-O-H-G-S$ now by $Brocard$ theorem on $H_{C}H_{B}S_{B}S_{C}$ ,we have $S$ is $ortocenter$ of $\triangle ARE$ so $RS$ is perpendicular $AE$ ,so $RE$ is perpendicular on $AE$ so $ALHK$ is $cyclic$ .

This post has been edited 3 times. Last edited by Garfield, Apr 9, 2016, 10:56 AM
Reason: latex

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Eray

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Eray

#124 h Apr 9, 2016, 11:03 AM • 3 Y

Y by armann, Adventure10, Mango247

Another proof for $AE\perp OH$ :

Call the circle $AH_BH_CH$ as $\omega_H$ , and the cricle $AX_BX_CO$ as $\omega_O$ .
The centers of $\omega_H$ and $\omega_O$ are the midpoints of $AH, AO$ respectively. Call them $H_1$ and $O_1$ , respectively.

Easy to see the power of points $A$ and $E$ are equal w.r.t $\omega_H$ and $\omega_O$ . Which means $AE\perp O_1H_1$ . But $O_1H_1\parallel OH$ , so $AE\perp OH$ as desired. $\blacksquare$

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mihajlon

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mihajlon

#125 h Apr 9, 2016, 11:46 AM • 2 Y

Y by Adventure10, Mango247

Here are the preliminary results:

As I mentioned before you can send us mail(until $9$ th April, $14:00$ $GMT$ ...) , if we gave you less points than your team deserve(by solution...)

This post has been edited 1 time. Last edited by mihajlon, Apr 9, 2016, 12:07 PM
Reason: TIME SET

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mihajlon

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mihajlon

#126 h Apr 9, 2016, 2:02 PM • 4 Y

Y by Garfield, adityaguharoy, Adventure10, Mango247

I hope all of you have fun participating on 1st USBs(p)O, thanks for participating and Good luck in the future!

Final results:

Attachments:

1st USBspO results.xlsx (35kb)

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abnunc

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abnunc

#127 h Apr 9, 2016, 3:07 PM • 3 Y

Y by adityaguharoy, Adventure10, Mango247

But why are there 3 gold medals, 1 silver and 2 bronze??

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mihajlon

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mihajlon

#128 h Apr 9, 2016, 3:17 PM • 2 Y

Y by Adventure10, Mango247

abnunc wrote:

But why are there 3 gold medals, 1 silver and 2 bronze??

because there are $3$ teams with max number of points, and since there are $3$ gold, it should be at least $1$ silver and thus $2$ bronze(and also it is first competition of this type and since not much contestants solved(or sent their solutions)
, so...)...

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Eray

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Eray

#129 h Apr 9, 2016, 4:26 PM • 1 Y

Y by Adventure10

Thank you Mihajlon and Nikolapavlovic! It was fun

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abnunc

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abnunc

#130 h Apr 9, 2016, 4:32 PM • 2 Y

Y by Adventure10, Mango247

Yeah, it really was

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mihajlon

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mihajlon

#131 h Apr 9, 2016, 4:33 PM • 2 Y

Y by Adventure10, Mango247

Eray wrote:

Thank you Mihajlon and Nikolapavlovic! It was fun

Thanks Eray!

abnunc wrote:

Yeah, it really was

Thanks abnunc!

And thanks for participating!

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adityaguharoy

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adityaguharoy

#132 h Apr 26, 2016, 7:21 AM • 3 Y

Y by Irfan, Adventure10, Mango247

Must be so

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