calculus
integration
geometry
analytic geometry
graphing lines
slope

Ok I need help figuring out how far my car traveled over a certian time period.  This is the information that I know.

My car went from 0mph - 53mph in 16 seconds.  That means that at Time 0 my car was traveling 53mph and at Time 16 my car was traveling 53 mph.

I need to figure out how many feet my car traveled in this time period.

This is how I worked it so far.

First I converted 53MPH to Ft/sec.

[u]53Miles.....5280 ft.....1hour........1min[/u]= 77.733 ft/sec
1hour.......1mile.......60mins.......60secs

I was thinking that I could solve this by making a graph and then calculating the area under the curve by setting up an integral.  Correct me if I'm wrong but if the X axis is sec and the Y axis is ft/sec then the area under the curve is ft because ft/sec X sec = ft.

So I set up a graph of the equation f(x) = 4.858x because with the points (0,0) and (16, 77.733) you get the slope of 4.858 and the graph goes straight through the orgin.

I then made the integral  (the S is an integral sign lol)

[size=150]S[/size] 4.858x and evaluated from 0-16

so..

((4.858(16)^2)/2) - ((4.858(0)^2)/2)
Which I evaluated to come out to 621.824 ft.  Which is 207 yards which seems right to me.

if anyone could help me at all, check my work and tell me if I'm even going in the right direction to begin with.

Thanks a bunch.

Well actually...
I don't know calculus at all, but if I interpreted the problem correctly, then you're setting up a graph with time of the x axis and speed on the y axis.  Then you find the area under that for x=0-16.
Look at it this way.  Your speed at the beginning is 0, and at the end, it's 77.73333..... ft/sec.  Therefore, assuming that you accelerated at a constant rate, we can say that your average speed was 77.73333.../2=38.8666...ft/sec.  Multiple that by the time, 16 seconds, to get 621.86666.... ft.
So you did it correctly; if you think about it, that is the correct for for an integral.  You just didn't use that many decimal places and thus got an answer just slightly off.

I have to use this in court for support.  Is this problem also right assuming that I didn't accelerate at a constant rate?

I'm thinking it would be because even if for the first 8 seconds I drove extremely slow, I would have to extremely fast to still reach the same velocity at 16 seconds.

If your using it in court, get an experienced professional mathematician to solve the problem and sign a declaration.

There are lots of problems with what you're saying.  It's easy to say that you accelerated at a constant rate, but given that you changed gears multiple times between 0 and 53 mph (probably twice), any graph of your velocity is going to have big dips from when your engine revs drop.

[quote]My car went from 0mph - 53mph in 16 seconds. That means that at Time 0 my car was traveling 53mph and at Time 16 my car was traveling 53 mph[/quote]

Theoratically speaking - Your average velocity can be anything between (Plese see the note below :)  0 and  53 mph and thus you could travel anywhere between  0 to 1244 feet. Uniform accelarion would give  average speed of 53/2 and distance  would be 622 feet.

Of course depending on your car,  (and assuming you did not pass out)put only certain maximum  accelation is possible, thus  it will reduce  the maximum distance, .. Assuming  you are not riding on a Saturn V rocket and have a max g of (1/2)   (say 0 to 60 mph in 5 seconds -  A real sports car) than  you travelled 11 seconds at 53 mph and 5 seconds at 26.5 mph (you do the math)

(Note - Mathmatically speaking,  the answer could be [b]almost any[/b] value -  (Well Physics would put certain restriction such as  you can not travel faster than light  :) etc.) if one assumes you went faster than 53 miles per hour than slowed down ... or for that matter went in revers ... etc :)

For any normal car, assuming you did not slam on the brakes or jammed on the pedal,  my estimate would be  about 600 feet to  900 feet  - with only information given above..

You can probablly make a case (If I was the judge etc..) if you want to prove  that you did not go more than 900 feet  (and may be  700 or 800 feet - depending on  what kind of car you have) but that's all unless you can provide some more data...

Assuming that your acceleration a was constant, then the velocity v after accelerating from rest is v = at and the path s traveled is $s = \frac 1 2 at^2 = \frac 1 2 vt$. If

$v = 53\ \text{mi/hr} = \frac{53\ \text{mi} \cdot 1760\ \text{yd/mi}\cdot 3\ \text{ft/yd}}{1\ \text{hr} \cdot 3600\ \text{sec/hr}} = \frac{53 \cdot 1760 \cdot 3}{3600}\ \text{ft/sec} =$

$= \frac{53 \cdot 44}{30}\ \text{ft/sec} = \frac{233.2}{3}\ \text{ft/sec} \doteq 77.7333...\ \text{ft/sec}$

$s = \frac 1 2 vt = 77.7333...\ \text{ft/sec} \cdot \frac{16\ \text{sec}}{2} = 621.8666...\ \text{ft}$

However, your acceleration did not have to be constant. Moreover, the test you did is not valid, especially if your fuel pump was not OK. In such case, the fuel pressure depends on ambient temperature in the fuel tank (this is an experience) and the fuel pressure decreases in hot weather or after the rear of the car is exposed to the sun.

Only once I tried to beat a speeding ticket, when I was a new immigrant doing manual jobs and the fine seemed just too high. In short, I asked the officer how many times he measured my speed and when he replied that once (as expected), I told the judge that it is impossible to say anything about the measured quantity based on just a single measurement. But the judge said that while he agrees that the method is unscientific, he is bound by precedents from higher courts that a qualified police officer found my speed to exceed the speed limit and I had to pay. Over the years, I got only 1 more speeding ticket (and 2 warnings with no fine), although I usually keep the posted speed limits only in a blizzard. Just watch for police cars the next time around or get a radar detector. In addition, I was able to beat some no seatbelt tickets and in the process, I observed, what officers and people with speeding tickets told the judge. The officer testifies how he first estimated the speed, what are the results of his tests in estimating speeds and how he tested the radar on that day. The judge dismissed the cases only if people claimed (under oath) that they were traveling to a doctor or to a hospital, either themselves or with their children. The only benefit of going to the court is that the police officer has to come as well and he may not show up. The case is then automatically dismissed. But if the officer shows up, be prepared to say something not too lame, otherwise you may upset the judge for wasting everybody's time.

Yetti

[quote="Awsamia"]The thing is that I was accused of going 53 in a 35.  I was making a right hard turn from 0 mph and the spot that the officer supposedly got me at is much closer then 600 feet.  At the time my fuel pump was rotting in my car and greatly reduced my acceleration.  I tested my 0-53 time from a right hand turn on a similar street and it took me 16 seconds with my foot all the way on the gas.[/quote]

Hi  what you need in this case is to measure the distance where the officer stopped you from the corner  (Take a photograph)  and use the formula:

For simplicity use all measurements in meters (or feet if you like - I will use meter here)

$v = \sqrt {2a  d }$

Where $d$ is distance, let me take it as  $200$ meters (for example)
(You may type  "600 feet in meters" in google and find the value in meters etc...)

$v$ would be maximum velocity you could have had.  (Remember   53 mph  = about 23 m/s) (Use google agian and type some thing like "53 miles per hour in meters per second"   use google!

fNow unless you have a very fast car  (Even in that case there would be tyre marks for any value of $a$ say greater than 5 m/s  (and that is really fast car)

You could be  going as fast as 70 mph (and leave very little tire marks if you had 600 feet to pllay with)
For normal car, (even the old car)  calculation show that you can reach 53 mph with as little  $a$ as  0.14 g   that a   car (old)  can  achieve ....

So , IMO,   following  can help:

1.   Some kind of proof that it  takes 16 seconds to reach 53 mph in your car  (that would    give a= 01.4 g which is belivable for an old car.
2.    Distance  less than 600 feet! (To beat  35 mph rap  you need  about 480 feet )
3.   lack of tyre mark (or rain on the day or anything which makes the road slipplery  would help too)

Inany case
[b]
Measure the distance  and measure the acceleration for your car by measuing accurately how long it takes you to for from 0 to 50 mph. [/b]

 Good Luck.

Related subject  (Not really )

By the way you have to  be real carful about using Physics in a court!  (I know I have tried it but that's another story) You must have heard of the case where a scientist  (who was charged with not stopping at a red light) argued:

[quote] Your honor, I saw a [color=green]green [/color]light due to "Dopler Effect". I was after all approaching the light   [/quote]

Judge did agree but asked him to calculate the speed  to see  a[color=red] red[/color] light  appear [color=green]green[/color]. And of course he was fined for speeding!!! :D  :)

This problem can be solved with physics, as well.

$53\; mph \approx \frac{212}{9}\; m/s$

Assuming a constant acceleration, we have $a=\frac{\frac{212}{9}}{16}=\frac{53}{36}$.

I'll use the equation $x=V_ot+\frac{1}{2}at^2 \Rightarrow x=\frac{1}{2}\displaystyle\left(\frac{53}{36}\right)(16^2)=\frac{1696}{9}\approx 188.45$ ft.

If you don't know physics, you can graph the velocity to time instead, then take the integral.

Like someone suggested earlier, it might be best to get a professional mathematician, perhaps from a local college to verify that you are using the correct method.You could go in there and make the best mathematical speech that you could and the judge might not understand a lick of it. If you can't get the signature, at least bring a book that has the formula in it or something, so don't have to trust you totally for the working of the problem.


geometry
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Read the title. That's it! You can organize in any format, for example, period by period would be nice. Also, it would be nice if you would mention your grade or what section of school you're in (ele. school, mid school or high school).

I'm in 7th grade.

For me, I didn't go to school today since I was sick :D But here it is from Friday:

Period 1: Orchestra
Alright, this is an intermidiate orchestra class and I've never held a violin in my hand. So I am not that good at this class but I've wanted to play the violin since 4th grade!

Period 2: Geography
I had forgotten some homework so I did it near my locker :P Aside from that, discussed maps.

Period3: Algebra 2
Just the usual, teacher writing on board, everybody taking notes and me playing with my pencil and then realizing that I missed some important homework information :)

Period 4: LA
Discussed about immigration. Completely off-topic.

Period 5: Lunch
Ate my lunch (which was thepla, which is indian flat bread, with chutney, it was really good!) than ran off to practice in the orchestra room. Got a lot better.

Period 6: Beginning Spanish
Worked on projects.

Period 7 Physical Science
Did this dumb dimensional analysis.

I'm in ninth grade. For some reason while almost all other school districts have middle schools for grades 6-8 and 9-12 in high school, our school district has junior highs for grades 7-9 and 10-12 in high school.

Period 1: Physics
It's a regular 12th grade class that I applied into and got in (along with about 20 other people I know from our district's gifted program). Our teacher rambled on incessantly about comets shattering celestial spheres and the Heaven's Gate cult, and things like that. We missed all the discussion we were supposed to have in class today.

Period 2: Social Studies
Took notes on the parts of geography.

Period 3: Spanish I
Review for the test tomorrow. Nothing too hard.

Period 4: Algebra 2
Test. On arithmetic and geometric sequences.  Finished in 10 minutes, double-checked and triple-checked answers, turned in quiz first, and doodled for the next 20 minutes. No difference from all other math tests since sixth grade.

Period 5: Language Arts
Worked on our projects.

Period 6: Computer Programming
Worked on our "vision statements" for computers and our future.

Period 1- The World of Music.  Bleh.  Basic stuff.
Period 2- Study Hall.  Proctor let me use my iTouch.
Period 3- Study Hall.  Different proctor let me download apps off iTunes.
Period 4- Precalculus Honors.  Mini-quiz.  Got a 100.
Period 5- Study Hall.  Finished up Precalculus homework.
Period 6- World History I.  Discussed very early people.  Like Old Stone Age Era.
Period 7- Lunch.  Yeah.
Period 8- English.  Discussed "Inherit the Wind" by Jerome Lawrence and Robert E. Lee.
Period 9- Study Hall.  Finished World History homework.  Left school 10 minutes early.

Period 1-Civics. have like all my friends there
Period 2-English. i dont like the teacher.
Period 3-Math. i think i have a C rite now no joke lol.
Period 4-Spanish. its fun and my friends sit at my table
Period 5-Band. i dislike it cuz its so competitive that its not even funny
Lunch- sit with my friends
Period 6-Gym. the teachers are mean and i have it rite after lunch
Period 7-Science. teacher is nice and have all my friends.

then i have to race to my locker to get my stuff and get the back seats in the bus

Period 1&2: Math. YAY! MATH! We discussed stuff (this is in table groups)... some involving the subject  :oops:
Period 3: Creative Writing... I was creative, I free wrote about polluted environment :D
Period 4: Math Enrichment, basically math. But an extra period for it :wink:
Period 5: Er... I think it was... OH RIGHT POWER USER! It's a computer class. We talked about US citizenship, the privileges and etc.
LUNCH: Was good
Period 6&7: Social Studies!! I got a sticker showing how good a student I am
Period 8: French, we learned how to count to 100 (in French.)

7th grade

P1: Civics
P2: Band
*P3: Language Arts (hate the subject, got a project in the 1st week of school and homework on first day)
P4: Science
P5: Health
P6: Geometry:

*: I have a split lunch. It means I have half of P3, go to lunch, and then go back to P3.

P1: reading
Got progress report found out that I epically failed ELA
P2:Math
Learned that homework is a 100 if you do it which is good cuz a got a 50 on a worksheet
P3: Band
I play clarient.
P4: Science
Played a jeporary like game I failed got 7th sigh
P5: ELA
Teacher corrected Progress report from a 75 to a 98 yeah.
As you can see my school has no excitement sigh.

in 9th grade.

Period 1: World History
learned about different types of maps
Period 2: Band
also a clarinet player
Period 3: Study Hall
Period 4: Computer Programming
the internet got disconnected, so we did nothing
Period 5: Wellness
learned slangs for some topic( :blush: )
Period 6: English
wrote about summer reading
Period 7: AP Stat
too easy(we learned about mean and median :( )
Period 8: AP Physics C
was also easy. We learned about motion diagram.

My first day of school happened about a week ago...but I can remember it as if it was yesterday....
*FLASHBACK*
"Hey now you're an All Star get the game on Go Play. Hey now, you're a Rock Star get the show on Get Paid!" ringed the alarm...
"Uggh..5 A.M.....I'm gonna snooze" I thought as I hit the snooze button on my cell phone...
-10 minutes later-
"Hey now you're an All Star get the game on Go Play. Hey now, you're a Rock Star get the show on Get Paid!" ringed the alarm...
"Bleh...10 more minutes..." I thought as I hit the button again..
(This happened 6 times)

6:00 A.M
"KING1(This isn't my name :P )!" screamed my mom
"Oh god..its 6...ahh" I screamed as I rushed into the bathroom to brush my teeth
The rest of the morning was a total blur..
Then...school..
I walked to my locker: LE49, typed in my combo and started pulling the tab down..locker didnt open!
Tried this about 4 times when the guy next to me said I had to push up...WOW!
I got my A.P Stats book and ran upstairs to my A.P stats class....I was late...
"Hello there!" said my teacher, "Whats your name?"
"Uh..KING1" I replied as I looked around the class to see a bunch of sophomores,juniors, and seniors..there was no freshman in the class other than me..."Great" I thought..
"Ok.....class welcome to A.P Stats!" said the teacher out loud, blah blah blah...etc. etc.
"I expect you to do you're homework in this class. I will not be checking your homework everyday, but I will occasionally collect you're homework. Dont try to be, (looks out the classroom), smart :censored: and not do your homework just because I dont check. It hurts you." said the teacher.
The rest of hte class continued as usual...boring stuff...
Next hour.. Honors Pre-Calc.. ugghy
"Hi class! How are you!" said the teacher out loud, "I love math! In fact, I wanted to be a Math teacher since I was in 8th grade, and I never changed my mind!"
There were some of my freshmen friends in class so yeah...that was cool..
"Oh yeah...class..we will be starting our first lesson today! Lets not waste time!", said my teacher, "your first quiz will be in two days!" said the teacher.

Blah blah....next hour...French 2!

"(Teacher enters classroom saying something in french really fast..)"
"Heh? Come on you guys answer my question!" said the teacher
*Crickets chirping*
"Ok..you guys need a lot of work in French.."

Blah Blah...next Chemistry!
I enter the class and see all juniors and seniors...and 1 other freshman!
"Hi everyone! What is Chemistry?" asks the teacher, "How about you young man?"
The teacher picks on a kid snoring in the back.....it was kinda funny..
"Huh? Oh..chemistry is...uh...pouring stuff in a bottle and watching bubbles come out?"
*People giggle*
Another guy raises his hand, then puts it down, then raises it, then puts it down..then finally puts it halfway in the air. Teacher picks him.
"Chemistry is making an amalgamation of diverse elements from the periodic table, and scrutinize the various ramifications such as effevervescence, color change and various other ramifications." said a kid
"Good job..but thats just part of science.." chuckled the teacher nervously
"Wha..? Why isnt that kid speaking english? asked the kid snoring in the back.
"Can anyone interpret that for him?" asked the teacher.
I slowly raised my hand..and the teacher picked me..great..
"Eh...he just said chemistry is mixing different elements, and observing their effect..like....uh...bubbling..color change...and some other results." I said
The class went on...boring stuff...and finally got back home..

Gah.  Jam-packed day today.

Period 1: Human Connections- Talked about Kanye West and Taylor Swift.  :P
Period 2: English- Discussed more Inherit the Wind, and a little bit of Of Mice and Men.
Period 3: Precalc Honors- Spent the day asking questions about homework. (rofl!)  Still 2 problems to discuss for Friday. :P
Period 4: Handbells- Woohoo!  I play the teeny-tiny ones!
Period 5: Latin 2- Boo.  Quiz on nouns.  But my teacher is the best!
Period 6: World History- Discussed random things which don't matter.
Period 7: Lunch- Uhhh... yeah, again.
Period 8: Biology- Quiz, completely failed it.
Period 9: Study Hall- Nothing.  Left on time today. :)

School
social studies:bleh ....project...fail partners
science- bleh book
rest of day goes on...math is the only fun thing.

We have tests called embedded assements every month for Language Arts, Math, and Science.

I have one tomorrow. :(

AP Stat (normal distribution, eh?), AP BC Calc, Russian, Economics are my major ones. The rest aren't that interesting.

Honors Language Arts workload is hardly Honors-level. We get three days to take notes on two articles.
In history a few days ago, our teacher gave us a portion of an actual AP US History multiple-choice exam (we studied US history last two years), and I got a 36/40. "We're preparing you for AP classes"...really?
Math is a drag. "Reviewed" linear equations. Still waiting for the curriculum to move past what I learned in fifth grade. I really should have skipped another math level.

In earthsci, I sat next to someone with a cold. I colored continents and tracked their movements since Pangaea broke apart. It was mind-numbingly boring.

In English, we talked about... a literature test we're going to have on A Raisin In The Sun.

In health, I learned that I am a control freak when it comes to projects and I have to check everything myself and do it just so, and it drives me insane when something isn't done just right but I feel mean when I point something out to my group members that needs to be changed.

I got second violins to pizzicato a quiet part in a piece in orchestra [i]just the right way[/i]. *wipes a tear from eye* I was... so proud. It was beautiful.  :winner_first:

Tech App: Did getting to know you quiz , when I finished I got on AoPS.
PE: Ran
Math: Easy percents and simple interest
Lunch: We sang random theme songs
Band: I think I went to Band Boot Camp
ELA: Took Benchmark

OMG!!! SCHOOL YESTERDAY WAS ABSOLUTELY HORRIBLE!!!!!

Hour of math
5 freaking hours of science review
1 hour of Art History and piano

Yesterday I was at my school for the bulk of the day.
7:45-1:00-took SAT and epic failed
1:00-2:30-Science Olympiad practice!  Unfortunately, the Subway order was so slow I had to leave before I could get it
4:00-9:30-My friend's birthday party.  SO FUN.  They just remodeled their basement with air hockey, pool, ping pong, and an awesome plasma TV.

in mah 4th period which is math these two guys started singing best of both worlds by hannah montana  :o then one of them started singing single ladies! it was so funni yet akward at the same time! he even did teh dance  :o   :o

Tech App: Went on AoPS
SS: Finished Timeline
PE: Ran mile
Math: I read chap. 10 of AoPS VOl 1 while the class learned about simple interest. Towards teh end two guys started singing random songs
ELA: Took Benchmark

Tech Apps: free time !
Band:played F major scale
Science: Learned about earthquakes
SS: Watched The Alamo
ELA: Took Benchmark, but I read AoPS vol. 1

I've been at a funeral Thursday-Friday in Spokane(now I'm back at home)  but this Tuesday I have MC Chapters!

i am in sixth grade

first period: Tech theater.
doing costume design try to act like i don't fail at drawing :rotfl:
Second period: SS African American history project
spent whole period messing around in library since i am done with research.
Third period: LA
working on poetry booklet
4th period: Lunch
5th period: Mythology
pretty boring already know everything so far.
6th period: Algebra 1 :first:
yay!! teacher puts notes on the bored and we copy them down and sometimes a philosophical discussions
7th period: Spanish
learning about breakfast and lunch.
8th period: science
learning about space

Tech Apps: Microsoft Word
Math: Surface area, it was sooooooooooooooooooo hard I finished in 5 minutes.
Band: My guy teacher wore a dress, and a girlly wig. This is not a joke.
Science: I don't remember.
ELA: CBA, but I read intro to c and p. I love the smaller AoPS books they are so portable and can easily be covered with book covers to make them look like textbooks.

10:30am - Woken up by sister, she alerts me that I will be home alone for half an hour
10:30.5am - I go back to sleep
11:30am - My dad tells me to wake up
11:32am - I go back to sleep
11:40am - My dad tells me to wake up
11:43am - I go back to sleep
11:45am - My dad tells me we're going out to lunch
11:45.5am - I rush out of bed
11:45am to 12:50pm - I get ready
12:55pm - We get into the car and my dad lectures me about taking so long to get ready
1:05pm - We pick up my mother at work
1:10pm - We find out the teriyaki place we wanted to go to is closed.
1:15pm - We find out the Japanese Buffet we wanted to go to is closed.
1:20pm - We settle for pho.
1:20pm to 1:50pm - I eat my pho.
1:52pm - We leave.
2:00pm - We drop my mother off at her workplace.
2:10pm - We get home
2:15pm to now - I'm on FTW/AoPS.

*note
I did not have school today
:)

Tech App: Word review
SS: Wrote an essay
PE: Did puzzles
Math: Helped everyone use a TI 84 plus to find are of cones
ELA: Rewrote essays

Tech App: Excel so boring
Band: Uper hard music
Science: Studied our DNA my DNA looks like 2 brains connected 2gether.
Lunch: Okay so I am known to be like a like preppy/ valley girl person who says like every 3 secs so someone like dared me to not say like for like the rest of the day..
Texas His.: Took Test.
Ela: When I had to say the word like the class said it for me!
I LIKE SURVIVED LIKE BEING LIKE SO TOTALLY UNME! I LIKE WAS UPER QUIET FOR LIKE 2 HOURS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

[size=200]LIKE![/size]

I'm in debate club, and today was the big day.
We went to a center and debated all day. After that, I did Homework, practiced piano, studied, ate, and now I'm here.


An isosceles triangle ABC has vertex angle x and its sides are sin x,  :sqrt: sin x, and :sqrt: sin x. What is its area?

actually these two questions are from a test on the arml site itself. (the 1999 and 200 tests are there in full.) also, if you keep searching, you will find some more problems. fierytycoon found another test on the nebraska arml team site some time ago (some of the questions are on this forum somewhere.) i also have heard of a book of arml problems, but i was unable to find it.

Im pretty sure they sell the newer book on amazon, and there is a link i think on the resources page here.  I think the older one is out of print.

[hide] use law of cos:

 sin^2(x)=sinx+sinx-2sinxcosx

 sin^2x=2sinx(1-cosx)         (sinx is not 0)

sin(x)/2=(1-cosx)

sin^2(x)/4=1-2cosx+cos^2(x)

sin^2(x)=4-8cosx+4cos^2(x)

1-cos^2(x) = 4-8cos(x)+4cos^2(x)

0= 5cos^2(x)-8cos(x)+3

0=[5cos(x)-3][cos(x)-1].  cos(x) isn't one.

so cos(x)=3/5, and sin(x) is then 4/5.  The area is 1/2*sqrt(sinx)*sqrt(sinx)*sinx=1/2*sin^2x=1/2*16/25=8/25.

is that right?  [/hide]

Which year was this from?

Just wondering about the books.

I bought the old book after it became out of print by e-mailing the people at ARML, they have extras stocked.

[quote]Im pretty sure they sell the newer book on amazon,[/quote]

But what is this new book you are talking about?

89-94.  This older book you are talking about is not this, right?  How much did it cost?

cats..., that's correct. It's from ARML 2000, and it's #7.

my ARML coach said yesterday that they're coming out with a newer book with contests from like 94-99, i think.  He said it should be coming out soon.

Is there any way you can use the formula 1/2*side A*side B*sin( angle included by side A and B)?  It seems like it would work well for this problem, but I can't figure it out.

A=1/2( :sqrt: sin(x)) :^2: *sin(x)=sin :^2: x
....

cats...  used that as his final step, after having solved for the sin and cosin of the angle using the law of cosines.  You would need another expression for the area to use that earlier in the solution.  However, another nice-looking area formula is difficult to come by in this particular problem.


geometry
rectangle
calculus
derivative
Pythagorean Theorem

A chord of length $\sqrt{3}$ divides a unit circle into 2 regions.  Find the rectangle of maximum area which can be inscribed in the smaller of these 2 regions.

Probably wrong but whatever
[hide]By Pythagorean theorem, the distance from the center of the circle to the chord is $\frac 12$. If set the width of the rectangle to $a$, then the length is equal to $\sqrt 3 - 2a \sqrt 3$.

The area is then $3a \sqrt 3 - 2a^2 \sqrt 3$ and the maximum area is then $\frac {\sqrt 3}{8}$.[/hide]

I got an answer using analytic geometry(I cheated, sorry). I think my calculation is wrong, since it's not clean as I expected.
[hide]$\frac{23-4\sqrt{19}}{25}$[/hide]

Haha... your guys' answers are so complex, mine is probably wrong.

I got $\frac 1 4$

edit:  wait... did something wrong... fix soon

This is so interesting! :)  Evertone got different answers for this simple(?) problem.

I used calculus to find the maximum of $f(x)=x(2\sqrt{1-x^2}-1)$ subject to the condition that $2\sqrt{1-x^2}-1>0$.

I got the function. Since I am very bad at calculus, I used Maple. Maple says,
\[1/8\,\sqrt {30-2\,\sqrt {33}} \left( 2\,\sqrt {{\frac {17}{32}}+1/32\,
\sqrt {33}}-1 \right) \]
is the maximum.

[quote="deej21"]Really?  Because I'm doing the same thing.  Although setting the derivative of that equal to 0 is pretty rough... You get a quartic, which I could complete the square and solve... but I kinda lost the drive to finish...

Oh, I tried a different way before that.  I was under the impression that it had to be a square... but now I'm not so sure.  Anyway, I know the rectangle's height would be $\sqrt{1-x^2}-\frac 1 2$, where $x$ is the x-value on the graph of the unit circle.  So if it was a square, $2x = \sqrt{1-x^2}-\frac 1 2$, but I solved that, and got x to be 4 something, which obvioulsy isn't right.  I must have done something wrong.  Although since it's inscribed in a circle (well, semi-inscribed), I don't know that a square would necessarily yield the largest area.[/quote]

There is no reason why a square would yield the largest area in this situation.

It seems to me that blahblahblah's answer is correct.

Now that that's settled, does someone want to prove that the rectangle does have to have sides parallel/perpendicular to the chord?

[quote="blahblahblah"]I get

$\frac {(\sqrt{33}-3)\sqrt{30-2\sqrt{33}}}{32}$

which is almost surely wrong, because it would be the single ugliest answer to an olympiad problem I've ever seen.[/quote]

I got the same thing. Yay!

Wow, I didn't know it would be that bad.  I had never actually figured out this problem. :?

I got the same result but I  can't avoid using Calculus.
The result  is probably correct as I have double checked the result by using Geometer Shetchpad to explore different possible positions of such a rectangle.
The remaining part of the solution is to argue that we only need to consider rectangles whose sides are parallel to the chord. ( At least I have assumed that in my calculation.)
This fact can be proved by using parallel transform.

I wonder if someone can solve this problem without using Calculus?


The digital sum of a number is found when the digits of the number are added and the digits of the sum are added again until only one digit is left. For example, the digital sum of $ 65536$ is $ 6\plus{}5\plus{}5\plus{}3\plus{}6\equal{}25 \rightarrow 2\plus{}5 \equal{} 7$.

What is the digital sum of $ {2008}^{{2008}^{2008}}$?


calculus
derivative
function
modular arithmetic
real analysis
real analysis unsolved

Prove or disprove:

$ \frac {\rm{d}^n}{\rm{d}x^n}_{x \equal{} 1}\frac {1}{x^2 \plus{} 1} \equal{} \frac {1}{2}, \minus{} \frac {1}{2}, 0$ if $ n \equiv_3 0,1,2$ respectively.

If $ n\equiv 3,0,1,2$ modulo what?  That last part doesn't make any sense to me.

I read it as:

if $ n \equal{} 0$ mod $ 3$, it should be $ \frac12$;

if $ n \equal{} 1$ mod $ 3$, it should be $ \minus{}\frac12$;

etc.

This is wrong I think.

I get that the fourth derivative at 1 is -3 and second derivative at 1 is 1/2.

The function which satisfies the condition you gave is

$ f(x) \equal{} \frac{1}{2(x^2\minus{}x\plus{}1)}$

I think.

Where did you find this problem?

I'm sorry, it's supposed to be

$ \frac {d^n}{dx^n}_{x \equal{} 1} \frac {1}{x^2 \plus{} 1} \equal{} \frac {1}{2}( \minus{} 1)^{k \plus{} 1}$ for $ n \equal{} 2k \plus{} 1$, and $ 0$ for positive even $ n$.
$ k$ is any non-negative integer.

Still does not work as second derivative at 1 is 1/2. (if my working is correct)

btw, the function I gave earlier is probably wrong.

Where did you get this problem? Did you make it up yourself?

Suggestion:

$ \frac1{x^2\plus{}1}\equal{}\frac1{2i}\left(\frac1{x\minus{}i}\minus{}\frac1{x\plus{}i}\right)$

Writing it in that form should make it easier to compute higher order derivatives. Evaluating those derivatives at $ x\equal{}1$ will then create the algebra problem of combining (negative) powers of $ 1\plus{}i$ and $ 1\minus{}i.$

By the way, the argument of $ (1\plus{}i)^{\minus{}n}$ is periodic of period 8. That seems completely incompatible with any formula stated mod 3 - maybe mod 4 or mod 8?

Here's what I get for $ \left.\left(\frac {d}{dx}\right)^n\frac1{x^2 \plus{} 1}\right|_{x \equal{} 1}:$

If $ n\equiv 0\pmod{8}$ or $ n\equiv 2\pmod{8},$ this is $ \frac {n!}{2^{\frac n2 \plus{} 1}}.$

If $ n\equiv 1\pmod{8},$ this is $ \frac { \minus{} n!}{2^{\frac {n \plus{} 1}2}}.$

If $ n\equiv 3\pmod{8}$ or $ n\equiv 7\pmod{8},$ this is $ 0.$

If $ n\equiv 4\pmod{8}$ or $ n\equiv 6\pmod{8},$ this is $ \frac { \minus{} n!}{2^{\frac n2 \plus{} 1}}.$

If $ n\equiv 5\pmod{8},$ this is $ \frac {n!}{2^{\frac {n \plus{} 1}2}}.$

[quote="Aryabhatta"]I get that the fourth derivative at 1 is -3 and second derivative at 1 is 1/2. [/quote]
If my formulas are correct, the fourth derivative should be $ \frac { \minus{} 4!}{2^3} \equal{} \minus{}3,$ and the second derivative should be $ \frac {2!}{2^2} \equal{} \frac12,$ so I agree with this.

The way I did it was:

$ f(x) \equal{} \frac {1}{1 \plus{} x^2}$

So

$ f(x)(1 \plus{} x^2) \equal{} 1$

By Leibniz's rule of differentiating product

$ \sum_{k \equal{} 0}^{n} \binom{n}{k} f^{n \minus{} k}(x)(1 \plus{} x^2)^{'k} \equal{} 0$

Where $ (1 \plus{} x^2)^{'k}$ is the $ k^{th}$ derivative of $ 1 \plus{} x^2$.

This gives us a recurrence between $ f^n(1)$, $ f^{n \minus{} 1}(1)$ and $ f^{n \minus{} 2}(1)$.

Ok, this looks correct guys, thanks. I did a stupid mistake.


[quote="ally-lee"]Hello!  :lol: We are professional of playing the wow games. If you are tired of power leveling, and you love still the games. Do not hesitate to visit our website http://www.lelelink.com or contact us at msn/email: seller@lelelink.com .[/quote]
Away, away with your shoddy advertising.
Bring us math!

He's advertising to help you on the World of Warcraft game in some way or another. I'm not sure how. A program to do stuff for you? Those are common.

It's probably just a group of people who play the game all day and are willing to go onto your account and level you up. Usually people who want to have a strong character but don't have the time to do it are the main customers for this. I wouldn't recommend it because it is really easy to have everything stolen this way.

And either way, this is AOPS, so advertise somewhere else.

advertising on AoPS? thats new :P
edit: isnt advertising illegal on a website unless the company pays the site or its from the server?

No, not much is illegal on the internet, aznmonkey, unless it cripples thousands of computers.

And this isn't the first time it's been done, and probably not the last.

Right so...end of discussion.

If you wanna talk about Warcraft, start another thread.


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
geometry proposed

Let $E$ be the further intersection of $FD$ with the circumcircle. Now, $\angle CEB=\pi-\angle CFB=\pi-\angle CAB$ and
\[\angle CDB=\pi-\angle BCD-\angle CBD=\pi-\angle DCB-(\angle CBA-\angle DBA)\]
\[=\pi-\angle DCB-(\angle CBA-(\pi-2\angle BAC))=\pi-(\pi-\angle ACB-\angle CBA - 2\angle BAC)\]
\[=\pi-\angle BAC\]
so $D$ is the orthocenter of $CBF$ (it's on the altitude from F to BC and views BC under the angle $\pi-\angle BFC$).
Then, CD is perpendicular to BF.

Using your nomenclature I have got to a point where you need to prove that $BC$ bisects       $\angle DCE$.Could you help me proceed further?
      Ashwath

And to do so I'll possibly need to prove that  the triangles $BDC$ and $BEC$ are congruent. I* have two of the three conditions , could someone help me find the third?
I have proved that $\angle BDC = \angle BEC$ and it is plain for all and sundry to see that $BC$ is a common side, but I really asm breaking my head as to how to get the third condition.
      Ashwath

[quote="ashwath.rabindranath"]And to do so I'll possibly need to prove that  the triangles $BDC$ and $BEC$ are congruent. I* have two of the three conditions , could someone help me find the third?
I have proved that $\angle BDC = \angle BEC$ and it is plain for all and sundry to see that $BC$ is a common side, but I really asm breaking my head as to how to get the third condition.[/quote]

Well, the missing condition is $DE\perp BC$. In fact, from < BEC = < BDC, it follows that the point E lies on a certain circular arc - namely, on the circular arc k which contains all points P such that < BPC = < BDC and such that the points P and D lie in different halfplanes with respect to the line BC. From $DE\perp BC$, it follows that the point E lies on the perpendicular to the line BC through the point D. Thus, the point E is a common point of the circular arc k with the perpendicular to the line BC through the point D. On the other hand, if we call G the reflection of the point D in the line BC, then we have < BGC = < BDC and $DG\perp BC$, so that we similarly obtain that the point G also is a common point of the circular arc k with the perpendicular to the line BC through the point D. But the circular arc k and the perpendicular to the line BC through the point D have only one common point; thus, G = E. Now, since the point G is the reflection of the point D in the line BC, the triangles BDC and BGC are congruent; since G = E, we thus obtain that the triangles BDC and BEC are congruent, and this is what you wanted.

Anyway, here is a simpler solution of the problem:

[quote="Mathx"]let $ABC$ be an acute triangle with $\angle BAC>\angle BCA$, and let $D$ be a point on side $AC$ such that
$|AB|=|BD|$. Furthermore, let $F$ be a point on the circumcircle of triangle $ABC$
such that line $FD$ is perpendicular to side $BC$ and points $F,B$ lie on different sides of line $AC$.
Prove that line $FB$ is perpendicular to side $AC$[/quote]

Let the altitude of triangle ABC from the vertex B intersect the circumcircle of triangle ABC at a point T (apart from B). Then, $BT\perp CA$; hence, if we succeed to show that T = F, then we will get $BF\perp CA$, and the problem will be solved. Thus, it remains to prove that T = F.

Since AB = BD, the perpendicular bisector of the segment AD passes through the point B. Also, this perpendicular bisector of the segment AD is perpendicular to the line CA (in fact, the line AD is the same as the line CA). Hence, this perpendicular bisector of the segment AD is the altitude of triangle ABC from the vertex B. Since we know that the point T lies on this altitude, we thus see that the point T lies on the perpendicular bisector of the segment AD. Hence, TA = TD, so the triangle ATD is isosceles. Thus, using directed angles modulo 180º, we have < TDA = < DAT. But < DAT = < CAT, and since the point T lies on the circumcircle of triangle ABC, we have < CAT = < CBT. Thus, < TDA = < CBT. In other words, < (DT; CA) = < (BC; BT). Hence,

  < (DT; BC) = < (DT; CA) + < (CA; BC) = < (BC; BT) + < (CA; BC) = < (CA; BT).

Now, $BT\perp CA$ yields < (CA; BT) = 90º, so that < (DT; BC) = 90º, and thus $DT\perp BC$. Hence, the point T lies on the perpendicular to the line BC through the point D. Also, we know that the point T lies on the circumcircle of triangle ABC, and, as we can easily see, the points T and B lie on different sides of the line AC (here, we use the condition that the triangle ABC is acute-angled). But there is only one point lying on the perpendicular to the line BC through the point D, lying on the circumcircle of triangle ABC, and lying on a different side of the line AC than the point B, and this point was called F in the problem. Hence, T = F. Thus, the problem is solved.

  darij


Let $ABC$ be a triangle, and let $D$ be a point on its side $BC$. Clearly, the line $BC$ is one of the two external common tangents of the incircles of triangles $ABD$ and $ACD$. Let $k$ be the other external common tangent of these incircles. Let this common tangent $k$ meet the line $AD$ at a point $K$. Find the locus of the point $K$ when the point $D$ varies on the line $BC$.

I didn't understand .Can anyone explain it.specialy $kvmedya.$.

I think I understand what problem payman_pm meant:

[color=blue]Let ABC be a triangle, and let D be a point on its side BC. Clearly, the line BC is an external common tangent of the incircles of triangles ABD and ACD. Let k be the other external common tangent of these incircles. Let this common tangent k meet the line AD at a point K. Find the locus of the point K when the point D varies.[/color]

And the answer is really surprising: The locus of the point K is the circle with center A and radius s - a, where $s =\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC.

In order to prove this, it is enough to show that AK = s - a for every point D on the side BC of triangle ABC.

And this is easy to verify:

Let the line k intersect the line BC at a point L. We can WLOG assume that the point L lies on the extension of the segment BC beyound B.

Let the incircle of triangle ABD touch the sides AD, BD and AB of this triangle at the points P, Y and P' and the line k at the point Y'. Let the incircle of triangle ACD touch the sides AD, CD and AC of this triangle at the points Q, Z and Q' and the line k at the point Z'.

Since the two tangents from a point to a circle are equal in length, we have AP = AP', BP' = BY, DY = DQ, LY = LY', KP = KY', AQ = AQ', CQ' = CZ, DZ = DZ', LZ = LZ' and KQ = KZ'.

Using this equations, we obtain

  AK + KL = (AP - KP) + (KY' + LY') = (AP' - KY') + (KY' + LY) = AP' + LY = (AP' + BP') + (LY - BP')
                = (AP' + BP') + (LY - BY) = AB + BL;

  AK - KL = (AQ - KQ) - (LZ' - KZ') = (AQ' - KZ') - (LZ - KZ') = AQ' - LZ = (AQ' + CZ) - (LZ + CZ)
               = (AQ' + CQ') - (LZ + CZ) = AC - CL.

Thus,

$2\cdot AK=\left(AK+KL\right)+\left(AK-KL\right)=\left(AB+BL\right)+\left(AC-CL\right)$
$=AB+AC-\left(CL-BL\right)=AB+AC-BC=c+b-a=\left(a+b+c\right)-2a$
$=2s-2a=2\left(s-a\right)$,

and it follows that AK = s - a. Proof complete.

A little remark about this proof: The careful reader probably has noted that the argumentation above works only if the line k intersects the line BC, i. e. if the line k is not parallel to the line BC. However, the case when k || BC can be either considered separately (hint: prove that KP = DQ), or it can be handled using the algebraicity or the continuity of the assertion of the problem.

[b]EDIT:[/b] The problem has also been posted (with different notations) at http://www.mathlinks.ro/Forum/viewtopic.php?t=58217 .

  darij

[quote="darij grinberg"]And the answer is really surprising: The locus of the point K is the circle with center A and radius s - a, where $\displaystyle s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC.

In order to prove this, it is enough to show that AK = s - a for every point D on the side BC of triangle ABC.[/quote]

Are you sure about the answer to the problem? You only proved that $K$ lies on that circle, not that the locus of $K$ is the circle. (sorry if I'm missing something)

[quote="perfect_radio"]Are you sure about the answer to the problem? You only proved that $K$ lies on that circle, not that the locus of $K$ is the circle. (sorry if I'm missing something)[/quote]

Well, as I have showed, the point K lies on the circle with center A and radius s - a, and of course, it lies on the segment AD with D tracing all the segment BC. So it is pretty clear that, while the point D is moving, the point K traces the minor arc of the circle with center A and radius s - a which is bounded by the sides AB and AC. This was pretty obvious and not the thing people usually care about in locus problems, so I didn't bother explicitely writing it down.

  Darij

By means of a well-known property ( the subject of the following interesting and useful lemma which prove easily) the proposed problem becomes more accessible. With the help of this lemma prove a great part from the problems in which appear circumscribed quadrilaterals, eventual degenerated in incircles or exincircles for certain triangles. In fact, Darij passed (in the his solution) through the proof of this lemma.

[quote][color=green][b]Lemma.[/b] Given are the triangle $ABC$ and a inner point $X$ for which define the intersections $E\in XB\cap AC$ and $F\in XC\cap AB$. Then the following statements are equivalently:
$1.\blacktriangleright\boxed {\mathrm {The\ quadrilateral\ } AEXF \mathrm {\ is\ circumscribed}}\ .$
$2.\blacktriangleright\boxed {\ XB-XC=AB-AC\ }$ (the geometrical locus of the inner point $X$ for which the quadrilateral $AEXF$ is circumscribed, is the intersection between the interior of the triangle $ABC$ and the hyperbole with the focus in the points $B$ and $C$) .
$3.\blacktriangleright\boxed {\ BF+FC=BE+EC\ }$ (the triangles $BEC$ and $BFC$ have same perimeter) .[/color][/quote] [color=darkblue][b] Proof of the proposed problem.[/b] Suppose $BC\cap k\ne \emptyset$ and define the intersections $P\in AB\cap k$ and $R\in BC\cap k$. Suppose w.l.o.g. that $B\in (RC)\ .$
$\blacktriangleleft$ Apply $(1^{\circ}\Longleftrightarrow 3^{\circ})$ to the triangle $DAR$ and the inner point $P\Longrightarrow AK+KR=AB+BR\ \ (1)\ .$
$\blacktriangleleft$ Apply $(1^{\circ}\Longleftrightarrow 2^{\circ})$ to the triangle $CAR$ and the inner point $K\Longrightarrow KA-KR=CA-CR\ \ (2)\ .$
$\blacktriangleleft$ $(1)\ \wedge\ (2)\ \Longrightarrow 2\cdot AK=AB+AC-(CR-BR)\Longrightarrow$
$AK=\frac 12\cdot (AB+AC-BC)=p-a$ (constant) .[/color]

[color=darkred][b]Some applications.[/b]
$1^{\circ}.\blacktriangleright$ Given is a ellipse with the great semiaxis $AB$. For two points $M$, $N$ which belong to ellipse so that the line $AB$ don't separates them, define the intersections $P\in AM\cap BN$, $R\in AN\cap BM$. Prove that the quadrilateral $MPNR$ is circumscribed.
$2^{\circ}.\blacktriangleright$ Given are a hyperbole with the focus $A$, $B$ and two points $M$, $N$ which belong to the same branch of the hyperbole. Define the intersections $P\in AM\cap BN$, $R\in AN\cap BM$. Prove that either the quadrilateral $AMBN$ is circumscribed or the quadrilateral $PMRN$ is circumscribed.
$3^{\circ}.\blacktriangleright$ Let $ABCD$ be a convex quadrilateral so that $\max \{A+B,A+D\}<180^{\circ}$. Define the intersections $I\in AC\cap BD$, $E\in AB\cap CD$ and $F\in AD\cap BC$. Prove that if $AB-CD=AD-BC=AC-BD$, then the quadrilaterals $BICE$ and $CIDF$ are circumscribed.
$4^{\circ}.\blacktriangleright$ See http://www.mathlinks.ro/Forum/viewtopic.php?t=76032[/color]

In the last reply I added three interesting applications of the above lemma $\uparrow\blacktriangle\uparrow$

If I let $X$ be the tangency point of the incircle of triangle $ABC$ and $BC$,can you show that $X$ and $K$ are symetric wrt. the line of the centers of that two smaller circles?


analytic geometry
geometry
geometry proposed

[color=darkblue][b]Notation.[/b] $[XYZ]$- the area of the triangle $XYZ$.

[b]Definition D1.[/b] The line $d$ separates the points $X,Y$ if and only if $X\not\in d$, $Y\not\in d$ and $XY\cap d\ne\emptyset\ .$

[b]Definition D2.[/b] The point $P$ has the barycentrical coordinates $(x,y,z)$ w.r.t. a given triangle $ABC$ if and only if :
$1.\blacktriangleright\frac{|x|}{[PBC]}=\frac{|y|}{[PCA]}=\frac{|z|}{[PAB]}$ and $x+y+z=1$ so that :
$2.1.\blacktriangleright x=0\Longleftrightarrow P\in BC$ a.s.o.
$2.2.\blacktriangleright x<0\Longleftrightarrow$ the line $BC$ separates the points $A,P$ a.s.o.
$2.3.\blacktriangleright x>0\Longleftrightarrow$ $P\not\in BC$ and the line $BC$ doesn't separate the points $A,P$ a.s.o.

[b]Remark.[/b] If the triangle $ABC$ is fixed, then for any point $P$ from the its plane
exist and are uniquely the real numbers $x,y,z$ so that $x\cdot\overline{PA}+y\cdot\overline{PB}+z\cdot\overline{PC}=\overline{0}\ .$

[b]Proposed problem.[/b] Given are a triangle $ABC$ and two points $E,F$ so that the line $EF$ doesn't cross the triangle $ABC$.
For a point $P$ with the barycentrical coordinates $(x,y,z)$ w.r.t. the  triangle $ABC$there is the relation : $x\cdot [AEF]+y\cdot [BEF]+z\cdot [CEF]=[PEF]$

[b]Particular case.[/b] See[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=105232

Your notation $[XYZ]$ shows geometric ares or algeraic areas ?

$[XYZ]$ is geometrical area (without sign).

[quote="Virgil Nicula"][color=darkblue]

[b]Proposed problem.[/b] Given are a triangle $ABC$ and two points $E,F$.
For a point $P$ with the barycentrical coordinates $(x,y,z)$ w.r.t. the  triangle $ABC$there is the relation :
$x\cdot [AEF]+y\cdot [BEF]+z\cdot [CEF]=[PEF]$

[b]Particular case.[/b] See[/color] http://www.mathlinks.ro/Forum/viewtopic.php?t=105232[/quote]

I think It should be algebraic areas, because try with $E,F$ are midpoint of $AB,AC$and $G(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ we see
$\frac{1}{3}.[BEF]+\frac{1}{3}.[CEF]-\frac{1}{3}.[AEF]=[GEF]$ while follow your proposed $\frac{1}{3}.[BEF]+\frac{1}{3}.[CEF]+\frac{1}{3}.[AEF]=[GEF]$

I am sorry ! I forgot say that the line $EF$ doesn't cross the triangle $ABC$. Again, sorry ! Thanks.

ok youe are right, we can solve it by outer product:
$2(\sum x[AEF]) = \sum x(\vec{AE}\wedge\vec{AF})=\sum x(\vec{AP}+\vec{PE})\wedge(\vec{AP}+\vec{PF})=2(x+y+z)[PEF]$ note $\sum \vec{PA}=\vec{0}$ when $EF$ doesn't cross $\triangle ABC$ then triangle $AEF,BEF,CEF$ and$PEF$ has the same direction then they positive, we have this equality!


What is the graph of Re(z)=1?

I know z=x+iy...and the real parts are x and y. does that mean (1,1)?  Or, since the x axis in a complex plane is the real line, then is it just a vertical line at x=1? I don't understand completely.

Yes, the real part of a complex number is just its $ x$-coordinate, so the graph of all complex numbers having real part 1 is just the vertical line $ x\equal{}1$.

[quote="limac"]Where you are specifically getting the (1,1) from? The graph of any complex number is of the form $ (\text{Re(z)},\text{Im(z)}),$ in your case.[/quote]I wouldn't say 'the graph', the proper way would be to say 'the coordinates of any complex number on the [b]complex plane[/b] is of the form $ (\text{Re(z)},\text{Im(z)})$.'


[quote="N.T.TUAN"]Let $a_{n}>a_{n-1}>...>a_{1}.$
By Lagrange's Identity http://mathworld.wolfram.com/LagrangesIdentity.html
We have $n\sum_{k=1}^{n}a_{k}^{2}-(\sum_{k=1}^{n}a_{k})^{2}=\sum_{1\leq i<j\leq n}(a_{j}-a_{i})^{2}\geq \sum_{1\leq i<j\leq n}(j-i)^{2}=\frac{n^{4}-n^{2}}{12}$, qed.[/quote]

Congratulation man. I was thinking a lot on this problem. It is a very nice solution.

[quote="cckek"]

Congratulation man. I was thinking a lot on this problem. It is a very nice solution.[/quote]

Thank you  :blush:


An aquarium contains 280 tropical fish of various kinds. If 60 more clownfish were added to the aquarium, the proportion of clownfish would be doubled.
How many clownfish are in the aquarium?

Let $ x$ be the initial number of clownfish.  We're told that $ \frac{2x}{280}\equal{}\frac{x\plus{}60}{340}$.  Hence $ 680x\equal{}280x\plus{}16800$.  Solving for $ x$ yields $ x\equal{}42$ so now there are 84 clownfish.


geometry
3D geometry
sphere
tetrahedron
circumcircle
conics
ellipse

The sphere inscribed in a tetrahedron $ABCD$ touches face $ABC$ at point $H$. Another sphere touches face $ABC$ at $O$ and the planes containing the other three faces at points exterior to the faces. Prove that if $O$ is the circumcenter of triangle $ABC$, then $H$ is the orthocenter of that triangle.

Suppose you have an arbitrary ellipse with foci $ E, F$ inscribed in a $ \triangle ABC.$ Its pedal circle $ (P)$ (centered at the midpoint $ P$ of $ EF$) cuts the triangles side $ BC$ at $ U, X$ the side $ CA$ at $ V, Y,$ and the side $ AB$ at $ W, Z.$ Pedal circle is locus of the feet of perpendiculars from foci $ E, F$ to the ellipse tangents, therefore, $ EU, FX \perp BC,$ $ EV, FY \perp CA,$ and $ EW, FZ \perp AB.$ Now you have 2 points $ E, F$ with pedal triangles $ \triangle UVW, \triangle XYZ$ and common circumcircle $ (P).$ The $ \triangle CUV \sim \triangle CYX$ are oppositely similar, because $ U, X, V, Y$ are concyclic, $ XY$ is antiparallel to $ UV$ WRT the $ \angle C.$ Circumcenter of the $ \triangle CUV$ is the midpoint $ E_c$ of $ CE$ on account of the right angles $ \angle CUE, \angle CVE$ and circumcenter of the $ \triangle CYX$ is the midpoint $ F_c$ of $ CF$ on account of the right angles $ \angle CYF, \angle CXF.$ Therefore, $ CE_c, CF_c$ are isogonal WRT the $ \angle C$ and so are $ CE, CF.$ Similarly $ AE, AF$ are isogonals WRT the $ \angle A$ and $ BE, BF$ WRT the $ \angle B.$

As for the ellipse pedal circle. Let $ a$ be a line and $ E, F$ points on the same side of this line. Where is the point $ X \in a,$ such that the sum $ EX \plus{} FX$ is minimum ? Reflect $ F$ in $ a$ into $ F'.$ $ EX \plus{} FX \equal{} EX \plus{} F'X$ and for a minimum, $ E, X, F'$ must be collinear. Therefore, $ a$ must bisect the angle $ \angle EXF$ externally. But $ a$ also must be tangent to an ellipse with foci $ E, F$ and passing through $ X,$ otherwise the ellipse would intersect $ a$ and points with smaller distance sum would exist on $ a.$ Therefore, if $ a$ is an arbitrary ellipse tangent at $ X,$ reflection $ F'$ of the focus $ F$ in $ a$ lies on $ EX.$ The distance $ p \equal{} EF' \equal{} EX \plus{} XF' \equal{} EX \plus{} XF$ is constant (equal to the ellipse major axis $ p$), therefore $ F'$ is on a circle $ (E)$ with radius $ p \equal{} EF'.$ $ P$ is the midpoint of $ EF$ (the ellipse center) and $ Q$ is the midpoint of $ FF'$ (foot of the perpendicular $ FF'$ to the ellipse tangent $ a$). $ PQ$ is a midline of the $ \triangle EFF',$ $ PQ \equal{} \frac {EF'}{2} \equal{} \frac {p}{2}$ (the ellipse semimajor axis), $ Q$ is on a circle $ (P)$ with radius $ PQ \equal{} \frac {p}{2}.$

Although I do not speak polish, I think I was able of translate the official solution of this problem, from the web page of the Polish Olympiad. It seems to read in this way:
Let K and L the points of tangency of the insphere with the plans ACD and BCD, respectively. Of the same manner, let P and Q the points of tangency of the excribed sphere with the plans ACD and BCD, respectively. The triangles KCP and LCQ are equal, because CK = CL, CP = CQ, KP = LQ.
We made a rotation with edge AC of the triangle CKP in a such way the point K goes to the point H; in this rotation point P transforms in P'.
Similarly, the triangle CLQ rotates (with edge of rotation  BC) in a such way that L goes over the point H; in this rotation, Q transforms in Q'.
We will finish the problem in the plane ABC:
Triangles HCP' and HCQ' are equal, and P'Q' is perpendicular to CH.
Quadrilaterals AOCP' and BOCQ' are rhombuses, and from this we deduce AB parallel to P'Q'. Therefore CH is perpendicular to AB.
In a similar way we prove BH perpendicular to AC.
So, H is the orthocenter of the triangle ABC.

I am so impressed at your great solution,yetti.
If I am just near you, I will kiss you three times. :wow:

To "28121941"
Thanks so much. I am happy at your nice solution,too. :blush:

Wow :love: !!! But really, I am an old guy (and so is the proof), so save it for the others.

[quote=28121941]As a matter of fact, the solution posted is not mine....in the polish web page the solutions are credited to "mek" , "wp" (both names are know by me) and a third person.
The problem is really nice, I think. I wonder if some of the contestants gave a solution like that of yetti...[/quote]
mek = Marcin E. Kuczma University of Warsaw
wp = Waldemar Pompe University of Warsaw
jwr = Jarosław Wróblewski University of Wrocław.


i don't think there is a solution. is there?

When you multiply the first equaton by 3 and the second one by 2, you get 12x - 18y = 15 for one, and 12x - 18y = 8. It is impossible.


The World Cup will take place in the morning of 30 October, where each mental calculator will have to solve tasks in six categories (addition of ten 10-digit numbers, multiplication of two 8-digit numbers, extracting the square root from a 6-digit number, calculating the day of the week and two tasks of surprise). There are several prizes for the best participants. The overall winner will be awarded the title "World Cup winner for mental calculation".

http://www.recordholders.org/en/events/worldcup04/

[quote="arthur"]great
l heard about a german who has a huge memory and is able to calculate mentally divisions of odd primes numbers...[/quote]

I saw him on TV.He was able to do divisions between numbers with 40 digits and more  :D


trigonometry
inequalities
geometry proposed
geometry

let ABC be a triangle
prove that
$\sum \frac{\sin ^2 A \sin ^2 B}{\sin ^2 C} \geq \frac94$.

The sum is over what? Moreover if it is a product the inequality does not hold (let A - > 0).
In another hand, the following inequality holds :
sin 2 (A) + sin 2 (B) + sin 2 (C) \leq 9/4.

Pierre.

No, I think he refers to a problem of Hojoo Lee, in fact to a variation of Iran 1996:

       [tex]\sum\frac{\sin^2A\cdot\sin^2B}{\sin^2C}\geq\frac{9}{4} [/tex]


calculus
integration
geometry
rectangle
complex analysis

I have tried to evaluate the following integral. It is not very diffcult but I still need help.

int {e^(1/z) dz} on the boundary of the rectangle centered at
(1+ 2i) of width 5 and height 6.
 I divided it into four contours where do I go from there?

Thank you

$\int\!\!\!\!\!\!\square \;e^{1/z}dz=\oint e^{1/z}dz$

Right?  So that's just $2\pi i$ since the residue at 0 is 1.  Perhaps you still wish to calculate it around the square contour.  Me, I'd be interested in how each of the 4 contours contribute to $2\pi i$.  How about this part:

$\int_{3.5-i}^{3.5+5i}e^{1/z}dz$

Well, you could extract the real and imaginary parts and use:

$\oint f(z)dz=\oint (udx-vdy)+i\oint (vdx+udy)$

but the components are messy and I don't think the antiderivatives are easily expressed so suppose we do  everything numerically for starters.  Here's how I'd do the contour over the right vertical line in Mathematica:

exp[x_,y_]:=ComplexExpand[Exp[1/(x+i y)]];

NIntegrate[exp[3.5,y] i ,{y,-1,5}]

Know why the i is in there right?

Now, do the other three in the right orientation, add them up and should get close to $2\pi i$

What does it really mean to do such a contour integration over the rectangle? Can it be made to have a physical/geometric meaning?


I must be missing something here... can't you just take (9-2-2-2-2) and add as many of them as you need to get whatever number?

i thought you can only use each of 2, 0, 0, 9

Yeah, one of each.  However, you can do things like

$ (\sqrt{9})!\equal{}6$.

would using the square root symbol use the two, even though it is typically not written down?

i could figure out a couple... but then i got lazy.............


[quote="can_hang2007"][quote="hedeng123"]Let $ x,$ $ y,$ $ z$ be positive real numbers. If $ k \ge 0,$ then
$ \frac {1}{{{x^2}\sqrt {{x^2} \plus{} k{y^2}} }} \plus{} \frac {1}{{{y^2}\sqrt {{y^2} \plus{} k{z^2}} }} \plus{} \frac {1}{{{z^2}\sqrt {{z^2} \plus{} k{x^2}} }} \ge \frac {{3 }}{\sqrt {1 \plus{} k}{{x}{y}{z}}}.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-60-unknown-author.html[/quote]

I am sorry, I can not access this link,Can you upload your solution?

thank you very much.

 :lol:

[quote="can_hang2007"][quote="hedeng123"][quote="can_hang2007"]Ok, you can download it here: http://cid-b2deb869e6bb1093.skydrive.live.com/self.aspx/Inequality%20Problems/Inequality%2060.rar[/quote]

I have been to your space,downloaded it inside the 60 Inequality Problems,unfortunately, when the extracted files, the computer prompts file has been damaged,what a pity. :([/quote]
I checked the file, and it is ok, dear hedeng123. I will guide you to read my file:

- First, rename the extension .RAR_ as .RAR.
- Second, use a djvu reader program to view files with extension .djvu.[/quote]

Thank you very much,I have done according to the steps,unfortunately, still the same,I changed a computer, is still the same.

I know where the problems lie in. :lol:

[quote="can_hang2007"]Actually, there is something happened to me and I dont want to upload my files on Mathlinks anymore. But as your interest, I upload my solution here. This is pdf file and I am sure that you can read it.
[quote="xzlbq"]dear V. Q. B. Can:
      you post to me, so I get your e_mail.
            BQ[/quote]
Dear xzlbq,

My e-mail is babylearnmath@yahoo.com. Nice to meet you.

Regards
Can[/quote]

thank you very much.Now I can read the files inside your space.


geometry
inequalities
analytic geometry
blogs
triangle inequality

2. Alice and Bruno are looking for an apartment in Ideal City. Alice works as an acrobat at amusement park $A = (-3, -1)$, and Bruno works as a bread taster in bakery $B = (3,3)$. Being ecologically aware, they walk wherever tey go. They have decided their apartment should be located so that the distance Alice has to walk to work plus the distance Bruno has to walk to work is as small as possible. Where should they look for an apartment?

3. In a moment of chivalry, Bruno decides that the sum of the distances should still be a minimum, but Alice should not have to walk any farther than he does. Now where could they look for an apartment?

4. Alice agrees that the sum of the distances should be a minimum, but she is adamant that they both have exactly the same distance to walk to work. Now where could they live?

5. After a day of fruitless apartment hunting they decide to widen their area of search. The only requirement they keep is that they both be the same distance from their jobs. Now where should they look?

have fun  :P

[hide]
2. Wouldn't the minimum distance just be anywhere on the line connecting $(-3,-1)$ and $(3,3)$?  Because if you use any other point by the triangle inequality the sum will be greater...

3. on the line, from the midpoint $(0, 1)$ until $(3,3)$

4. at the midpoint of the two points, $(0,1)$.

5. Let the point be $(x,y)$.  So using the distance formula we have $\sqrt{(x+3)^2+(y+1)^2}=\sqrt{(x-3)^2+(y-3)^2}$.  Squaring, expanding, and simplifying we have anything on the line $3x+2y=2$.

Edit: I forgot to do the other questions...woops.  I put answers for them now.
[/hide]

isn't obvious that the minimum total distance is a line between the 2 points,
then from the midpoint to alice is so she walks less,
then the midpoint for them going the same distance,
the locus of points equidistant from 2 points is the perpendicular biscector of the line connecting the 2 points

hmm ...  i think you two may be misunderstanding the point of taxicab geo

taxicab is basically coordinate geometry but using only the lattice points

so basically think of this problem as your on a grid of lattice points, and your pretty much going from dot to dot

meanwhile here's another problem... this one i was wondering if anyone could do it and check with my answer

The telephone company wants to set up pay-phone booths so that everyone living within twelve blocks of the center of town is within four blocks of a pay phone. How few booths can they get by with, and where should they be located?

[hide]

i got 9... and basically if u draw the town (just a square with side length $12$), you draw in 9 squares, and the center of those squares is where the telephone booths should be

[/hide]

wouldn't it be a circle with r = 12, and how many circles with r = 4 will cover the same area
that is more difficult...i am not sure what you would do, if you have 4 small circles, such that their centers form a square, and they are tangent, that might be the most efficient way to cover area

actually in taxicab geometry, the locus of points from say point A, is actually a square, not a circle

where can i learn more about this? i have never encountered anything like this before

well i have a book on it... umm... i kno 10000th users has a blog and he posts a lot of taxicab geo on there too

other than that... just try searching random places? i dont think it's that well known of a topic

interesting, i might pursue this, but i still have a few chapters in AoPS 1 and the whole other book, and i also have school working me to the bone, so i will weigh the importance of this...it is pretty rare


A total of $mn$ people are placed into $m$ groups each having $n$ people. How many distinct ways can we do this?

Not sure if there's a simple closed form of this...not sure if it's right either.

[hide]$\frac{{mn\choose n}{mn-n\choose n}{mn-2n\choose n}\cdots{2n\choose n}}{m!}$[/hide]

[quote="Rzeszut"][hide]It is just $\binom{mn}{\underbrace{n,\ldots,n}_{m}}= \frac{(mn)!}{\left(n!\right)^{m}}$. [/hide][/quote]

But don't you have to divide by $m!$ because the groups of people are indistinguishable?

if we have mn distinguishable balls and n indistinguishable urns...

see stirling numbers of the second kind

Is there an intuitive way to approach this without using the explicit formula for stirling numbers of the second kind?


I learnt this method on the internet but I really can't remember the link.
Anyway, this trick is quite interesting and let me share it with all of you.

"What does 777 represent?"
"Luckiness!"

"And what is the meaning of the number 13?"
"Unluckiness!"

"Use a calculator, multiply 777 to your age, then multiply the result by 13, what is your answer?"

  :o  :o  :o  :)  :)  :)

[quote="luimichael"]You are so young, DonkeyKong.  :)[/quote]
Actually I am not..
I am an 8th grader..so this is my last year I can participate in MathCounts.. :(
One thing I regret is that I started from 7th, not 6th...but oh well..

121212....
xD
i remember this from like 2nd grade, but it was 7 * 13 * 11= 100100 and i would amaze everbody else...
someone else taught it to me..

I'm $ 0.0659340659$ years old. Thus, I received $ 666$

technically im 121212 soon to be 131313 (on nov. 8!!)

Who's rating all the posts one? I rated the first few back up, but I ran out of rates.

I only rated the last few, because it's really unnecessary to post "I got 141414!" or something like that. In fact, it's quite annoying to view topics hoping to find something useful or at least interesting, and instead finding a bunch of almost identical posts.

[size=150][color=green][i]"Able was I ere I saw Elba.''[/i][/color][/size]  [b]  by Napoleon Bonaparte[/b]

Talking about palidrome, the above famous phrase immediately appears in my mind.

Do you know what is the longest palidrome word (a single word which is the mirror image of itself)?

[quote="DonkeyKong"][quote="Fraenkel"][hide]
Is it TATTARRATTAT?
[/hide][/quote]
That is a sound..
If that's the case, mine should work too.
It's HAHAHAHAHAHAAHAHAHAHAHAH (laughter) :lol:[/quote]

:lol: I totally agree with you!
The word I gave was found off the internet. It was in James Joyce's Ulysses.

Another interesting one I found was
[hide] REDIVIDER [/hide]

Hmm... is it okay if it is a place name?

[hide]Wassamassaw (South Carolina) [/hide]

* I found a site where it contains whole bunch of palindrome words and phrases and sentences.
   It's [url]http://www.rinkworks.com/words/palindromes.shtml[/url]


A satellite of mass 5M orbits the earth in a circular orbit. At one point in its orbit,  the satellite explodes into two pieces, one of mass M and the other of mass 4M. After the explosion the mass M ends up travelling in the same circular orbit, but in opposite direction. After explosion the mass 4M is in

A) bound orbit
B) unbound orbit
C) partially bound orbit
D) insufficient data

Please provide complete logic.

I believe, but am not entirely positive, that the other mass has to also be in the same orbit, but in the opposite direction as the M piece.  When the satellite of 5M orbits, it has some angular momentum so when it explodes, there has to be conservation of angular momentum, so it will have to orbit as well at some angular velocity.  But I could be wrong.

shud b (A) perhaps!!!
the part wud come in higher orbit
orbital vel is $ \sqrt{gR}$
by consv of linear momentum ,we get vel. of $ 4m$ as $ \sqrt{3/2gR}$which is less than escape speed

sorry !!
correction
i made a calculation mistake
v comes out to be $ \sqrt{\frac{9gR}{4}}$
which is   greater than escape speed --sso body escapes in a hyperbolic path..

answer shud be (B)


Find all integer sequences of the form $ x_i, 1 \le i \le 1997$, that satisfy $ \sum_{k\equal{}1}^{1997} 2^{k\minus{}1} x_{k}^{1997}\equal{}1996\prod_{k\equal{}1}^{1997}x_k$.

Here is the hint from the Hungary-Israel Mathematical Olympiad book:

[hide="Hint"]
Note that if $(x_1,\dots,x_{1997})$ is a solution, then so is $\left(\frac{x_1}{2},\dots,\frac{x_{1997}}{2}\right)$.
[/hide]


modular arithmetic
superior algebra
superior algebra unsolved

Let $ S$ be a finite family of algebraic numbers and suppose that for all positive integers $ k$, the sum of $ k$-th powers of the elements of $ S$ is an integer. Show that all elements of $ S$ are actually algebraic integers.

Let $ a_{i},\ i=\overline{1,n}$ be the algebraic numbers in question, and let $ K=\mathbb Q\left(a_{i},\ i=\overline{1,n}\right)$ be the extension of $ \mathbb Q$ obtained by adjoining the $ a_{i}$. It suffices to prove that all non-archimedean valuations of $ K$ give non-negative values when evaluated on the $ a_{i}$. Let $ \nu$ be such a valuation. Instead of $ K$, we can just as well work in the completion of $ K$ with respect to $ \nu$. Denote this completion by $ K$ also, since this is the only field I'll be working with from now on. The problem has now become:

Let $ K$ be a finite extension of some $ p$-adic field $ \mathbb Q_{p}$, with normalized valuation $ \nu$ and ring of integers $ O$. Also, let $ a_{i}\in K,\ i=\overline{1,n}$ be such that for all $ k\in\mathbb N$, the sum $ s_{k}=a_{1}^{k}+\ldots+a_{n}^{k}$ belongs to $ O$. Then all the $ a_{i}$ are integers.

Take $ \pi\in K$ with $ \nu(\pi)=1$. Every $ a_{i}$ can be written as a quotient between some $ b_{i}\in O$ and a power of $ \pi$; we assume that the $ a_{i}$ are [i]not[/i] all integers. After bringing everything to a common denominator, the hypothesis then implies that $ \pi^{k}\ |\ t_{k}=\sum b_{i}^{k},\ \forall k\in\mathbb N\ (*)$ and at least one of the $ b_{i}$ is not divisible by $ \pi$. It makes no difference in $ (*)$ if we keep only such $ b_{i}$, so we may as well assume that none of the $ b_{i}$ are divisible by $ \pi$. Taking $ k$ to be $ |O/\pi O|p^{u}$ for a sufficiently large positive integer $ u$ gives the desired contradiction: for all $ i$ we'll have $ b_{i}^{k}=1\pmod{\pi^{t}}$ (with $ t$ increasing as $ u$ does), so $ t_{k}=n\pmod{\pi^{t}}$ if $ t$ is larger than $ \nu(n)$, in which case $ \pi^{k}$ cannot divide $ t_{k}$ unless it divides $ n$; of course, this can't happen for large $ u$.


calculus
integration
logarithms
trigonometry
calculus computations

Let $ I = \int^{\pi /2}_{0}\frac{\ln(\sin x)}{\sin x}\,dx$.

Can someone provide an hint on how to compute this integral? Thanks.

Do you mean in the proper sense? Or even in the Principal Value sense?

I got the primitive down to showing $ \int \frac{u}{\sqrt{1-(e^{u})^{2}}}$ where u= ln ( sin x)

Hi Gib Z,
The Cauchy principal value doesn't exist here.
If $ I(u)=\int_{u}^{\pi/2}\frac{ln(sin(x))dx}{x}$ then $ I(u)\sim\frac{-ln(u)^{2}}{2}$ when $ u\rightarrow{0+}$.


(1) find all function of type f(x)*f(-x) = f(x^2).

I really don't understand your questions  :?: , $ f$ is from where to where ?

We can define the function nearly arbitrarily on $ {\bf R}_{> 0}$ (only condition is that if $ f(x) \equal{} 0$ then we must also have $ f(x^2) \equal{} 0$), set $ f(0) \equal{} 0$ or $ f(0) \equal{} 1$, and this determines $ f$ on the negative reals by $ f(\minus{}x) \equal{} \frac{f(x^2)}{f(x)}$ provided $ f(x) \neq 0$, $ f(\minus{}x)$ arbitrary otherwise.


I've put the code at the bottom in a file called aops.tex and included in a file with \include{aops} but when I do this: [code]\begin{solution}{5.1.13}[/code] or some other number with dots in the number, it just prints the string before the dot as the problem number. What do I do to make it realize that the dot and the following digits are also parts of the number?

Thank you!

Billy

[code]\newcounter{qnumber}

\newcommand{\realname}[1]{\newcommand{\printrealname}{#1}}
\newcommand{\username}[1]{\newcommand{\printusername}{#1}}
\newcommand{\userid}[1]{\newcommand{\printuserid}{#1}}

\setlength{\headheight}{0.75in}
\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\voffset}{-.5in}
\setlength{\headsep}{10pt}
\setlength{\textwidth}{6.5in}
\setlength{\headwidth}{6.5in}
\setlength{\textheight}{8in}
\lhead{Student: \printrealname\ \\Username: \printusername\ \\ID\#: \printuserid}
\lfoot{}
\cfoot{}
\rfoot{Page \thepage\ of Problem \theqnumber}
\renewcommand{\headrulewidth}{0.5pt}
\renewcommand{\footrulewidth}{0.3pt}
\setlength{\textwidth}{6.5in}

\newenvironment{solution}[1]{\setcounter{qnumber}{#1}\setcounter{page}{1}}{\eject}
[/code]
(yes, that is somewhat adapted from the USAMTS style sheet, but I don't use it on the USAMTS.)

may be you need to use a different counter for each level

You can't set a LaTeX counter ("qnumber" is this example) to be something like "5.1.13" -- it has to be an integer.

As Ledurt suggested, if you want to do something like this, you'll have to have separate counters for each level (in other words, you'll have to have separate counters to keep track of the "5", "1", and "13").


(Canada 1996)
Εαν $ a,b,c$ είναι ρίζες της $ x^3 \minus{} x \minus{} 1$, να βρείτε την τιμή της παράστασης :
$ A \equal{} \frac {1 \minus{} a}{1 \plus{} a} \plus{} \frac {1 \minus{} b}{1 \plus{} b} \plus{} \frac {1 \minus{} c}{1 \plus{} c}.$

Well.
Αφού δεν βλέπω ενδιαφέρον(ή δεν υπάρχουν juniors) δίνω ένα οφθαλμοφανές

[hide="hint"]Vieta’s Formulas :wink: [/hide]

Εγω δεν θα την ελεγα Junior, εκτος αν λυνεται και πιο απλα.
Βαζω μια λυση στα γρηγορα επηδη δεν εχω καθολου χρονο. Ελπιζω να μην μου εχει ξεγυγει κατι.

Θετουμε $ x \equal{} \frac {1 \minus{} a}{1 \plus{} a}$ και κυκλικα. Ειναι $ abc \equal{} 1$, αρα θετοντας $ a \equal{} \frac {k}{m}$ και κυκλικα και χρησημοποιοντας το πολυωνυμο Lagrange λαμβανουμε $ xy \plus{} xz \plus{} yz \equal{} \minus{} 1$.

Ειναι ακομα $ a \equal{} \frac {1 \minus{} x}{1 \plus{} x}$ και κυκλικα με $ abc \equal{} 1, a \plus{} b \plus{} c \equal{} 0$ οποτε αντικαθιστοντας στην εξισωση και χρησημοποιοντας την ταυτοτητα του Euler εχουμε:
$ 2\sum{\frac {1}{x \plus{} 1}} \equal{} 3$ και μετα τις πραξεις $ x \plus{} y \plus{} z \minus{} 3xyz \minus{} (xy \plus{} xz \plus{} yz) \equal{} \minus{} 3$

Τελος απο $ \frac {(1 \minus{} x)}{(x \plus{} 1)}\frac {(1 \minus{} y)}{(y \plus{} 1)}\frac {(1 \minus{} z)}{(z \plus{} 1)} \equal{} 1$ μετα τις πραξεις λαμβανουμε $ x \plus{} y \plus{} z \equal{} \minus{} xyz$

Απο τις παραπανω ευκολα προκειπτει:
$ 4(x \plus{} y \plus{} z) \equal{} \minus{} 3 \plus{} (xy \plus{} yz \plus{} yz) \equal{} \minus{} 4 \Rightarrow x \plus{} y \plus{} z \equal{} \minus{} 1$

Πιστεύω ότι είναι για αρχιμήδη των μικρών,και νομίζω ότι βγένει $ 1$.
Αυτή είναι η λύση μου
$ a\plus{}b\plus{}c\equal{}0,ab\plus{}bc\plus{}ca\equal{}\minus{}1,abc\equal{}1$

O παρανομαστής είναι:
$ (a\plus{}1)(b\plus{}1)(c\plus{}1)\equal{}ab\plus{}a\plus{}b\plus{}1\plus{}abc\plus{}ca\plus{}cb\plus{}c\equal{}1$
Άρα αρκεί να βρούμε τον αριθμητή:
$ \sum_{cyc}(1\minus{}a)(1\plus{}b)(1\plus{}c)\equal{}\sum_{cyc}1\plus{}b\minus{}a\minus{}ab\plus{}c\plus{}cb\minus{}ca\minus{}abc\equal{}3\plus{}a\plus{}b\plus{}c\minus{}ab\minus{}bc\minus{}ca\minus{}3abc\equal{}1$
Eλπίζω και εγώ να μην έχω καμία βλακεία....

Πάλι με Vieta αλλά λίγο διαφορετικά:

$ \frac{1\minus{}a}{1\plus{}a} \plus{} \frac{1\minus{}b}{1\plus{}b} \plus{} \frac{1\minus{}c}{1\plus{}c} \equal{} \minus{}3 \plus{} 2\left(\frac{1}{1\plus{}a} \plus{} \frac{1}{1\plus{}b} \plus{} \frac{1}{1\plus{}c} \right)$

$ a,b,c$ ρίζες του $ x^3 \minus{} x \minus{} 1$

άρα $ (1\plus{}a),(1\plus{}b),(1\plus{}c)$ ρίζες του $ (x\minus{}1)^3 \minus{} (x\minus{}1) \minus{} 1 \equal{} x^3 \minus{} 3x^2 \plus{} 2x \minus{} 1$

άρα $ 1/(1\plus{}a),1/(1\plus{}b),1/(1\plus{}c)$ ρίζες του $ x^3 \minus{} 2x^2 \plus{} 3x \minus{} 1$

άρα $ 1/(1\plus{}a) \plus{} 1/(1\plus{}b) \plus{} 1/(1\plus{}c) \equal{} 2$ και $ \frac{1\minus{}a}{1\plus{}a} \plus{} \frac{1\minus{}b}{1\plus{}b} \plus{} \frac{1\minus{}c}{1\plus{}c} \equal{} 1$.

Οντως 1 βγαινει, σιγουρα εχω χασει καποιο προσημο στις πραξεις.


projective geometry
geometry proposed
geometry

Through a point $P$ exterior to a given circle pass a secant and a tangent to the circle. The secant intersects the circle at $A$ and $B$, and the tangent touches the circle at $C$ on the same side of the diameter through $P$ as the points $A$ and $B$. The projection of the point $C$ on the diameter is called $Q$. Prove that the line $QC$ bisects the angle $\angle AQB$.

O==center(ABC); AO = BO ==> <BAO = <ABO; PC^2 =PQ*PO = PA*PB ==> AQOB cyclic ==> <AQP = <ABO = <BAO = <BQO = a. <CQA = 90-a = <CQB.

Let the lines CQ and AB meet at X.

From the definition of the line CQ, it directly follows that the line CQ is the polar of the point P with respect to our circle. Hence, by the harmonic property of a polar, the point X, where this line CQ meets the line AB, is the harmonic conjugate of the point P with respect to the segment AB. In other words, the points P and X divide the segment AB harmonically. Hence, the rays QP and QX divide the angle formed by the rays QA and QB harmonically. But on the other hand, we know that the rays QP and QX are perpendicular (in fact, the ray QP is the diameter of the circle through the point P, and the ray QX is the ray CQ). It is well-known that if two perpendicular rays divide an angle harmonically, then these two rays are the two angle bisectors (internal and external) of the angle. Hence, the rays QP and QX are the two angle bisectors of the angle AQB. Since the ray QX is the ray QC, it follows that the ray QC bisects the angle AQB. Proof complete.

This is in fact Darij's solution, re-posted.

This easy and nice problem is a immediate consequence of the properties of a harmonical division. See and other problems to http://www.mathlinks.ro/Forum/viewtopic.php?t=46146