No Olympiad Problems? [Resolved]

What happened to the new problems for this week?

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Comments

Oh man..I even woke up early to do the new problems today..

Thaddeus Abiy - 8 years, 2 months ago

same over here.. sad. :(

Ahmed Abid Abbash - 8 years, 2 months ago

Hi everyone,

This was a glitch and has been corrected. You can see new Olympiad problems now. We are very sorry for the mistake.

Happy problem solving!

Silas Hundt Staff - 8 years, 2 months ago

why are some of them recycled though??Ive seen them before..or is it just me

Thaddeus Abiy - 8 years, 2 months ago

Yes, i too saw at least 2 repeated questions.

Rohan Rao - 8 years, 2 months ago

@Rohan Rao – They didn't even bother to change the number of people it's been solved by..I don't like remembering problems I like solving them..Please brilliant,If you are going to recycle problems,At least change the numbers

Thaddeus Abiy - 8 years, 2 months ago

@Thaddeus Abiy – I think they must have been working on the new problem bank this week and thus didn't have time to create/present new problems.

David Altizio - 8 years, 2 months ago

Yeah...

Tan Li Xuan - 8 years, 2 months ago

They are probably behind, or saving something because of the revamp of curriculum mathematics that is being done. So I bet they are working very hard this week! Lets find problems other places. If you have any good problems you would like to share, please post them as a reply so that everyone can have problems to do while waiting.

Here, An auditorium has a rectangular array of chairs. There are exactly 14 boys seated in each row and exactly 10 girls seated in each column. If exactly 3 chairs are empty, find the maximum number of chairs in the auditorium.

Brock West - 8 years, 2 months ago

If you wanna keep this going, then here's a GREAT problem from the 1983 ARML competition:

In an isosceles triangle, the altitudes intersect on the inscribed circle. Compute the cosine of the vertex angle.

David Altizio - 8 years, 2 months ago

Thanks for posting! :)

Ahaan Rungta - 8 years, 2 months ago

Should I post a solution for this one? Its a fairly simple exercise in trigonometry, but there will be people still trying to solve this one.

I arrived at the answer 1/9 (assuming you are taking the cosine of the 'unique' angle)

Gabriel Wong - 8 years, 2 months ago

@Gabriel Wong – It says 'vertex angle', so yea your answer is correct.

Don't post the solution though; while its a relatively simple problem I still like it for having such an interesting condition.

David Altizio - 8 years, 2 months ago

Under the assumption that boys and girls cannot share a chair, let number of rows and columns be r and c. (r >= 14; c >= 10)

Then rc = 3+14r+10c

rc-14r-10c-3 = 0

rc - 14r - 10c +140 - 143 = 0

(r-10)(c-14) = 143

Now, the possibilities for (r,c) are (1,143), (11,13), (13,11) and (143,1). Checking all 4, the number of chairs in each is (11)(157), (21)(27), (23,25), (153,15).

Clearly (23)(25) > (21)(27) and (11)(157)<(153)(15). (if a>c>d>b>0 and a+b = c+d, ab < cd) Also, (153)(15) > (23)(25) obviously.

The maximum number of chairs is thus (153)(15) = 2295 (with 2292 chairs filled up)

Gabriel Wong - 8 years, 2 months ago

Why doesn't, say, r=21 and c=27 work?

David Altizio - 8 years, 2 months ago

@David Altizio – was wrong initially, edited

Gabriel Wong - 8 years, 2 months ago

why ? defining X = r-10, Y = c-14, then that means X = 1,Y = 143 -> r = 11, c 157 -> rc = 1727 X = 11, Y = 13 -> r = 21, c = 27 -> rc = 567 X = 13, Y = 11 -> r = 23, c = 25 -> rc = 575 X = 143, Y = 1 -> r = 153, Y = 15 -> rc = 2295 (This is the max number of chairs)

Raymond Christopher Sitorus - 8 years, 2 months ago

@Raymond Christopher Sitorus – I think the answer would be rc and not rc-3 though, since the three chairs that are empty are still chairs in the array.

David Altizio - 8 years, 2 months ago

@David Altizio – thx David for correcting

Raymond Christopher Sitorus - 8 years, 2 months ago

try this: 1=6 2=12 3=18 4=24 5=30 6=??

Jon-jon Castro - 8 years, 2 months ago

1,because 1=6,so 6=1 :)

Tan Li Xuan - 8 years, 2 months ago

@Tan Li Xuan – 6 = 6, obviously

Zi Song Yeoh - 8 years, 2 months ago

Yes 6=6 obviously but if the equal to mean's multiply by 6 its 36

Thaddeus Abiy - 8 years, 2 months ago

1 is correct

Jon-jon Castro - 8 years, 2 months ago

@Jon-jon Castro – 1 = 6 is illogical

Zi Song Yeoh - 8 years, 2 months ago

@Zi Song Yeoh – because 1=6,so 6=1

Tan Li Xuan - 8 years, 2 months ago

I got the same problem... Hope they fixed it soon.

André Macedo - 8 years, 2 months ago

There will be a lot to look forward to; new challenges and a huge database will be up tomorrow or Tuesday. Peter T. posted this earlier. I can't find a link at the moment, sorry!

Ahaan Rungta - 8 years, 2 months ago

I hate it.. I hope they can fix it as soon as possible..

Neil Tinaytina - 8 years, 2 months ago

Hopefully it's just because they're upgrading the techniques trainer.

Clifford Wilmot - 8 years, 2 months ago

Hopefully

Vishwa Iyer - 8 years, 2 months ago

But atleast they could send the weekly problems of one of the Olympiad sections. I'll get bored....

Garvil Singhal - 8 years, 2 months ago

Advitiya Brijesh - 8 years, 2 months ago

I hope they will bring new things.I will wait with patience

shaheed vh - 8 years, 2 months ago

Why arent there any qns today? I thought they post qns here every Monday?!?!

YuJiahuan LuvRafflesfriends - 8 years, 2 months ago

They're up now.

Kenneth Chan - 8 years, 2 months ago

problems are back yepiii!!!!!!!!!!!!!!!!!!!!!

Tejas Kasetty - 8 years, 2 months ago

i cant see my trignometry & calculus problems since 2 weeks...

Nikhil TR - 8 years, 2 months ago

can you??

Nikhil TR - 8 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$