AMC? Mathematical Competition? (2 Viewers)

[Archman]

No, it was definitely 540 - and I didn't even use hexagons. Further proof that it was not 1080 is given in my diagram.

Question 19 had me stumped, but then I thought to myself "well, let's guess. Take the number 10. It's got 3 integers - 2,3,5. Maybe it'd work. And yes, it is even a possible solution. But I ignored it and went back to it latter when I'd done the other questions.

It turns out that a + 1/(b+1/c)=37/16 works when the values 2,3,5 are placed into this.

However, the way I worked it out was that 37/16 must be like 32/16 + 5/16, and comparing that to the LHS, gives a=2, 1/(b+1/c) = 5/16. So take the reciprocal of both sides to get b+1/c = 16/5, which is 15/5 + 1/5, comparing that to the new LHS gives b = 3, 1/c = 1/5. Reciprocal: c = 5.

na, 1080. try with real coins

(i hate the perimeter question)

 

[ngai]

lol i was waiting for u to post here..
couldn't convince them myself
so how did u go?

 

[Archman]

bah, 147 or 150. (stupid perimeter question)

how bout u

 

[turtle_2468]

haha archman Too much maths eh... (just joking!)

 

[Iota]

But unless my memory fails me, 1080 wasn't an included answer, was it?

Do you happen to have a solution, Archamn or ngai?

Pardon me for saying you were incorrect.

 

[ngai]

perimeter, whcih one was that?...a section one?
lol i'm just 123, better not find any more sillies

well if u slide the coin around, u'll get 540, but if u roll the coin around ull get 1080
haven't got a good solution except 4 10c coins

 

[Archman]

turtle the man himself finally speaks (yay)

anyway might just du this for reference
last 10:
21: B. 2, try intersect 2 equilateral triangles, just look at how many different exterior angles for the hex will be sufficient, not hard to c that if one of them is a then the next is 120-a, then a, the 120-a then a then 120-a

22: product a - product (a-1) = 113
expand: sum ab - sum a = 112
so sum ab = 135
2*sum ab = 270
sum a^2 = (sum a)^2 - 2*sum ab = 259
so its E

23: err not much to say on this one, stare hard enuf and u'll c its D

24: multiples of 5 have to go, otherwise it will end in 0 so 20 of them is gone
now we have 1,2,3,4,6,7,8,9 11,12,13,14,16,17,18,19 etc
look at each group of 4, the last digit of product of each group is 6
since 6 times 6 has last digit 6, the last digit of the whole thing will be 6
so 20 is not sufficient
but if we get rid of 8
1*2*3*4*6*7*9 has last digit 2, since the rest of stuff have last digit 6,
2*6 gives last digit 2, good enough.
so answer is B 21

25. let midpt of SR be X, RQ be Y, the pt of tangent from UT is Z, let YT = a, let SR be 4
now UX = UZ, TY = TZ
so UR^2 + RT^2 = (UZ+ZT)^2
1^2 + (2-a)^2 = (1+a)^2, solve for a and u get
a = 2/3 so RT = 4/3, so RQ/RT = 3
so its A: 1/3

26: the question is equivalent of how many left to right paths (ignore the dots)there are in the diagram below, im sure u can count them in your own time: A 35
....../
...../\
..../\/
.../\/\
../\/\/
./\/\/\
/\/\/\/

27: well all possible distances between 2 of those mentioned points are rt2, rt4, rt6 and rt8
if the largest side of the triangle is rt8, the other side cant be rt2 and rt2 or rt2 and rt4 by simply looking at a diagram. if the other sides are anything else, the end result is a right or acute triangle.
now if longest side is rt6, we CAN get a rt2,rt2,rt6 triangle, and the angle is 120
if longest side is rt4 or rt2, we cant get obtuse angles.

so it is C 120

28: let roots be p and q with p>q
now if there are 3 integers between them, 4>p-q>=2, (try plot it on a number line and u see y)
well use quad. formula, we know difference of roots is rt(b^2 - 4ac) = rt(b^2 - 8)
solve to get 24>b^2>=12 so b can only be +-4 (they both work too)
so its C: 2

29. consider points A(x,y), B(x-1,y), C(x, y-1), D(x-3, y-4)
now the sum is the sum of distances of origin to those 4 pts.
so we hafta pick an O that minimize the sum. since ABCD is a convex quad.
obviously picking the intersection of the diagonal will be a good choice.
if we dont, say we pick P, but by triangles inequality:
PA+PD >= AD, PB+PC >= BC, so its not better than the intersection
so the required answer is AD+BC = 5+rt2, D

30. now firstly at least one color has to occur once. (assume not, take the pair with same color and have the closest distance, there cant be two of the same between them, and wateva is inside cant be on the outside, so there is only one of those).
now if one color occurs k times, there exist k-1 colors to fill in the gaps between them and those k-1 color only occur once each. (proof similar to earlier, look at each gap separately)
so yes both ppl can make a string of 8, since one color hafta occur once, then one has to occur at least 3 times (by pigeon holes), but that means another color hafta occur once. so the last color also occur 3 times. well not too hard to c that it wont work with for colors having respecitvely 1,3,1,3 occurrances.
so max is 7 and both can achieve it: rygbgyr, ryrgbgr


hmm some explanation may not be too clear, post if u think so and i'll try explain a bit better.

 

[Xayma]

perimeter, whcih one was that?...a section one?
lol i'm just 123, better not find any more sillies

well if u slide the coin around, u'll get 540, but if u roll the coin around ull get 1080
haven't got a good solution except 4 10c coins

Was the perimeter one the rectangle 25x its width to its length?

The answer was 13:5

Consider it to be 25units by 1 unit.

Then it has a perimeter 52units.

A square of equivalent would be 5x5 (25 units square)

So the ratio is 52:20
13:5.

Oh yeah and for 30:

ryrgbgr can also be expressed as ryrgrbr

 

[Rorix]

yes i knew it was 5 + sqrt2 =) damn steph

 

[acmilan]

Does anyone mind posting the answers for 1-20 (no explanations necessary)

 

[Archman]

Oh yeah and for 30:

ryrgbgr can also be expressed as ryrgrbr

no it cant... in ryrgrbr: the second r occurs between the first and the third, but theres a fourth r outside first and third, so its banned(well strictly speaking anyway since theres nothing said about same colors in the rules)




anyway the diagram is fixed in the solns


and for first 20:
d,b,e,a,a,
b,a,b,c,e,
c,c,a,c,b,
b,c,d,a,e

 

[QuaCk]

if in the last section i scored 2 right after attempting 5 questions.
what will i get as a mark for the last section?

5*3 + 2*8 = 31?

seems harsh as it is only 1mark above those who did not attempt them.

maybe it's there for protection ... so people don't fluke an award
:|

 

[Rorix]

uh. if you managed to attempt 3 questions that you got wrong, exactly how are you not trying to fluke it/?

 

[QuaCk]

well my answers were always one option next to the real answer if that justifies my efforts

 

[ngai]

lol if that justifies ur efforts, then a 5 yr old colouring every single C has 60% chance not to be fluking for each question

 

[paper cup]

oh gawd, I can't believe this - I spent about 5 mins on that stupid a+1/(b+1/c) question, worked it out correctly and messed up the addition...2+3+5 = 11 -_- what species of dumbass am I?

 

[Grey Council]

aarrrghhhh!

Stupid Archman! Now I've got yet ANOTHER question wrong.
nnnooooooooooo!!!!!
my chances for a HD are slipping!


*sulks*

blah, ah well.
lol, t'was good of you to post up the solutions Archman. very nice of you.

anyway, cherryblossom, your the extremely amusing type of dumbass. A stupid dumbass like me has to hope there are more of you smart but amusing dumbass's in the world, so I can get my HD...

I think I'm on 101 now. Hrm, last year me gotted 88, and a narrow missing out on HD.
*kicks stupid tricky questions*
1080 my arse. humph, Archman, can you post up a solution to that one? Or anyone, for that matter. Turtle?

 

[Teoh]

...Stupid lying question...

 

[Grey Council]

lol! how could you get that wrong?

just had to write out the cases, wouldn't have taken more than half a minute.
lol, but it was tricky, so fair enough. hehehe

 

[Rorix]

i think i'm 113 ah well