Ok, I'll have a bunch of challenging questions from the AMC (Australian Mathematical Competition) I did today. I'll post one at a time so as not to confuse the posters and myself. The questions I suppose will be of ascending difficulty, those which I wasn't able to solve.

1. There's a given equation; . There was a sketch along, that of an inverted parabola, which had a positive y-intercept and the turning point was on the y-axis.

Which is true?
a) a + b + c = 0
b) a + b - c < 0
c) -a + b - c > 0
d) a + b + c < 0
e) There is not enough information.

I ruled out a) and d), since there is a solution other than 0 when putting x = 1.
The others, I'm at a lost.

Thanks for replying

Survivorboi, wanna make an attempt? I'm sure you'll be interested too to know how to solve the problems I'll post

Unknown008

Oh, I get it finally! wow, I didn't think I had to think like that lol!

Ok, I'll skip no. 20 for later, that was pretty a 'logical' question.

21.
A palindromic number is a 'symmetrical' number which reads the same forwards and backwards. For example, 55, 101 and 8668 are palindromic numbers. There are 90 four-digit palindromic numbers. How many of these four-digit palindromic numbers are divisible by 7?
a. 7
b. 9
c. 14
d. 18
e. 21

How am I supposed to find that in about 2 minutes? I would like to know the 'trick'

galactus

If I am thinking about this correctly, 1001 is palindromic and divisible by 7.

1001, 2002, 3003, ......

If we had 770, we get the others.

1001, 1771, 2002, 2772, etc.

I believe there are 18 if we count all them.

Check me out on this on.

Unknown008

Terrific! Amazing! Thanks yet again galactus! I had to spread the rep...

For the next number. I'll post the pic tomorrow. The circles are touching each other so that if their centres are joined, three centres would form an equilateral triangle.

22. What is the area in square centimetres, of the parallelogram that would fit snugly around 6 circles, each of radius 3 cm, as shown in the diagram?
a. 108
b.
c.
d.
e. 216

I tried to get the area through , then double that yo give

The base is given by (3*4) + 2(extensions)

Extension =

Slant length = 6 + 2(extensions)

Area =

=

I can't seem to get my mistake.. I'll check in tomorrow, bedtime here.

EDIT: Ok here's the pic.

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galactus

Break it up into various rectangles and they add to 216.

Shift the parallelogram to the right by 'pushing' on the top and transform it into a rectangle. Easier to envision that way.

morgaine300

Not sure it's any higher than what we've got here. I knew how to do a lot of this stuff by time I finished 10th grade, and the rest (at least what you've posted so far) by 11th when I got to higher algebra/trig. 12th I was taking analytic geometry (which at the time I thought was loads of fun - cough!) and some intro calculus (which I hated). Of course, not a high percentage of people were taking the courses I was - I was doing the "college prep." Which is kinda funny since the college I work at gives college credit for stuff I took in high school. My high school may have been a little over-ambitious.

Now as for that time limit... took me an hour to figure out #8. LOL.

This doesn't seem like enough info. Was this:

given with the problem?

EDIT: Never mind... That 5^x... equation in there was not showing up for me in your original post. That's darn weird.

Don't understand the way they word some of these. Do they mean the number of possible eyebrows you could make?

I made the mistake of paying attention to what you were doing and trying to follow it. I was about to attempt it like what galactus was doing, and then when I saw what you were doing I gave up and decided I didn't know how to do it. That's what I get. (I usually trust myself better than that.)

As for the rest, I've given up. They're getting into too many shapes. I don't even remember what an "acute" angle is. And circles - forget it. I can find the area and that's it. So this is starting to get way beyond my memory. A palindromic? I don't even remember ever knowing that. And as soon as I saw the word tengent I quit reaading.

morgaine300

I apparently also quit knowing how to type... sheesh.

galactus

I assume 'tengent' is meant to be 'tangent'. A tangent isn't anything complicated. It's just a line that touches a circle or other curve at some point.

For instance, the tangent to the curve x^2 at x=1 is an example in the graph below.

Also, a palindromic number is just that. Same as a palindromic word. It's the same backwards as forwards.

1001, 2002, 3223, and so on

And Napoleon's famous line, "Able was I ere I saw Elba" is a palindrome. He was exiled to the island of Elba in case you wonder what that means.

I have noticed a lot of terms in math scare people because they think it's complicated, but it's not.

How about a 'rectangular parallelopiped'?. That's just another name for a box.

As for high levels of math, check out a Chinese version of the same test. Then you can say, 'sheesh'.

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Unknown008

I didn't know that was possible... I mean, that when you 'push' the upper left vertex, making the new shape as a rectangle will keep the actual area.

Ok, lemme rephrase my previous comment then: Mauritius has a so low level of Maths!

I have heard that somewhere. I think in my vectors class. That's a prism with a base in the shape of a parallelogram, right?

Ok, now for the number 23.

In 3009, King Warren of Australia suspects the Earls of Akaroa, Bairnsdale, Claremont, Darlingdust, Erina and Frankston are plotting a conspiracy against him. He questions each in private and they tell him:
Akaroa: Frankston is loyal but Erina is a traitor
Bairnsdale : Akaroa is loyal
Claremont : Frankston is loyal, but Bairnsdale is a traitor.
Darlingdust : Claremont is loyal but Bairnsdale is a traitor.
Erina: Darlingdust is a traitor.
Frankston: Akaroa is loyal.

Each traitor knows who the other traitors are, but will always give false information, accusing loyalists og being traitors and vice versa. Each loyalists tells the truth as he knows it, so his information can be trusted, but he may be wrong about those he claims to be loyal. how many traitors are there?
a. 1
b. 2
c. 3
d. 4
e. 5

I got d. 4 but my friend told me otherwise...

24.
Four circles of radius 1 cm are drawn with their centres at the four vertices of a square with side length 1 cm. The area, in centimetres, of the the region overlapped by all four circles is
a.
b.
c.
d.
e.

i don't even know where to start this...

galactus

I figured #24. I think you have a typo on c. It should be . That is the solution.

If you do not have a drawing, make one with a ruler and a compass so that it is accurate.

You see a little square with bulging sides in the center of the original square. That is the area of overlap for the 4 circles.

It is kind of hard to explain what I did to arrive at the solution.

I found the intersection of two of the circles with centers (0,0) and (0,1):



They intersect at

Subtract half the width of the square, 1/2, and we have a side of a triangle we can use Pythagoras on.

. That is the side length of the 'bulgy' square. But we have to find the area of those bulges which are 4 circular segments.

These areas can be found from the formula for a circular segment and multiplying by 4:



The area of the square is

Add them:

There are many ways to tackle this.

I made this graph with my downloaded graphing utility. It is unconstrained, otherwise, it would look more like circles than ellipses, and the center region would look like a square with bulging sides. I also drew it with a compass and rule on paper.

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morgaine300

Yes, tangent. Was having trouble typing. If I don't remember how to do anything with circles and curves, I'm not likely to remember how to do anything touching it. That was also the kind of thing I disliked in high school.

I've never been scared by math or terms. I just don't remember a lot of stuff I learned over 30 years ago. There's a big difference. I tutor math - just not this stuff.

I was referring to the typos. I'm picky about that sort of thing and don't usually leave them. :-)