Middle School Marathon

Math Champion

No PoP, the vertex of a parabola is its minimum/maximum point. Give both x and y coordinates.

**Posted:** Sat Jul 11, 2009 3:37 pm

_________________
~Math Champion
If you don't believe that this post was helpful, please message me and tell me how I can improve my posts.

AIME15

You guys should stop relying on calculus and buy Intro Algebra Razz

The vertex occurs when $x=\frac{19}{2}$ . Substituting this yields $\frac{361}{4}-\frac{361}{2}+47=\boxed{\frac{173}{4}}$ .

NP: $a$ and $b$ are positive integers smaller than $10$ such that $\frac{ab}{a+b}=\frac{12}{7}$ . Find all possible values of $a^b$ .

**Posted:** Sat Jul 11, 2009 3:51 pm

_________________
GOALS: MC School: 46 | Chapter: 46 | State: 46 | National: 46 | AMC 8: 25 | AMC 10: 150 | AMC 12: 150 | AIME: 15 | USAMO: 42 | TST: 63 | IMO: 42 |

PowerOfPi

How did you get the $x$ value?

**Posted:** Sat Jul 11, 2009 5:04 pm

_________________
Sign up for MOCK AMC!

RoFlLoLcOpT

Complete the square

**Posted:** Sat Jul 11, 2009 5:14 pm

AIME15

Complete the square for $ax^2+bx+c=0$ to derive a formula for $x$ . Now, answer my problem Razz

(yes, they're unrelated).

**Posted:** Sat Jul 11, 2009 5:28 pm

_________________
GOALS: MC School: 46 | Chapter: 46 | State: 46 | National: 46 | AMC 8: 25 | AMC 10: 150 | AMC 12: 150 | AIME: 15 | USAMO: 42 | TST: 63 | IMO: 42 |

Math Champion

Actually, you could find the x value by using $x=-\frac{b}{2a}$

**Posted:** Sat Jul 11, 2009 6:01 pm

_________________
~Math Champion
If you don't believe that this post was helpful, please message me and tell me how I can improve my posts.

AIME15

I'm showing him how to, not just giving a formula Wink

Now, solve my problem Razz

**Posted:** Sat Jul 11, 2009 6:10 pm

_________________
GOALS: MC School: 46 | Chapter: 46 | State: 46 | National: 46 | AMC 8: 25 | AMC 10: 150 | AMC 12: 150 | AIME: 15 | USAMO: 42 | TST: 63 | IMO: 42 |

Math Champion

Just list the different cases that add up to 7 or 14, with product 12 or 24, respectively. For 7, there is (3,4) and (4,3). For 14, there is (2,12) and (12,2), and so you can find the 4 values of $a^b$ .

**Posted:** Sat Jul 11, 2009 6:16 pm

_________________
~Math Champion
If you don't believe that this post was helpful, please message me and tell me how I can improve my posts.

cgyao15

maybe forgetting ur own advice? Smile

NP
Compute
$\ 2008(1+2+3+....+2009)-2009(1+2+3+...+2008)$
Is there another way to not brute force?

**Posted:** Sun Jul 12, 2009 4:22 am

_________________
Check out my new avatar made by Izzy. PM her to get one!

isabella2296

The sum of the first 2009 numbers is $\frac{2009 \times 2010}{2}$ , so we can express the first part as $2008 \cdot \frac{2009 \times 2010}{2}$ . The second expression is $2009 \cdot \frac{2008 \times 2009}{2}$ . Thus, with some simplifying, etc., we get 2017036.

New problem: Square ABCD is reflected over the line CD to make A'B'CD. If AB = 4, what is the length of AB'?

**Posted:** Sun Jul 12, 2009 4:33 am

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Nine-Point Center

picture74 · Poincare Conjecture Offline Joined: 06 Jul 2009 Posts: 125

**Posted:** Sun Jul 12, 2009 6:08 am

King! · Riemann Hypothesis Offline Joined: 27 Mar 2009 Posts: 352

Actually the answer is: $4\sqrt {5}$

NP----

1. $3$ Woman and $7$ Men are split into two groups of $5$ . There has to be at least one woman and one man in each group. How many groups can be formed?

2.How ways can you split them into two groups?(Same conditions, except the number of people in each group can vary)

**Posted:** Sun Jul 12, 2009 7:56 am

_________________
♣Hm♣ Add some zest to the humdrum of your everyday life!(Mafia games in my blog)

isabella2296

Before the NP's, a solution would certainly be nice. Smile

**Posted:** Sun Jul 12, 2009 8:04 am

_________________
Czech out $\mathfrak {Encyclopedia } \mathfrak { Mathematica}$
Latest entry: Nine-Point Center

King! · Riemann Hypothesis Offline Joined: 27 Mar 2009 Posts: 352

Oh sorry

Solution

**Posted:** Sun Jul 12, 2009 8:09 am

_________________
♣Hm♣ Add some zest to the humdrum of your everyday life!(Mafia games in my blog)

AIME15

Solution to 1

Current problem is P2.

**Posted:** Sun Jul 12, 2009 10:46 am

_________________
GOALS: MC School: 46 | Chapter: 46 | State: 46 | National: 46 | AMC 8: 25 | AMC 10: 150 | AMC 12: 150 | AIME: 15 | USAMO: 42 | TST: 63 | IMO: 42 |

Math Champion

Guys, also, starting now, if you could number your problems, that would be great, just to get a running tally of how many problems we've gone through. P2 is Problem 17, btw.

**Posted:** Sun Jul 12, 2009 11:58 am

_________________
~Math Champion
If you don't believe that this post was helpful, please message me and tell me how I can improve my posts.

brainomega

solution

18.

**Posted:** Sun Jul 12, 2009 12:20 pm

King! · Riemann Hypothesis Offline Joined: 27 Mar 2009 Posts: 352

**Posted:** Sun Jul 12, 2009 12:42 pm

_________________
♣Hm♣ Add some zest to the humdrum of your everyday life!(Mafia games in my blog)

FlyAgaric

**Posted:** Sun Jul 12, 2009 1:12 pm

_________________
Fly Agaric (fli-a-ger-ik) n. 1. type of mushroom which is bright red with small white dots; consumption may lead to death or extreme, short-term growth

Maybach

I don't know, once someone answers the question we can work on this new problem.
Click to reveal hidden content

**Posted:** Sun Jul 12, 2009 1:59 pm