really hard problems for math genious?

Question

1. In convex hexagon ABCDEF, all six sides are congruent, A and D are right angles, B and C and E and F are congruent. The area of the hexagonal region is 2116((square root of 2) + 1) Find AB.

2. Find the number of ordered pairs of positive integers (a,b) such that a + b = 1000 and neither a nor b has a zero digit.

3. When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face F is greater than 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is 47/288 . Given that the probability of obtaining face F is m/n where m and n are relatively prime positive integers, find m+n

These are not my homework. they are problems from hmmt, if you can solve these, you should be proud of yourself.

Answer 1

1) In convex hexagon ABCDEF, all six sides are congruent, A and D are right angles, and angles B and C and E and F are congruent. The area of the hexagonal region is
2116[(√2) + 1]. Find AB.
_________

Calculate the number of degrees in a hexagon.

180(6 - 2) = 180*4 = 720°

Two of the angles are right angles. The other four angles are all equal. Calculate their measure.

(720 - 2*90)/4 = (720 - 180)/4 = 540/4 = 135°

This information allows us to draw an approximate sketch. It is a hexagon with two 90° angles and four 135° angles. From the sketch we see that we can add one point P in the interior of the hexagon such that we can form:

Square ABPF
Rhombus BCDP
Rhombus DEFP

The two rhombi both contain two 45° angles and two 135° angles. Let the length of one side be x. Calculate the area.

Area = (Area ABPF) + (Area BCDP) + (Area DEFP)
Area = x² + x(x cos45°) + x(x cos45°)
Area = x² + x²/√2 + x²/√2 = x² + x²(√2)
Area = (√2 + 1)x² = 2116(√2 + 1)
x² = 2116
x = 46

AB = x = 46
___________

2) Find the number of ordered pairs of positive integers (a,b) such that a + b = 1000 and neither a nor b has a zero digit.

I assume we have no leading zeroes.

Without restrictions we have 999 ordered pairs of positive integers (a,b) such that a + b = 1000.

1 + 999
thru
999 + 1

The number of numbers where the 1's digit is zero are:
9 + 9*10 = 9 + 90 = 99

To get a sum of 1000, these numbers will be added to other numbers also ending in zero.

The number of numbers where the 10's digit is zero and the 1's digit is not zero are:

9*9 = 81

To get a sum of 1000, these numbers will only be added to numbers not containing a zero.

The number of ordered pair of positive integers (a,b) such that a + b = 1000 and neither a nor b has a zero digit is:

999 - 99 - 2*81 = 999 - 99 - 162 = 738
___________

3) When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face F is greater than 1/6, the probability of obtaining the face opposite is less than 1/6, the probability of obtaining any one of the other four faces is 1/6, and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is 47/288 . Given that the probability of obtaining face F is m/n where m and n are relatively prime positive integers, find m+n

Four of the sevens will result from two sides being rolled, each with probability 1/6. The remaining two sevens will be rolled from sides containing the probabilities:

m/n + (2/6 - m/n) = 1/3
m/n + (1/3 - m/n) = 1/3

P(7) = 4(1/6)² + 2[(m/n)*(1/3 - m/n)] = 47/288
4/36 + (2/3)(m/n) - 2(m/n)² = 47/288
1/9 + (2/3)(m/n) - 2(m/n)² = 47/288
(2/3)(m/n) - 2(m/n)² = 47/288 - 1/9 = 5/96
2(m/n)² - (2/3)(m/n) = -5/96
2(m/n)² - (2/3)(m/n) + 5/96 = 0
(m/n)² - (1/3)(m/n) + 5/192 = 0

[(m/n) - 1/8] [(m/n) - 5/24] = 0

m/n = 1/8, 5/24
but m/n > 1/6

Therefore the only solution is:
m/n = 5/24

m + n = 5 + 24 = 29
__________

Answer 2

These questions also appeared on the 2006 Alternate AIME (1, 7, and 5 respectively) so they are definitely not homework. I took the 2006 AIME my junior year and used these for practice the past year so I was able to recognize them. Good luck with HMMT!

1) Call the side length of the hexagon x. This is what we wish to find. You can split the hexagon into 2 triangles and a rectangle (whose areas are easier to compute) by adding lines BF and CE. I'll use brackets [ ] to denote area. We know that [ABC] + [CDE] + [BCEF] = 2116(1 + sqrt(2)). See if you can express those 3 areas in terms of x and then solve.

2) First note that there are 999 solutions without restrictions. There are two ways of attempting this that I can think of. One involves casework (how many digits do a and b have) and the other approach involves the complementary counting and the principle of inclusion-exclusion (PIE). I will use the second approach.

We can use PIE to find the number of solutions such that at least 1 one a or b has a zero digit (complementary counting). This value which I will call x is equal to

x = (# ordered pairs with "a" having a zero) + (# ordered pairs with "b" having a zero) - (# ordered pairs with both "a" and "b") having zeros.

We still haven't done any counting, but I will let you have a crack at that (keep in mind that the final answer is 999-x). Think about this one and see if you can finish from here. I get 738 as the answer.

3) Let's first consider the probability of getting a 7 with a pair of fair dice. Well, the first roll can be anything which occurs with probability 6/6 = 1, and the second roll must show the opposite side of the the first which occurs with probability 1/6. Thus the probability in the case of fair die is 1(1/6) = 1/6. This is useful because we learn something about the tactics used in computing such a probability.

Now for the problem at hand. We can get a sum of 7 in 2 ways: (1) roll F and the opposite of F or (2) roll one of the other 4 sides and then roll the opposite side. The probability that we seek, P(sum 7) = P(1) + P(2). See if you can finish from here. I get that the probability of obtaining face F is 5/24 so m+n = 29.

Hint for 3: Let the probability of obtaining face F, P(F) = 1/6 + x. Then, what is the probability of obtaining the face opposite F in terms of x? Note that you don't have to do this. You can let P(F) = y and get the prob. of obtaining the opposite face in terms of y.

Answer 3

q2
zero digit. is that 0 placed at ones (zero ending) or zero anywhere in the numbers?

if 0 digit = 0 ending,
answer
= (1000+1)C1 - (100+1)C1
= 1001 - 101
= 900

if 0 digit = 0 anywhere in number,
answer
= (1000+1)C1 - (100+1)C1 - 9*2*9
= 1001 - 101 - 162
= 738

q3
sum of 7 came from these sets
(1,6) (6,1)
(2,5) (5,2)
(3,4) (4,3),

P(sum 7) = 47/288
= 4(1/6)^2 + 2(m/n)(1/3 - m/n)
= 1/9 + 2m/3n - 2(m/n)^2
47 = 32 + 192m/n - 576(m/n)^2

let m/n be c,
576c^2 - 192c + 15 = 0
(24c -3)(24c -5) = 0
m/n = c
= 3/24 or 5/24
= 1/8 or 5/24
F > 1/6, so F = 5/24 = m/n
m+n = 5+24 = 29

Answer 4

Ahahaha, I need help too. How old are you btw? I'm like 12 but I'm taking Algebra 1 I guess that don't count as a genius, or rather, I'm only gifted in the area of math. Sorry, I can't help, but maybe you could help me?

Answer 5

The answer is 2

Answer 6

hey ur problem cannot be answered.
coz....
1) i didn't understood what r u trying to tell by 1st Question.
2) its too long to be answered. try asking one at a time.