For this week's blog post, I want everyone to understand how to solve ALL of the problems from the competition you had last Saturday. For those of you who didn't attend the competition, you have an additional homework of solving all the problems on all the tests. This post is due at 3:00 pm by Wednesday, November 23rd. I know this deadline is somewhat early but you have no school so you can devote more time for this. Just for convenience, I am labeling the Sprint round problems 1 - 30, the target round problems 31 - 38, and the team round problems 39 - 48. In addition, each student this time has a "buddy" who will solve nearly the same problems. Pairs of buddies are divided by a space. The first "buddy" to post on the blog will do a regular post while the second "buddy" will have to post and include critiques to his/her partner's solutions (if any). By the way, even if you can't attend the meeting, this blog post is still do and I will check it.
Alice : 1, 13, 19, 25, 31, 37, 43
Christine : 2, 13, 19, 25, 31, 37, 43
Ming : 3, 14, 20, 26, 32, 38, 44
Allan : 4, 14, 20, 26, 32, 38, 44
Maya : 5, 15, 21, 27, 33, 39, 45
Stephanie : 6, 15, 21, 27, 33, 39, 45
Jacky : 7, 16, 22, 28, 34, 40, 46
Sagaar : 8, 16, 22, 28, 34, 40, 46
Christopher: 9, 17, 23, 29, 35, 41, 47
Vivian : 10, 17, 23, 29, 35, 41, 47
Julianna : 11, 18, 24, 30, 36, 42, 48
Anindit : 12, 18, 24, 30, 36, 42, 48
Start posting! The faster you post to the blog, the better!
I CLAIM FIRST COMMENT:
ReplyDeleteI apologize for any incorrect solutions.
6. Simplify: 111+221+331-11-21-31.
This problem is pretty self explanatory.
Rewrite this as:
111 – 11 + 221 – 21 + 331 – 31
100 + 200 + 300
600
Answer: 600
15. How many ordered triples of positive integers (a,b,c) satisfy a+b+c=10?
Assume a < b < c.
1, 2, 7
1, 3, 6
1, 4, 5
2, 3, 5
Answer: 4
21. Find the sum of the number of vertices, edges, and faces in an octahedron.
This problem relies on your knowledge of an octahedron (http://upload.wikimedia.org/wikipedia/commons/0/07/Octahedron.svg)
Faces: 8
Edges: 12
Vertices: 6
8 + 12 + 6 = 26
Answer: 26
27. x=1 is a solution to the equation 2x^3+6x^2-x-7=0. What is the sum of the other two solutions?
For all equations ax^n + bx^n-1 + cn^x-2 + …. = 0:
x1 + x2 + x3 +….. xn = -b/a
^(sum of the roots)
In 2x^3+6x^2-x-7=0, n = 3, a = 2, b = 6.
x1 + x2 + x3 = -b/a
x1 + x2 + x3 = -6/2
Since 1 root is 1:
1 + x2 + x3 = -3
x2 + x3 = -4
Answer: -4
33. The sum of 5 consecutive integers is 265. What is the value of the largest of these integers?
Let x = the largest of the 5 consecutive integers
x – 4 + x – 3 + x – 2 + x – 1 + x = 265
5x - 10 = 265
5x = 275
x = 55
Check: 51 + 52 + 53 + 54 + 55 = 265
Answer: 55
39. A two-digit integer is randomly chosen. What is the probability that it is divisible
by 11? Express your answer as a common fraction.
There are 90 two-digit integers. (10-99)
There are 9 two-digit integers divisible by 11.
Therefore, the probability that a two-digit number is divisible by 11 is 9/90 or 1/10.
Answer: 1/10
45. The perimeter of an equilateral triangle equals the perimeter of a square. What
is the ratio of the area of the triangle to that of the square? Express your answer
as a common fraction in simplest radical form.
Let the perimeter of the equilateral triangle and the perimeter of the square = 12x (to make things easy)
Since an equilateral triangle has 3 sides, each side is 4x. By drawing a perpendicular bisector from a vertex to the opposite base, we can find using the Pythagorean theorem that the height of the triangle is 4√3 x (x is not included in the root sign). Therefore, the height of the triangle is:
A = bh
A = 4√3 x * 4x
A = 16√3 x^2
Since a square has 4 sides, each side is 3x.
A = bh
A = 3x * 3x
A = 9x^2
Ratio of area of triangle to area of square:
16√3 x^2 / 9x^2
16√3 / 9
Christopher-
ReplyDelete#9:The operation /\ is defined by a/\b = ab+a+b/b. Find 9/\(6/\3).
Solution: According to the definition, 6/\3 is 6(3)+6+3/3 which is 9, and 9/\9 is 9(9)+9+9/9 which is 11.
Answer: 11
#17: A rectangular prism is drawn with edges parallel to the three coordinate axes. If two vertices of the prism are (-2,3,0) and (6,7,9), then what is the volume of the prism?
Solution: Since the edges are parallel to the three coordinate axes, we can find the dimension of the prism simply by multiplying the differences of each coordinate. 6-(-2) = 8, 7-3 = 4, and 9-0 = 9, so the volume of the prism is 8*4*9, which is 288.
Answer: 288
#23: For integers a and b such that a<b, denote a O b as the sum of the integers from a to b, inclusive. Find the value of 2001 O 2011/1001 O 1011. Express your answer as a common fraction.
Solution: The sum of the integers from 2001 to 2011 is the same as the sum of 5.5 pairs of 4012 each. Likewise, 1001 O 1011 is 5.5 pairs of 2012. The fraction would then be 4012*5.5/2012*5.5, which is simplified to 1003/503.
Answer: 1003/503
#29: If x + 1/y = 9. y + 1/z = 7, and xyz = 1, then find the value of z + 1/x. Express your answer as a common fraction.
Solution: If x + 1/y = 9, then x = 9-1/y. Substituting this into the equation xyz = 1 yields 9yz-z = 1. If y + 1/z = 7, then y = 7-1/z. Substituting this into 9yz-z = 1 yields 62z-9=1. Solving this equation reveals that z is 5/31, and plugging this value into y + 1/z = 7 reveals that y is 4/5. Plugging this value into x + 1/y = 9 reveals that x is 31/4. So, z + 1/x = 5/31 + 4+31 = 9/31.
Answer: 9/31
#35: Adam multiplies a number by the square of its reciprocal. Bob multiplies the square of the same number by the reciprocal of the number. Amazingly, they get the same result. Find Adam's original number, given that it is a positive integer.
Solution: If the original number is denoted as a/b, then ab^2/ba^2 = ba^2/ab^2. This means that this number equals its reciprocal, and the only number that satisfies this condition is 1.
Answer: 1
#41: Let f(x) = 98/12-z. For how many values of k is f(k) an integer?
Solution: 98 has a prime factorization of 2*7^2, which means that it has six positive factors and six "negative factors". In order for f(k) to be an integer, 12-k must equal one of these twelve values. So, there are 12 values of k.
Answer: 12 values
#47: If x^2 = x + 1, find the value of x^4 - 2x^3 + 2x^2 - x + 12.
Solution: If x^2 = x + 1, then 1 is equivalent to x^2 - x. This means that x^2 - x - 1 = 0. Using the quadratic formula, we find that x = 1 +or- sqrt(5)/2. Plugging this into the polynomial x^4 - 2x^3 + 2x^2 - x + 12 yields the value of 14, which is only one value because whether the radical is positive or negative, the sum of the radicals will equal zero. Here are the values for each term in the polynomial:
x^4 = 14 +or- 6(sqrt(5))/4
-2x^3 = -(16 +or- 8(sqrt(5))/4)
2x^2 = 12 +or- 4(sqrt(5))/4
-x = -(2 +or- 2(sqrt(5))/4)
12 = 12
x^4 - 2x^3 + 2x^2 - x + 12 = 8 +or- 0(sqrt(5))/4 + 12 = 14
Answer: 14
I CLAIM THIRD :p
ReplyDelete11) The Problem started with: If 6a=5b, 11b=15c, and ja=kc, where j and k are relatively prime positive integers, then find j+k.
Basically, we have to find a in terms of b. To do this, we can multiply the first equation by 11/5 to get 66a/5 = 11b. If you subtract the 2 equations from each other, you get:
66a/5 = 15c. Multiply (not divide, this is what I missed on the actual test) both sides by 5 to get 66a = 75c. Divide both sides by 3 to find that 22a = 25 c. 22 and 25 have no common factors, so you j must be 22 and l must be 25.Therefore, j+l = 27.
18) This problem started with: If the equation x+4x=k has exactly one distinct real solution x, then what is the sum of all possible values of k?
The way I did it on the actual test was really stupid as I just guessed and checked till I found the lowest possible value.
However, if you've done Algebra 1, you will know that K is basically y. You will also know that the parabola formed by graphing the equation has a minimum point. The x value of this point is the average of the roots (can be proved with calculus, or so dad says). the average of the roots is -b/2a (which can be proved). If you substitute this value into the equation, you will the y value of the lowest point. After doing this, you find that the lowest possible value of k is -21. (sorry for the messy explanation, I need paper to do this right)
24) What is gcd(1001, 70)LCM[1001, 70]
This problem is really hard if you don't know the trick. The trick is that the GCD * LCM of 2 numbers is the product of the two numbers. So, after doing that, you find that the answer is 70700.
30) Wayne flips a fair coin 4 times and Maria flips a fair coin 6 times. What is the probability that they flip the same number of heads? Express your answer as a common fraction.
For this problem, what you have to do is calculate the probability of each event happening. Therefore:
Probability that 0 head is flipped:
1 *1 / 64*32 = 1/1024
Probability that 1 head is flipped:
6 *4 / 64*32 = 24/1024
Probability that 2 head is flipped:
6*15 / 64*32 = 90/1024
Probability that 3 head is flipped:
6 *4 / 64*32 = 24/1024
Probability that 4 head is flipped:
1 *1 / 64*32 = 1/1024
add all the probabilities together to get 13/128
36. 6. How many positive factors does 2^2 * 3^3*4^4*5^5*6^6 have?
This seems like an easy case of finding the factors at first and thats exactly what i did wrong on the competition. Instead, what you have to do is take into account that 4 is actually 2^2 and 6 is actually 2*3. so the prime factorization of the number actually becomes 2^16 * 3^ 9 * 5*5. use the factor trick to get 1020 factors
42 (assuming this is team #4) for this problem ,if you know the trick its really easy. So, let x = sqrt(12+sqrt(12 + sqrt(12....)). you can then set up this equation: x = sqrt(12 + x). Solve for x to get x = -3, 4. However, -3 can't work because roots cant be negative.
48: IDK
@ stephanie (i mean seri) for the one about a, b, and c, it doesn't really say anwhere that you have to assume that a<b<c, so what you have to do is use the sticks and stones counting method to get 9C2 = 36 ways
ReplyDelete-Jacky-
ReplyDelete7)There are 10 that are not green and you pick 2 balls so the equation is:
(10/19)*(9/18)
(10/19)*(1/2)
10/38
Ans: 5/19
16)20>x>40
x/3=R2
Possibilities:20,23,26,29,32,35,38
x/7=R4
Possibilities:25,32,39
There is only 32 in both sets of numbers so
Ans: 32
22)I didn't solve.....
28)I didn't solve either.....
34)x^2-3x+41<=100
x^2-3x-59=0
Use the quadratic equation:
[-b(+-)sqrt(b^2-4ac)]/2a
[-(-3)(+-)sqrt((-3)^2-4(1)(-59))]/2(1)
[3(+-)sqrt(9+236)]/2
[3(+-)sqrt(245)]/2
(3(+-)15)/2
18/2,-12/2
9,-6
9+6+1
You add one because you have to include 0
Ans: 15
40)ABC is similar to XYZ and <A is a right angle so <X is a right angle.
XY=12 and XZ=5 and both are the legs of the right triangle so
12^2+5^2=c^2
144+25=c^2
169=c^2
c=13
Ans: 13
46)There are 350 people, 180S, 201F, 157G, 104SF, 71FG, and 84GS.
(108+201+157)-350
466-350
116
(104+71+84)-116
259-116
Ans: 143
5. There are 7 letters in the word “arrange”, including two “A’s” and two “R’s”.
ReplyDelete7! / 2!*2! = 1260
Answer: 1260
15. This question asks how many different sets of three positive integers can have a
sum of ten, and the order of the three integers matters. These are the possible combinations of a, b, and c. The number in the brackets shows the number of ways that the three numbers can be ordered.
1, 1, 8 (3)
1, 2, 7 (6)
1, 3, 6 (6)
1, 4, 5 (6)
2, 2, 6 (3)
2, 3, 5 (6)
2, 4, 4 (3)
3, 3, 4 (3)
3+6+6+6+3+6+3+3 = 36
Answer: 36
21. For this problem, you have to know what an octahedron looks like.
It has 8 faces, 6 vertices, and 12 edges.
8+6+12=26.
Answer: 26
27. 2x^3+6x^2-x-7
The formula for finding the sum of the roots is -b/a
-b/a = -6/2
The sum of the three roots is -3.
However, one of the roots is given as 1, and the question asks to find the sum of the other two roots.
-3-1=-4
Answer: -4
33. Let x equal the smallest number
x + x+1 + x+2 + x+3 + x+4 = 265
5x+10=265
5x=255
x= 51
x+4 = 55
Answer: 55
39. There 90 two-digit numbers, so the denominator of our fraction will be 90.
There are 9 two-digit multiples of 11 (11, 22, 33, 44, 555, 66, 77, 88, 99), so the numerator of the fraction will be 9.
9 / 90 = 1 / 10
Answer: 1/10
45. I assumed that each side of the equilateral triangle was 4 and each side of the square was 3 (so that both perimeters would be equal to 12). The formula for finding the area of an equilateral triangle is √3 a^2/4. So in this case, the area of the equilateral triangle would be 16√3 / 4, which equals 4√3. The area of the square is 3*3 = 9.
The ratio of the area of the triangle to the area of the square is 4√3 / 9.
Answer: 4√3 / 9
#1
ReplyDeleteIt's pretty self explanatory
27*27=729
Answer:729
#13
The problem tells you that the area of the circle, pi*r^2=17. To find the area of the square you can use r^2 because the problem says that the side of the square is the same as the radius of the circle. Thus getting the answer 17/pi.
Answer: 17/pi
#19
The problem tells you that the radius of the cone is = to the diameter of the cylinder. So you know that the radius of the cone is 2 times of the cylinder. The problem also states that the volumne of the cone is equal to the volumne of the cylinder.
To find the volumne of the cylinder you use the formula: pi*r^2*h*(b*h)
To find the volumne of the cone you use the formula: 1/3pi*r^2*h(1/3b*h)
Then you put the two equations together to get: pi(r)^2*h=1/3(2r)^2*pi*k*h
Then you cross out r^2 on both sides to get the equation: h=4/3k*h
Then you flip flop so you get the equation 3/4h=kh. Then cross out h on both sides to get the answer k=3/4
Answer: k=3/4
#25
You know that the base of the original triangle is 12 and the height is 5 and the hypotnuese is 13. You can use a ratio to find the height like 4/13=x/5.
Using that you can use the equation to figure out ADE. You can use the equation: a=1/2bh
a=1/2*2(20/13)
a=1/2(40/13)
a=20/13
Answer=20/13
#31
You can first figure out how many Kevin can crush in 15 minutes.
You first divide 15 by 3 to see how much you have multiply 100.
15/3=5 and 5*100=500
Then you have to figure out how many cans Zach can crush in 15 minutes.
Since 15 isn't divisible by 6 you multiply 15*2=30. Then you divide 30/6=5
5*100= 500. Then you have to divide 500/2 because you multiplied 15 by 2 to make it easier. Making it 250
To get the final anwer you add 500+250=750
Answer: 750
#37
You can try a few numbers but x+y/y+x = 1 so the answer is 0
Answer:0
#43
This problem you can first find ou what 9*8*7; 10*9*8; 11*10*9; 12*11*10; 13*12*11; 14*13*12; 15*14*13
9*8*7=504
10*9*8=720
11*10*9=990
12*11*10=1320
13*12*11=1716
14*13*12=2184
15*14*13=2730
After figuring that out you can add the sums up and add them all up. After you add the sums up you get the number 10164.
The final step is to divide 10164/6 because all numbers were supposed to be divide by 1*2*3=6. The answer you get is 1694
Answer: 1694
Julianna :) -
ReplyDeleteProblems 11, 18, 24, 30, 36, 42, 48:
#11- If 6a=5b, 11b=15c, and ja=kc, where j and k are relatively prime positive integers, then find j+k.
If you multiply both sides of 6a=5b by 11, you get 66a=55b. If you multiply both sides of 11b=15c by 5, you get 55b=75c. Substituting the second equation into the first one, you get 66a=75c. In this equation, j=66 and k=75. However, 66 and 75 are not relatively prime. They can be simplified down to j=22 and k=25. j+k=22+25=47.
ANSWER- 47
#18- If the equation x+4/x =k has exactly one distinct real solution x, then what is the sum of all possible values of k?
After multiplying both sides of the equation by x and then moving all the terms to one side, you get the quadratic equation x^2-kx+4=0. Because there should only be one real solution, the discriminant must equal 0. D=b^2-4ac, so the discriminant here is D=k^2-16=0.
k^2-16=0
(k+4)(k-4)=0
k=-4 or k=4
-4+4=0
Therefore, the sum of all possible values of k is 0.
ANSWER- 0
#24- What is gcd(1001,70)*LCM[1001,70]?
When a problem asks for the gcd, it is basically asking for the gcf. Prime factorizing is the best way.
1001=7*11*13
70=2*5*7
The gcf is 7, and the LCM is 7*11*13*2*5=10010. LCM*gcf=10010*7=70070.
ANSWER- 70070
THE SOLUTION TO PROBLEM 30 SHALL BE INCLUDED IN A LATER POST ONCE I FIGURE IT OUT -.-"
#36 (Target Round #6)- How many positive factors does 2^2*3^3*4^4*5^5*6^6 have?
2, 3, and 5 are already prime, but 4 and 6 are not. After factorizing those, you get a total prime factorization of 2^16*3^9*5^5. To find the number of factors, you add one to each exponent, then multiply them all.
(16+1)(9+1)(5+1)=17*10*6=1020 factors.
ANSWER- 1020 factors
#42 (Team Round #4)- What is the value of √(12+√(12+√(12...?
Let the original expression be x. If you square this, you get x^2=12+√(12+√(12+...). Once you subtract the original equation from the squared one, you get x^2-x=12.
x^2-x-12=0
(x+3)(x-4)=0
x=-3 or x=4
However, the square root of a number cannot be negative. Therefore, it is 3.
ANSWER- 3
I have no idea how to do #48. My apologies.
(it's me again :D)
ReplyDeleteAn easier way to do number 24 would be to just multiply the two numbers. 1001*70=70070.
Allan : 4, 14, 20, 26, 32, 38, 44
ReplyDeleteProblem 4:
Express 36/14 as a common fraction.
A common fraction is where both the numerator and the denominator are both integers.
We need to put 36/14 in its simplest form.
36/14 = 18/7
Problem 14:
What is the area of a regular hexagon with side length 4? Express your answer in simplest radical form.
You can find the area of a regular hexagon by either dividing it up into 6 equilateral triangles or using the formula, (3√3/2 )r^2.
A=(3√3/2 )r^2
A=(3√3/2 )4^2
A=24√3
Problem 20:
The inequality x^2+ 6x- 12k is satisfied for all real numbers x. What is the largest possible value of k?
I will assume that the subtraction sign is the inequality sign since this equation is missing an inequality symbol (I may be wrong).
We can write x^2+6x=(x+3)^2-9.
The minimum value of (x+3)^2=0.
Therefore, the minimum value of x^2+6x=-9.
For the inequality to be valid 12k>-9.
So k>-3/4
Problem 26:
A right cylinder with radius 3 and height 8 is inscribed in a sphere. What is the surface area of the sphere? Express your answer in terms of π.
An easy way to do this problem would be to draw out a cylinder inscribed within a sphere and label parts. In order to find out the diameter of the sphere, you must use the Pythagorean Theorem since the diagonal of the cylinder is the diameter of the sphere.
a^2+b^2=c^2
6^2+8^2=c^2
100=c^2
c=10
The diameter of the sphere is 10, therefore the radius is 5.
Problem 32:
In rhombus ABCD, angle A is 30 degrees. If AB=6, what is the area of ABCD?
Drawing a rhombus would make the problem easier.
The easiest way to solve this problem would be to "cut" a triangle of one side of the rhombus and "attach" it to the other side create a rectangle.
Since angle A is 30 degrees, we can use the special property for rhombuses with 30 degrees.
In other words, the height of a rhombus will be one half of one side of the rhombus.
h=1/2S
h=1/2(6)
h=3
A=bh
A=3*6
A=18
Problem 38:
In equilateral triangle ABC, points A1 and A2 lie on BC such that BA1=A1A2=A2C=BC3, points B1 and B2 lie on CA such that CB1=B1B2=B2A=CA3, and points C1 and C2 lie on AB such that AC1=C1C2=C2B=AB3. The intersections of AA1 with BB2 and CC2 are denoted by X1, X2, respectively, the intersections of BB1 with CC2 and AA2 are denoted by Y1, Y2, respectively, and the intersections of CC1 with AA2 and BB2 are Z1, Z2, respectively. Find the area of hexagon X1X2Y1Y2Z1Z2, given that AB=20. Express your answer in simplest radical form.
To start off, draw a diagram, and you'll realize that with the hexagon formed, you can inscribe an equilateral triangle.
I don't know how to do the rest of the problem, sorry. All I know is that you need to find the area of the hexagon by dividing it into an equilateral triangle and three congruent obtuse triangles (when combined, they make up the hexagon) and calculating the obtuse triangles' area and the equilateral triangle's area separately and then adding together.
If I figure out a way, I'll post it later sometime.
Problem 44:
If x^2+1/x^2=k, where k>=2, then what are the two possible values for x+1/x.
Express your answer in simplest radical form in terms of k.
x^2+1/x^2 = (x+1/x)^2-2
(x+1/x)^2=k+2
x+1/x= +/- √(k+2)
Hi people :)
ReplyDeleteSo, this is how to do number 10 (48) on the team round...
#48- An ant is located on the origin of the Cartesian plane. During each second, the ant moves from its current location (x,y) to one of the points (x-1,y), (x,y-1), (x,y+1), or (x+1,y). Let p be the probability that after 20 seconds, the ant is located at the point (7,9). Round 10^8p to the nearest integer.
Basically this is saying that the ant moves 20 steps, and is asking for the probability of it landing on (7,9) if it moves directly up, down, left, or right every step. There aren't actually many different ways for it to get from (0,0) to (7,9). They are:
9 up, 9 right, 2 left
11 up, 2 down, 7 right
10 up, 1 down, 8 right, 1 left
However, these can be arranged in different ways. To find how many ways each path can be rearranged, use the "word rearranging method," as I like to call it, that you use to rearrange words (for example, how many ways can the letters in the word ARRANGE be rearranged?"). Do this with each path. For the first path, you solve 20!/9!2!9!=9237800. For the second one, you solve 20!/11!2!7!=6046560. For the third one, you solve 20!/10!1!8!1!=16628040. Once you add these big number all up, you get a total of 31912400 number of ways to get from (0,0) to (7,9). To find the probability, divide this by the number of total paths it could take from (0,0) to any other point. Each time, it has 4 different directions it can take. It takes 20 steps, so the total number of paths is 4^20. Divide 31912400 by 4^20, and you get approximately 0.000029024. Fortunately, the problem tells you to multiply this by 10^8 and round to the nearest integer. Therefore, the answer is 2902.
ANSWER- 2902
P.S. These huge numbers may seem overwhelming, but remember that you are allowed to use calculators for the team round. :)
MathLeague Competition
ReplyDelete~Christine T^T
~Problems 2, 13, 19, 25, 31, 37, 43
2) Evaluate: 1+3+5+7+9+11+13+15+17+19.
Answer: 100
For this problem, you can use the formula for sum of n natural numbers, n^2.
10^2
=100
Therefore the answer is 100.
13) I guess I agree with Alice.
Answer: 17 pi
19) I agree with Alice.
Answer: k=3/4
25) I agree with Alice
Answer: 20/13
31) I agree with Alice
Answer: 750
37) I agree with Alice
Answer:0
43) I agree with Alice
Answer:1694
Here are two of the ones people don't know...
ReplyDelete#30 (Anindit is wrong)- Wayne flips a fair coin 4 times and Maria flips a fair coin 6 times. What is the probability that they flip the same number of heads? Express your answer as a common fraction.
Anindit was close on this one, except the 6th row of the Pascal's triangle was messed up. The 4th and 6th rows of Pascal's triangle are very useful in this problem. How you do it is this:
Probability of getting 0 heads:
1 * 1/16 * 64=1/1024
Probability of getting 1 head:
4 * 6/16 * 64=24/1024
Probability of getting 2 heads:
6 * 15/16 * 64=90/1024
Probability of getting 3 heads:
4 * 20/16 * 64=80/1024
Probability of getting 4 heads:
1 * 15/16 * 64=15/1024
Adding all these probabilities up, you get a total of 210/1024, which can be simplified to a final answer of 105/512.
ANSWER- 105/512
I'm going to do #38 next, but blogger won't allow me to because it has to be "at most 4,096 characters" or something like that.
#38 (Target Round 8)- In equilateral triangle ABC, points A1 and A2 on BC such that BA1=B1B2=B2A=CA/3, and points C1 and C2 lie on AB such that AC1=C1C2=C2B=AB/3. The intersections of AA1 with BB2 and CC2 are denoted by X1, X2, respectively, the intersections of BB1 with CC2 and AA2 are denoted by Y1, Y2, respectively, and the intersections of CC1 with AA2 and BB2 are Z1, Z2, respectively. Find the area of hexagon X1X2Y1Y2Z1Z2, given that AB=20. Express your answer in simplest radical form.
ReplyDeleteJust as a warning, I think it is practically impossible to do this problem PLUS target round #7 in 6 minutes, and that the few people who did it are Einstein-level geniuses. Just a precaution.
Anyway, first off, if you're actually going to try to solve this my/Ashish's way, you should get some graph paper and a ruler and stuff, even though you're not allowed to during the real contest. But whatever. Draw the triangle and everything in it as given in the problem. Set point A as (0,0) and point B as (20,0) because the side length of the triangle is 20. Triangle ABC is an equilateral triangle, so from here, you should be able to find that point C is at (10,10 root 3). Also, from the conditions given in the problem, you can figure out that points A1 and A2 trisect segment BC, points B1 and B2 trisect segment AC, and points C1 and C2 trisect segment AB. Once you know this, the coordinates of these points are easy to find. They are:
A1(50 / 3,10 root 3 / 3)
A2(40 / 3,20 root 3 / 3)
B1(20 / 3,20 root 3 / 3)
B2(10 / 3,10 root 3 / 3)
C1(20 / 3,0)
C2(40 / 3,0)
(I recommend you copy these down first.) Those who are in pre-algebra, algebra, or geometry should know how to find the equation of a line. If you don't, you should review/learn how to, because it will probably help in the near future. Now find the equation of each of the line segments mentioned in the problem, in slope-intercept form (y=mx+b). They should be:
AA1: y=root 3 / 5 x
AA2: y=root 3 / 2 x
BB1: y=-root 3 / 2 x + 10 root 3
BB2: y=-root 3 / 5 x + 4 root 3
CC1: y=3 root 3 x - 20 root 3
CC2: y=-3 root 3 x + 40 root 3
There is more, just wait.
Now that you have these equations, you can find the coordinates of their intersection points, aka the vertices of the hexagon. To do this, set the equations whose intersection point you want to find as being equal to each other. After all the calculations, you should get:
ReplyDeleteX1(10,2 root 3)
X2(12.5,2.5 root 3)
Y1(12,4 root 3)
Y2(10,5 root 3)
Z1(8,4 root 3)
Z2(7.5,2.5 root 3)
At this point, you should connect some of the points to get the following line segments on your graph: X2Y2, Y2Z2, Z2X2, and X1Y2. Now you should see that the hexagon is made up of one triangle (triangle XX2Y2Z2) and 3 congruent obtuse triangles (X1X2Z2, X2Y1Y2, and Y2Z1Z2). The triangle can be found to be equilateral with a side length of 5 after you use the distance formula. Calculate the length of segment X1Y2. You should get a distance of 3 root 3. You should also be able to find that the height of the equilateral triangle with base X2Z2 is 2.5 root 3. Because the height of the equilateral triangle with base X2Z2 and the height of the obtuse triangle X1X2Z2 with base X2Z2 directly underneath the larger triangle make up segment X1Y2, you can determine that the height of the obtuse triangles are all 0.5 root 3 (remember that all the obtuse triangle are congruent by symmetry). Now, finding the area of the hexagon should be easy. The area of an equilateral triangle is s^2 root 3 / 4, where s is the length of one side, so the area of triangle X2Y2Z2 is 5^2 root 3 / 4 = 25 root 3 / 4. All the obtuse triangles have a base of 5 and a height of 0.5 root 3. Use the normal triangle area formula to find that each obtuse triangle has an area of (1/2)(0.5 root 3)(5), which is 2.5 root 3 / 2. Multiply this by 3 (there are three obtuse triangles) to get an area of 7.5 root 3 / 2. This is the combined area of all 3 obtuse triangles. Finally, add this area to the area of the equilateral triangle that you found before, and you get a grand total of 40 root 3 / 4, which can be simplified down to a final answer of 10 root 3.
ANSWER- 10 root 3
See I told you the people who can solve this problem plus the one before it in 6 minutes are genius whiz kids...
Hope you understood. I'm terrible at explaining stuff. Please post another comment if you have any questions. :)
And by the way, there is another solution that is supposedly faster but harder to understand...it's the way that they show in the solutions packet. I might post it after I understand it...
@unknown:
ReplyDeleteI just saw this...
For that problem it does say ORDERED pairs so I assume it means that they can't be the same. ?