unc16 solutionsbyrcv

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Solutions to questions 1 through 6(a), by Rocke Verser February 1, 2016

1. Even on the easiest problem on a contest, I often make a sketch and/or copy the information from the question. I find it helps me to internalize the problem, and to reduce "dumb" mistakes. Since this is a right triangle, I immediately think of the Pythagorean Theorem.

2 2 2

1250 56

c b a

= +

= + +

Solution 1: Pythagorean Triplets. (A nice hint from the name of the question.) Combine the last two equations written above.

25 625

1250

2 2 2

= =

= +

c c

The most well-known Pythagorean Triplet is the 3-4-5 triangle. If we scale this triangle by a factor of 5, we will have the correct hypotenuse. That would give us a 15-20-25 triangle. But the sum of sides is

15+20+25=60. And that doesn't meet the other specification of the problem.

The first series of Pythagorean Triplets continues with odd sides: 3-4-5, 5-12-13, 7-24-25, 9-40-41, ... . The 7-24-25 triangle has the correct hypotenuse. Check the sum of the sides, 7+24+25=56. This meets all the criteria of the question. Now, we can answer the question: "What is the area?" Using the formula for the area of a triangle, we get the final answer:

84 7 12 7 24

2 1 2

1 × = × × = × =

= base height A

Solution 2: Algebra. First, we solve for c. Then we solve for a+b. A technique that sometimes helps with right triangle problems that don't seem to give enough information is to square the sum or difference of two sides.

25 625

1250

2 2 2

= = = +

c c

(

)

961 2

31 31

56 25

56

2 2

= + +

= +

= + +

b ab a

b a

c b a

We already found a2 +b2 =c2 =625. Subtracting the last two equations leaves us with the area of 4 congruent

triangles. 2 961 625 336 ₂1 84

2

1 × = =

⇒

= −

= base height ab

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2. The factorial of n is the product of all of the positive integers from n down to 1. The first step is to break the expression down starting from the innermost parentheses.

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quotients involving factorials are often easier to divide when they are already factored. Second, this question is specifically asking for the "prime factorization". So, anything we multiply will have to be factored, later.

Probably the most straightforward way to solve this is to write out all the factors in the numerator and all the factors in the denominator and "cancel" out factors.

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This is a perfectly fine solution method. A subtle enhancement is to group the numerator into 4 sets of 6 consecutive integer factor. Each group of 6 factors in the numerator is evenly divisible by 6!. (Think about why that's true, as we cancel factors.)

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) (

)

The rightmost fraction completely cancels. Rearranging the factors in the other denominators:

19

Now, everything is completely factored, except 4=22, so we can directly write the final answer:

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3. First, if the question had not provided a reasonably nice sketch, that is the very first thing I would draw. Even with the given perspective sketch, I still need to draw and label a face of the cube and an equilateral triangle that replaced a corner.

The area of the face of the cube was originally the area of a square of side-length 1. But we shaved off four isosceles right triangles, each with an unknown side length of x. So, the remaining area of the face of the square

is 2 ₂1 2 2

2 1 4

1 − × x = − x . The hypotenuse of the isosceles right triangle has the same length as the side-length of

the equilateral triangles we created in place of the corners, s= 2x.

A student may know the area of an equilateral triangle in terms of its side-length. If not, divide the equilateral triangle vertically into a 30-60-90 triangle. The ratio of the side-lengths of any 30-60-90 triangle is 1: 3:2. All of the side-lengths can be found by proportions if any of the three side-lengths are known. Or, if these proportions are not known, we can use the Pythagorean Theorem. The length of the hypotenuse is s. The length of the short side is ₂1s_{(because we just cut the equilateral triangle in half). The unknown side (the}

height) is found by the Pythagorean Theorem. With the height now known, we can find the area of the equilateral triangle.

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s h

s h s

2 3 1

2 4 3 2

2 4 1 2

2 2 2 2 1

= =

− =

= +

2 60 60 60

4 3

2 3 2

1 2 1

s A

s s A

height base

A

=

× × =

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The requested area, A, is equal to the area of eight equilateral triangles. It is also equal to area of the octagonal face of the cube. Let's find both areas in terms of s2.

2 2

3 2

4 3 8

s A

= × =

2 2 2

2

1 2

2 2 1

s A

x s

x A

− =

= =

− =

Finally, let's equate the two areas, solve for s2, and then substitute to find A. Although this problem didn't explicitly ask, some contests prefer that you rationalize the denominator.

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)

1 3 2

3 2

1 3 2

1 1 1 3 2

1 3 2

2 2

2 2 2

+ =

× =

+ =

= +

− =

A

s A

s

s s s

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11 3 2 12 1 3 4

3 2 12

1 3 2

3 2 3 4

1 3 2

3 2

2 2

− = − ⋅ − =

− − ⋅ =

− − × + =

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4. Solution 1: Number Sieve. (A hint from the name of the question.) If a student has seen how the Sieve of Eratosthenes is used to quickly and reliably find the prime numbers less than 100, a similar strategy will work for this problem. First, we write the numbers from 1 to 99 or make a grid representing the numbers from 1 to 99. Then, for each factor given in the question, we put a tick mark on top of (or in the grid) for every number that is divisible by the specified factor.

0 1 2 3 4 5 6 7 8 9 0 / / // / /// / /// //

10 // //// // // /// ////

20 /// // / ///// / / // ///

30 //// /// / / // ///// / /

40 //// //// // /// / ///// /

50 // / // //// / //// / /

60 ///// / /// /// / /// // /

70 /// ////// / // // / ///

80 //// // / ///// / / / ///

90 ///// / // / / / ///// // //

Most of the tricks that make the Sieve of Eratosthenes fast (and clever) don't apply here. You have to

laboriously mark each factor of each number. You have to be careful not to obscure one tick mark with another. And if you lose track of where you are, you may have to start from the beginning.

Once the tick marks are done, find the numbers that are ticked exactly twice. I would suggest you write down those numbers (for partial credit). I found 18 such numbers. They are 4, 9, 10, 14, 15, 21, 27, 35, 44, 50, 52, 68, 75, 76, 81, 92, 98, and 99.

Solution 2. Logical Reasoning. This method may be less tedious and more fun. But, I had a difficult time convincing myself I had not missed any categories.

• Powers of 3: (9, 27, 81)

• Powers of 2: (4)

• Odd prime (>9) multiples of 9: 9 × (11)

• Odd prime (>9) multiples of 4: 4 × (11, 13, 17, 19, 23)

• Odd multiples of 2: 2 × (5, 7, 52, 72)

• Odd multiples of 3: 3 × (5, 7, 52)

• Odd multiples of 5: 5 × (7)

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5. Solution 1. Probability tree. Probability trees are a safe and effective way to solve this kind of problem. However, if the problem only wants a tiny bit of information, we sometimes do a lot of extra work to complete the entire tree. Nevertheless, that's the method we will use for this solution.

There are several different ways to draw and label a probability tree. In my sketch, we start at the top and work our way to the bottom. The top row represents the start of Day 1. The next row is the start of Day 2. And the bottom row is the start of Day 3. Horizontally, the left column represents zero rocks in the valley. Each subsequent column represents one more rock in the valley. The numbers in the nodes (circles) represent the probability that we are in that state at the start of the specified day. For example, we are guaranteed to have 1 rock in the valley on the start of Day 1, so we put "1" in the circle corresponding to "Day 1, 1 rock". Each edge (line segment) in our tree shows how the status can change from one day to the next. The numbers that label each edge represent the probability that the event will occur. So the edge that connects the node at Day 1 to the leftmost node at Day 2 indicates there is a 1/10 probability that we will go from 1 rock in the valley to zero rocks in the valley. The sum of the edges that leave any circle must always total 1. The numbers in each node can be found by summing the products of each probability along an incoming edge by the value in the

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The edge leaving "Start of Day 2 / 0 rocks" shows a probability of 1. Once there are zero rocks in the valley, there will always be zero rocks in the valley. The edges leaving "1 rock" are given directly by the problem.

The edges leaving "2 rocks" require some additional thought. Each rock is independently vaporized, rolled back down into the valley, or split into two rocks. The probability that both rocks are vaporized is 0.12=0.01. The probability that both rocks are duplicated is 0.42=.16. One rock can be vaporized and one rock can be rolled back down into the valley (resulting in 1 rock in the valley) in 2 different ways. So, the probability this occurs is 2×0.1×0.5=0.10. One rock can be duplicated and one rock can be rolled back down into the valley (resulting in 3 rocks in the valley) in 2 different ways. So, the probability this occurs is 2×0.4×0.5=0.40. And finally, one rock can be vaporized and one rock can be duplicated in two different ways (2×0.1×0.4=0.08) or both rocks can be rolled back down into the valley (0.5×0.5=0.25). So the probability we go from 2 rocks to 2 rocks is 0.08+0.25=0.33.

Let's check our probabilities of leaving "Start of Day 2 / 2 rocks". 0.01+0.10+0.33+0.40+0.16 = 1.00, which is the correct total.

Next, we sum the probabilities entering the nodes along "Start of Day 3". For the node with zero rocks, we have 0.1×1+0.5×0.1+0.4×0.01=0.1+0.05+0.004=0.154. For the node with one rock, we have

0.5×0.5+0.4×0.10=0.25+0.04=0.29. For the node with two rocks, we have

0.5×0.4+0.4×0.33=0.2+0.132=0.332. For the node with three rocks, we have 0.4×0.40=0.16. And for the node with four rocks, we have 0.4×0.16=0.064.

Let's check the total probabilities inside each node along "Start of Day 3". 0.154+0.290+0.332+.160+0.064=1.000, which is the correct total.

The chance of having 2 rocks in the valley at the start of Day 3 is 0.332 or 33.2%.

If you are confident in your skills with probability trees, we weren't required to compute the complete

probability tree. In fact, we only needed to compute the edges and nodes leading to the node at the start of Day 3 with 2 rocks. (4 edges and 4 nodes.)

Solution 2. Logical Reasoning. We can list all of the ways of ending up with 2 rocks:

Roll the rock into the valley (5/10 probability). Then duplicate the rock (4/10 probability). 20/100.

Duplicate the rock (4/10 probability). Then vaporize one, duplicate the other (4/10*1/10*4/10 * 2 ways) = 32/1000; or roll both rocks back down into the valley (4/10*5/10*5/10=100/1000).

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6(a). We can simplify the problem by ignoring Sisyphus and the valley and the days. The question simply wants to know the probability that the number of rocks never reaches 0, when starting with 1 rock. Zeus picks up a rock at his leisure and vaporizes it (with probability 1/10), returns it to the pile (with probability 5/10), or splits the rock in two (with probability 4/10). If there are no rocks remaining, Zeus rolls over and goes to sleep.

If Zeus returns the rock to the pile, nothing has changed, so we can further simplify the problem by ignoring that action, and considering only the relative probability of the other actions.

Zeus picks up a rock at his liesure. He vaporizes it (with probability 1/5) or splits the rock in two (with probability 4/5). (The probabilities of vaporization:duplication = 1:4, as before the simplification.)

Solution via State Diagram:

We can represent the situation with a State Diagram. There is a single state for every possible number of rocks in the pile. Zero rocks is State S0. One rock is State S1. Etc. For each state, we have a variable nk which represents the total number of times the system has entered that state from the beginning to the end of time. See attached illustration.

We can write formulas to relate the variables, n, with each other. For example, if you are in State S1, Zeus will vaporize that one and only rock with probability 1/5. So, the number of times we enter State S0 is exactly equal to 1/5 the number of times we enter State S1. We enter State S1 once when the system is initialized. We also enter State S1 when we were in State S2 and Zeus vaporizes one of the two remaining rocks. So, the net

number of times we enter State S1 is exactly

5 1 2 1

n

n = + . At each other State, Sk we can enter the State when

Zeus splits a rock while we were in State Sk-1 or when Zeus vaporizes a rock while we were in State Sk+1. We have an infinite number of simple linear equations and an infinite number of unknowns. We could make an arbitrary guess and solve the system of equations iteratively. Or we could make a good guess and prove that our guess satisfies the system.

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Let me redraw the State diagram ever so slightly:

Now, notice that state S1 looks almost the same as States S2 and higher. What if we guess that it *really* is the same as State S2? Let's set the input entering State S2 from the left equal to the input entering State S1 from the

left:

4 5 5 4

1= n₁ ⇒ n₁ = . Let's solve for n₂:

4 5 4

20 -25 n 5 1 4 5 5

1 2 2 ₂

1 = + ⇒ = + ⇒ = =

n n

n . Now, it is a

simple matter to confirm , 1 4

5 _≥

= k

n_k . Then we solve for

4 1 5

1 0 = =

n

n . The final answer is

4 3 1−n₀ = .

6(a). Solution Estimate via Probability Tree

See the adjoining probability tree. The leftmost column shows the probability the Zeus has vaporized the only remaining rock. The next column shows the probability that one rock remains. At each step, Zeus adds a rock or destroys a rock, so the number of rocks remaining alternates from odd to even. (Unless it was already zero.)

The total probability at each stage must equal 1. So, before Zeus picks up the solitary rock, the probability that 1 rock is present is exactly 1. (Represented by the number in the circle at the top of the tree.)

After Zeus handles a rock for the first time, we are on the next row. The total probability is 0.2+0.8=1. There is a 1/5

probability that Zeus destroyed the rock, and there is an 4/5 probability that there are now 2 rocks.

At the next stage, Zeus has either destroyed one of the two rocks remaining or he has split one of the two rocks remaining, and now we have three rocks. The total probability is 0.2 (no rocks remained from the previous step) + 0.16 (one rock remains) + 0.64 (three rocks are in the pile) = 1.00.

After the fourth step, the probabilities for 0, 1, 3, or 5 rocks are 0.232 + 0.0512 + 0.3072 + 0.4096 = 1.0000. After the ninth step, the probability that there are no rocks remaining is 0.248171008.

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6(a). Solution via Probability Tree

This is going to be an infinitely wide and deep probability tree. We need a formula. So, I'll begin by counting the number of paths that get us to each place where Zeus has destroyed the last rock. See attached diagram. The numbers in the circles

represent the number of paths from the root of the tree to that particular node. This was performed using standard

MATHCOUNTS path counting methods.

Note the 14s in the circles near the bottom. That indicates there are 14 paths from the root of the tree to those particular nodes. The lower-left of the two nodes is where Zeus has just destroyed the final rock. No matter how we get from the root to this node, Zeus has created 4 additional rocks and he has vaporized 5 rocks. Since Zeus has 14 ways to do this, the formula for the probability

that we reach this node is 0.001835008

5

which number you may recognize from the previous solution.

The sequence we have found begins with 1, 1, 2, 5, 14. Not surprisingly, these are the Catalan Numbers. (Also known as mountain numbers. If we rotate the diagram counterclockwise 90 degrees, we can imagine climbing a mountain, possibly with ups and downs, and eventually descending back to earth. C(0)=1, C(1)=1, C(2)=2, C(3)=5, C(4)=14.

Now we can write a general formula for each of the nodes where Zeus has just destroyed the last rock:

n

The total probability that all rocks will eventually be destroyed, is given by the infinite sum:

∑

∞

In general, the contest awards more points for a closed-form solution. However, as this is an infinite sum and the problem asks for "the probability", I am skeptical that much credit would be awarded without a closed form solution. Although I am not a grader, I would recommend a student should spend some of their 3 hours

attempting to reduce the infinite sum to a number. At the very least, I think a student should provide a convincing argument that the sum converges to a valid probability (a number between 0 and 1).

If we let

( represent each term in the summation, above, we can find the ratio between

consecutive terms. Using the formula for C(n), above, a little algebra quickly shows that

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Since the ratios of every term in our sequence are less than the ratios in a geometric series, we can find the upper bound on our summation, using the formula for the sum of a geometric series. (Although this may look too scary to do without a calculator, if you know your powers of 2, only the final term we show really requires any hand calculation (a power of 2, decimal shifted, times 14, divided by 9).

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No single step, above, is exceptionally hard. Putting it all together is rather brutal. I think most students who could put all this together will probably find an easier solution. However, with a few more (non-trivial) steps, the summation can be reduced to a closed form, but I will leave that as an exercise for the very dedicated reader.