Can anyone help me with some really hard problems?

marmot83457

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Hi my teacher assigned me some math problem due very soon but i could not figure out how to solve them so can anyone please help me with these? thank you very much

Problem 1
An increasing arithmetic sequence with infinitely many terms is determined as follows.
A single die is thrown and the number that appears is taken as the first term. The die is thrown again and the second number that appears is taken as the common difference between each pair of consecutive terms. Determine with proof how many of the 36 possible sequences formed in this way contain at least one perfect square.

Problem 2
George has six ropes. He chooses two of the twelve loose ends at random (possibly from the same rope), and ties them together, leaving ten loose ends. He again chooses two loose ends at random and joins them, and so on, until there are no loose ends. Find, with proof, the expected value of the number of loops George ends up with.

Problem 3
Let r be a nonzero real number. The values of z which satisfy the equation
R^4z^4 + (10r^6 - 2r^2)z^2 - 16r^5z + (9r^8 + 10r^4 + 1) = 0 are plotted on the complex plane (i.e. using the real part of each root as the x-coordinate and the imaginary part as the y-coordinate). Show that the area of the convex quadrilateral with these points as vertices is independent of r, and find this area.

Problem 4
Homer gives mathematicians Patty and Selma each a di_erent integer, not known to the other or to you. Homer tells them, within each other’s hearing, that the number given to Patty is the product ab of the positive integers a and b, and that the number given to Selma is the sum a + b of the same numbers a and b, where b > a > 1. He doesn’t, however, tell Patty or Selma the numbers a and b. The following (honest) conversation then takes place:
Patty: “I can’t tell what numbers a and b are.”
Selma: “I knew before that you couldn’t tell.”
Patty: “In that case, I now know what a and b are.”
Selma: “Now I also know what a and b are.”
Supposing that Homer tells you (but neither Patty nor Selma) that neither a nor b is greater
than 20, find a and b, and prove your answer can result in the conversation above.

Problem 5
Given triangle ABC, let M be the midpoint of side AB and N be the midpoint of
side AC. A circle is inscribed inside quadrilateral NMBC, tangent to all four sides, and that circle touches MN at point X. The circle inscribed in triangle AMN touches MN at point Y , with Y between X and N. If XY = 1 and BC = 12, find, with proof, the lengths of the sides AB and AC.
 
Of the numbers 1 through 6, only 4 pairings are perfect squares.

16, 25, 36, and 64.
 
Homer gives mathematicians Patty and Selma each a di_erent integer, not known to the other or to you. Homer tells them, within each other’s hearing, that the number given to Patty is the product ab of the positive integers a and b, and that the number given to Selma is the sum a + b of the same numbers a and b, where b > a > 1. He doesn’t, however, tell Patty or Selma the numbers a and b. The following (honest) conversation then takes place:
Patty: “I can’t tell what numbers a and b are.”
Selma: “I knew before that you couldn’t tell.”
Patty: “In that case, I now know what a and b are.”
Selma: “Now I also know what a and b are.”
Supposing that Homer tells you (but neither Patty nor Selma) that neither a nor b is greater
than 20, find a and b, and prove your answer can result in the conversation above.

This is similar to one I ran across some time ago and should help you solve your own.

The math teacher asks Alan and Beth to each write down a positive number other than zero on a piece of paper. Both are to keep their numbers hidden from one another. After taking the two numbers from Alan and Beth, he calculates the sum and product of both numbers, writing each computation on separate pieces of paper. He then throws one of the papers away and shows the remaining one to Alan, Beth and Chris, another student wwho just happened to be present. The computed number shown to both students is 24.

The teacher now asks Alan if he knows what Beth's number is. He says no.

The teacher then asks Beth if she knows what Alan's number is. She says no.

The teacher then turns to Chris and asks if he knows either of Alan's or Beth's numbers. Chris says he knows Beth's number but not Alans.

Can you deduce just how Chris figured out what Beth's number was?


The first thing that Chris did was to review in his mind just what factual information he had to start with.
...1) Both Alan and Beth know only their individual numbers and the number 24 shown to them by the teacher.
...2) Chris has knowledge of only the number 24.
...3) Most importantly, none of them knows whether the number 24 is the sum or product of Alan's and Beth's numbers.
...4) Chris assumes that each of them is a top notch math student and each is able to derive the same information and draw the same conclusions from any subsequent conversations.

All realize that their are only 31 possible pairs of numbers that either add up to 24 or multiply out to 24. They are

A...B......A...B......A...B.....A..B (Alan or Beth)
1+23.....9+15.....17+7.....1x24
2+22....10+14....18+6.....2x12
3+21....11+13....19+5.....3x8
4+20....12+12....20+4.....4x6
5+19....13+11....21+3.....6x4
6+18....14+10....22+2.....8x3
7+17....15+9......23+1...12x2
8+16....16+8 ................24x1

When the teacher asked Alan if he knew Beth's number, he replied no.

Beth and Chris immediately know that Alan's number is not, 5, 7, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 or 23.

Why? If Alan's number was 5, for instance, he would know immediately that Beth's number was 19 as none of 5, 7, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 or 23 are factors of 24.

Beth and Chris can also conclude that Alan's number is not 24 for, if it was, Alan would immediatley know that Beth's number was 1 because Beth's number cannot be zero..

At this point, Beth and Chris both realize that Alan's number is one of the 7 remaining factors of 24, i.e., 1, 2, 3, 4, 6, 8 or 12.

Beth and Chris now know that the only possible pairings of numbers for Alan and Beth still possible are now
A...B...............................A...B
1+23...............................1x24
2+22...............................2x12
3+21...............................3x8
4+20...............................4x6
6+18...............................6x4
8+16...............................8x3
12+12...........................12x2

When the teacher asked Beth if she knew Alan's number, she also replied no.

By means of the same logic just outlined, Alan and Chris conclude that Beth's number is not 5, 7, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23 or 24.

Chris then eliminates every possibility that contains any of these numbers leaving
A...B...............................A...B
.......................................2x12
.......................................3x8
.......................................4x6
.......................................6x4
.......................................8x3
12+12...........................12x2

At this point, it is important to remember that each person has come to the same conclusions about each others numbers without knowing what the other's number is. Thus, these 7 pairings are the only possible solutions at this point.

In examining the 7 equations, Chris knows that if Beth's number was 2, he would know that Alan's number was 12, because he knows that Alan's number is not 22.

Similarly, Chris knows that if Beth's number was 3, he would know that Alan's number was 8, because he knows that Alan's number is not 21.

Similarly, Chris knows that if Beth's number was 4, he would know that Alan's number was 6, because he knows that Alan's number is not 20.

Similarly, Chris knows that if Beth's number was 6, he would know that Alan's number was 4, because he knows that Alan's number is not 18.

Similarly, Chris knows that if Beth's number was 8, he would know that Alan's number was 3, because he knows that Alan's number is not 16.

Therefore, the only remaining possibilities are
.A....B...............................A...B
12+12...............................2x12

Clearly, Beth's number is 12.

Chris now reasons that if Alan's number was 12, Beth's number could be either 2 or 12, which leads to his acknowledgement that he did not know Beth's number.

Chris then reasons that if Beth's number was 12, Alan's number could be either 2 or 12, which leads to her acknowledgement that she did not know Alan's number.

Chris then reasons that if Alan's number was 2, Beth's number could only be 12. But since Alan said he could not determine Beth's number, it could not be 2 and must be 12.
 
Well, marmot83457,

It is now going to be very difficult to submit a response that is "your own work". It is hoped that you are able to discern the difference between cheating and not cheating.
 
Question to the local cowboys on the "6 ropes":
is the probability of ending up with only 1 loop = (10/11)(8/9)(6/7)(4/5)(2/3) ?

Like:
1st move: pick an end
2nd move: you're ok with 10 ends (only the other end of the rope you
picked is a no-no), so 10/11

So you're now effectively left with 5 ropes...
 
From the source of the above six exercises:

The USAMTS said:
[T]he USAMTS allows students a full month to work out their solutions....Students may use any materials - books, calculators, computers - but all the work must be their own. The USAMTS is run on the honor system - it is an individual competition....We wish to foster not only insight, ingenuity and creativity, but also the virtue of perseverance....
Since helping this student cheat on the first round of questions will only lead to him failing spectacularly in later rounds (while possibly denying a place in those later rounds to a student who actually tried to do these exercises on his own), perhaps we ought to respect the "honor system" and help "foster" "the virtue of perseverance" by refraining from posting further assistance until after the October 3rd submission deadline...? And then, only if the poster shows some progress of his own...?

Eliz.
 
ya ya, agree; asking this for myself; don't think a yes/no answer will help him...
 
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