Dr. Steve's Challenge Questions for an 800 in math

Test Preparation SAT Preparation
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<p>There was a problem presented earlier by one of the posters (conquerer7) to this thread that I felt needed a reply. The problem was this:</p>

<p>“If you choose two random real numbers between 0 and 1, what’s the expected value of the smaller one? (extension: what if you choose n numbers?).”</p>

<p>I hope I’m not intruding, so I promise just one post.</p>

<p>The problem requires a good understanding of probability theory and relatively advanced calculus to solve so it’s definitely not a SAT problem. What’s intriguing is the simplicity of the answer. For the case of two (independent) random numbers it’s 1/3, and for “n” random numbers it’s most likely 1/(n+1).</p>

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<p>Calculus works, but it’s tough for n > 2. The n=2 case isn’t as bad because you can ‘avoid’ the calculus. The calculus solution involves integrating (x)(x/2) where x is the first number: x is the probability of the second number being smaller, and x/2 is its average value, given that it’s smaller. But a simpler solution is to just graph the two numbers like they’re x and y! The set of possible pairs is a 1x1 unit square, and if we let y be the smaller value, our possible values are half of that square, split along the diagonal. The average value of the smaller is the average height of a point in the resulting triangle, which, since it’s the centroid, is at y = 1/3.</p>

<p>The most devastatingly simple solution of all involves symmetry. Imagine a circle with circumference 1. We randomly select three points on it. At our first point, we ‘cut’ the circle and unroll it, giving us the interval [0,1]. (the first point determines where 0 is) The second and third points are the two random numbers. Now our question is, what, on average, is the distance from the ‘smaller’ of those points to the first point? By symmetry, each of the points is on average 1/3 away from the rest, so the smaller value must be 1/3rd on average!</p>

<p>(I’m running on low sleep; apologies if this doesn’t make any sense)</p>

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<p>A cube is inscribed in a cone of radius 1 and height 2 so that one face of the cube is contained in the base of the cone. What is the length of a side of the cube?</p>

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<p>The slope of the slanted side of the cone=2/1=2
Let the side of the cube be x,
(2-x)/(x/2)=2
2-x=x
x=1</p>

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<p>My classmates asked me a question about probability which is quite intriguing.
In a chamber containing n people, who slaughter each other one by one rounds.
Each round one can only choose to target another person in the chamber randomly (no suicide), and the person who is targeted has no chance of escape, i.e. 100% dead.
For instance, there are four persons in the chamber, A, B, C and D.
First round A targets B, B targets A, C targets B, D targets A,
the ones who are targeted die in the first round (A and B).
In the second round C and D, who survived the first round, must target each other and both of them die. None survives after the game.
But in another situation, in the first round A targets B, B targets C, C targets A and D targets A, only D survives.
Find the probability so that one and only one person survives after the slaughtering, in terms of n.
He said he obtained the answer in 20 minutes.</p>

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<p>Probability that one person will survive is 1/n. Correct?</p>

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<p>I don’t have the answer, but I can assure it is incorrect.
When n=2, the probability should be 0, since the two persons must attack each other.</p>

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<p>For more specific case:
When n=3, there are three persons A, B, C in the chamber.
The condition for only one survivor is that, two of the three must target each other.
Otherwise the case would become either A->B, B->C, C->A or something like A->C, B->C, C->A => A->B, B->A, and all three die.
Let X<->Y symbolizes X and Y attack each other
P(A<->B)=P(B<->C)=P(C<->A)=(1/2)^2=1/4
P(A<->B or B<->C or A<->C)=P(A<->B)+P(B<->C)+P(A<->C)-P(A<->B and B<->C)-P(A<->B and A<->C)-P(B<->C and A<->C)+P(A<->B and B<->C and A<->C)
=1/4+1/4+1/4-(3)(1/4)^2+(1/4)^3
=3/4-3/16+1/64
=37/64
I don’t know if I am correct, it’s not a modal answer.</p>

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<p>I reached P(n)=1-P(n-1){1-[(n-3)/(n-2)]^(n-1)}, not yet derived a general formula.
Anyone tries to show your thoughts and ways?</p>

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<p>You’re over-complicating it.</p>

<p>Assume there are 3 people in the chamber. A, B, and C. We want C to escape unscathed. So, in order for that to happen, the following must happen:</p>

<p>A must choose to kill either B or C. Assuming both have the same probability of being chosen, the chance that C is not chosen is 1/2. </p>

<p>Now moving onto B. He can choose to kill either A or C. Assuming that both have the probability of being chosen, the chance that C is not chosen, is again 1/2.</p>

<p>And of course C will not choose to kill himself.</p>

<p>So the chances that C will escape unscathed, are 1/2 x 1/2 = 1/4</p>

<p>Another example: 5 people. A, B, C, D, and E.
A can choose to kill B, C, D, or E. We want E to stay alive. He has a chance of 3/4 of not being chosen.</p>

<p>B can choose to kill A, C, D, or E. We want E to stay alive. That’s another chance of 3/4 of not being chosen.</p>

<p>Same with C.</p>

<p>Same with D.</p>

<p>3/4 x 3/4 x 3/4 x 3/4 = 81/256.</p>

<p>My point: This is a very stupid question. (no offense)
If you want a formula though, it would probably look something like this:
(n-2/n-1) ^ (n-1)</p>

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<p>

</p>

<p>This is not correct. Let me rewrite the original question so it doesn’t get lost in the thread:</p>

<p>A cube is inscribed in a cone of radius 1 and height 2 so that one face of the cube is contained in the base of the cone. What is the length of a side of the cube? </p>

<p>Here’s a hint: slice the cone from the vertex to the base so that it cuts through the diagonal of the square base of the cube. What does this cross section look like?</p>

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<p>SirWanksALot, you seem to have misread the question twice. You obtained the chance of an arbitrary person surviving one round, which, in this situation, is not a useful quantity. Most of the time the game will last for more than one round, and the number of people remaining for the second round may vary.</p>

<p>warsoverign, I can’t see why your recurrence is true. Can you show how you got it? How could P(n) depend on P(n-1) like that if at least two people have to die each round?</p>

<p>I honestly have no idea how anybody could do this in 20 minutes; the situation seems extremely complex. Just figuring out the chances of so many people dying in any round seems to be very difficult.</p>

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<p>Is the answer to the cone question 2/sqrt(2)?</p>

<p>Since the radius is 1, then cutting the cone in half from the vertex of the squares would make a base hypothesis of 2 and a 45, 45, 90 triangle. So for that reason, the answer would be 2/sqrt(2). Hopefully I didn’t misinterpret the question.</p>

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<p>Ummmm…is it 2^8=256</p>

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<p>2/sqrt(2) is not the right answer. When you slice the cone as I suggested the cross section is a rectangle inscribed in an isosceles triangle. What are the dimensions of the rectangle in terms of x, the side length of the square? What are the base and height of the triangle? And how do you solve for x?</p>

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<p>Let f be a function defined on the points P = (a,b) with integer coordinates both between 0 and 2, inclusive. Suppose that f (A) + f (B) + f (C) + f (D) = 0 whenever ABCD is a square. Show that f (Q) = 0 , where Q = (1,1).</p>

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<p>Ha! Thought this question looked familiar: <a href=“American Mathematics Competitions | Mathematical Association of America”>American Mathematics Competitions | Mathematical Association of America;

<p>Your modification is great: It’s slightly easier now, but anyone who solves it will have the fundamental insight behind the general solution.</p>

<p>Excellent question! Thanks for posting it.</p>

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<p>Good catch satsb. The putnam exams are a great source of excellent problems, but they are generally way too difficult. If you look through this thread you’ll find one other “watered down” putnam problem. I may post more like that. The challenge is to find versions that are difficult but doable.</p>

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<p>I have absolutely no clue how to solve this…</p>

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<p>@Jeffery</p>

<p>Have you tried to start by drawing some pictures?</p>