25th VTRMC, 2003, Solutions

  1. The probability of p gains is the coefficient of (1/2)p(1/2)n - p in (1/2 + 1/2)n. Therefore, without the insider trading scenario, on average the investor will have 10000(3/5 + 9/20)n dollars at the end of n days. With the insider trading, the first term (3/5)n becomes 0. Therefore on average the investor will have

    10000(21/20)n - 10000(3/5)n

    dollars at the end of n days.

  2. We have

    -ln(1 - x) = x + x2/2 + x3/3 + ... = Si = 1$\scriptstyle \infty$(xn)/n.

    Therefore

    (1 - x)ln(1 - x) = - x + x2/2 + x3/6 + ...    
      = - x + Si = 1$\scriptstyle \infty$xn + 1(1/n - 1/(n + 1)) = - x + Si = 1$\scriptstyle \infty$xn + 1/(n(n + 1)).    

    Dividing by x, we deduce that

    Sn = 1$\scriptstyle \infty$xn/(n(n + 1)) = 1 + ((1 - x)ln(1 - x))/x

    for x =/=  0, and the sum is 0 for x = 0.

  3. Let I denote the 2 by 2 identity matrix. Since A = A-1, we see that A2 = I and hence the eigenvalues l of A must satisfy l2 = 1, so l = ±1. First consider the case det A = 1. Then A has a repeated eigenvalue ±1, and A is similar to

      (
     r s
     0 r
    		)

    where r = ±1 and s = 0 or 1. Since A2 = I, we see that s = 0 and we conclude that A = ±I.

    Now suppose det A = - 1. Then the eigenvalues of A must be 1, - 1, so the trace of A must be 0, which means that A has the form

      (
     a b
     b - a
    		)

    where a, b are complex numbers satisfying a2 + b2 = 1. Therefore b = (1 - a2)1/2 (where the exponent 1/2 means one of the two complex numbers whose square is 1 - a2). We conclude that the matrices satisfying A = A' = A-1 are ±I and

      (
     a (1 - a2)1/2
      (1 - a2)1/2 - a
    		)

    where a is any complex number.

  4. Set R = e2pi/7 = cos 2p/7 + isin 2p/7. Since R =/= 1 and R7 = 1, we see that 1 + R + ... + R6 = 0. Now for n an integer, Rn = cos 2np/7 + isin 2np/7. Thus by taking the real parts and using cos(2p - x) = cos x, cos(p - x) = - cos x, we obtain

    1 + 2cos(2p/7) - 2cos(p/7) - 2cos(3p/7) = 0.

    Since cosp/7 + cos 3p/7 = 2cos(2p/7)(cosp/7), the above becomes

    4cos(2p/7)cos(p/7) - 2cos(2p/7) = - 1.

    Finally cos(2p/7) = 2cos2(p/7) - 1, hence (2cos2(p/7) - 1)(4cos(p/7) - 2) = - 1 and we conclude that 8cos3(p/7) - 4cos2(p/7) - 4cos(p/7) = - 1. Therefore the rational number required is -1/4.

  5. Since / ABC + / PQC = 90 and / ACB + / PRB = 90, we see that / QPR = / ABC + / ACB. Now X, Y, Z being the midpoints of BC, CA, AB respectively tells us that AY is parallel to ZX, AZ is parallel to XY, and BX is parallel to YZ. We deduce that / ZXY = / BAC and hence / QPR + / ZXY = 180. Therefore the points P, Z, X, Y lie on a circle and we deduce that / QPX = / ZYX. Using BZ parallel to XY and BX parallel to ZY from above, we conclude that / ZYX = / ABC. Therefore / QPX + / PQX = / ABC + / PQX = 90 and the result follows.

  6. Set g = f2. Note that g is continuous, g3(x) = x for all x, and f (x) = x for all x if and only if g(x) = x for all x. Suppose y e [0, 1] and f (y) =/= y. Then the numbers y, f (y), f2(y) are distinct. Replacing y with f (y) or f2(y) and f with g if necessary, we may assume that y < f (y) < f2(y). Choose a e (f (y), f2(y)). Since f is continuous, there exists p e (y, f (y)) and q e (f (y), f2(y)) such that f (p) = a = f (q). Thus f (p) = f (q), hence f3(p) = f3(q) and we deduce that p = q. This is a contradiction because p < f (y) < q, and the result follows.

  7. Let the tetrahedron have vertices A, B, C, D and let X denote the midpoint of BC. Then AX = $ \sqrt{1 - 1/4}$ = $ \sqrt{3}$/2 and we see that ABC has area $ \sqrt{3}$/4. Let R, S, T, U denote the regions vertically above and distance at most 1 from ABC, BCD, ABD, ACD respectively. Then the volumes of R, S, T and U are all $ \sqrt{3}$/4. Since these regions are disjoint, they will contribute $ \sqrt{3}$ to the volume required.

    Let Y denote the point on AX which is vertically below D. Then Y is the center of ABC (i.e. where the medians meet), in particular / YBX = p/6 and we see that BY = 1/$ \sqrt{3}$. Therefore DY = $ \sqrt{1 - 1/3}$ = $ \sqrt{2/3}$ and we deduce that ABCD has volume

    (1/3)*($\displaystyle \sqrt{3}$/4)*$\displaystyle \sqrt{2/3}$ = $\displaystyle \sqrt{2}$/12.

    Next consider the region which is distance 1 from BC and is between R and S. We need the angle between R and S, and for this we find the angle between DX and DY. Now DX = AX = $ \sqrt{3}$/2 and DY = $ \sqrt{2/3}$. Therefore XY = $ \sqrt{3/4 -2/3}$ = 1/(2$ \sqrt{3}$). If q = / YDX, then sinq = XY/DX = 1/3. We deduce that the angle between R and S is p/2 + q = p/2 + sin-11/3. Therefore the region at distance 1 from BC and between R and S has volume p/4 + (sin-11/3)/2. There are 6 such regions, which contribute 3p/2 + 3sin-11/3 to the volume required.

    For the remaining volume, we shrink the sides of the tetrahedron to zero. This keeps the remaining volume constant, but the volumes above go to zero. We are left with the volume which is distance 1 from the center of the pyramid, which is 4p/3. Since 3p/2 + 4p/3 = 17p/6, we conclude that the volume of the region consisting of points which are distance at most 1 from ABCD is $\displaystyle \sqrt{3}$ + $\displaystyle \sqrt{2}$/12 + 17p/6 + 3sin-1(1/3) $\displaystyle \approx$ 11.77. Other expressions for this are $\displaystyle \sqrt{3}$ + $\displaystyle \sqrt{2}$/12 + 13p/3 - 3cos-1(1/3) and $\displaystyle \sqrt{3}$ + $\displaystyle \sqrt{2}$/12 + 13p/3 - 6sin-1(1/$\displaystyle \sqrt{3}$).



Peter Linnell
2003-11-20