15th VTRMC, 1993, Solutions
- We change the order of integration, so the integral becomes
∫01∫0√yey3/2 dxdy = ∫01y1/2ey3/2 = [2ey3/2/3]01 = (2e - 2)/3
- Since f is continuous,
∫0xf (t) dt is a differentiable
function of x, hence f is differentiable. Differentiating with
respect to x, we obtain
f'(x) = f (x). Therefore
f (x) = Aex
where A is a constant. Since f (0) = 0, we see that A = 0 and we
conclude that f (x) is identically zero as required.
- From the definition, we see immediately that
fn(1) = 1 for all
n≥1. Taking logs, we get
ln fn+1(x) = fn(x)ln x.
Now differentiate both sides to obtain
f'n+1(x)/fn+1(x) = fn'(x)ln x + fn(x)/x. Plugging in x = 1, we find that
f'n+1(1)/fn+1(1) = fn(1) for all
n≥1. It follows that
f'n(1) = 1 for all
f'n+1(x) = fn+1(x)fn'(x)ln x + fn+1(x)fn(x)/x,
f''n+1(x) = f'n+1(x)fn'(x)ln x + fn+1(x)fn''(x)ln x + fn+1(x)fn'(x)/x + fn+1'(x)fn(x)/x + fn+1(x)fn'(x) - fn+1(x)fn(x)/x2.
Plugging in x = 1 again, we obtain
f''n(1) = 2 for all
- Suppose we have an equilateral triangle ABC with integer
u = and
v = . Then by expressing the
cross product as a determinant, we see that
|u×v| is an integer. Also
|u|2 is an integer, and
|u×v| = |u|2sin(π/3) because
∠BAC = π/3. We
sin(π/3) = √3/2 is a rational number, which
is not the case.
- For | x| < 1, we have the geometric series
∑n=0∞xn = 1/(1 - x). If we integrate term by term
from 0 to x,
∑n=1∞xn/n = - ln(1 - x), which is also
valid for | x| < 1. Now plug in x = 1/3: we obtain
∑n=1∞3-n/n = - ln(2/3) = ln 3 - ln 2.
- Suppose f is not bijective. Since f is surjective, this means
that there is a point
A0∈R2 such that at least two
distinct points are mapped to A0 by f. Choose points
B0, C0∈R2 such that
A0, B0, C0 are not collinear.
Now select points
A, B, C∈R2 such that
f (A) = A0,
f (B) = B0 and
f (C) = C0. Since f maps collinear points to
collinear points, we see that A, B, C are not collinear. Now given
two sets each with three non-collinear points, there is a bijective
affine transformation (i.e. a linear map composed with a
translation) sending the first set of points to the second set. This
means that there are bijective affine transformations
g, h : R2 --> R2 such that
g(0, 0) = A,
g(0, 1) = B,
g(1, 0) = C,
h(A0) = (0, 0),
h(B0) = (0, 1),
h(C0) = (1, 0).
k : = hfg : R2 --> R2 fixes
(0, 0),(0, 1),(1, 0), and has the property that if P, Q, R are
collinear, the so are
k(P), k(Q), k(R). Also there is a point
(a, b)≠(0, 0) such that
k(a, b) = (0, 0). We want to show that
this situation cannot happen.
Without loss of generality, we may assume that
ℓ denote the line joining (1, 0) to (0, b). Then
k(ℓ) is contained in the x-axis. We
claim that k maps the horizontal line through (0, 1) into itself.
For if this was not the case, there would be a point with
coordinates (c, d ) with
d≠1 such that
k(c, d )= (1, 1). Then
if m was the line joining (c, d ) to (0, 1), we would have k(m)
contained in the horizontal line through (0, 1). Since m
intersects the x-axis and the x-axis is mapped into itself by
k, this is not possible and so our claim is established. Now
ℓ meet this horizontal line at the point P. Then we have
that k(P) is both on this horizontal line and also the x-axis,
a contradiction and the result follows.
- The problem is equivalent to the following. Consider a grid in the
xy-plane with horizontal lines at y = 2n + 1 and vertical lines at
x = 2n + 1, where n is an arbitrary integer. A ball starts at the
origin and travels in a straight line until it reaches a point of
intersection of a horizontal line and a vertical line on the grid.
Then we want to show that the distance dft
travelled by the ball is not an integer number of feet. However
d2 = (2m + 1)2 + (2n + 1)2 for some integers m, n and hence
d2≡2 mod 4. Since d2 = 0 or 1 mod 4, we have a contradiction and
the result is proven.
- The answer is 6; here is one way to get 6.
In this diagram, pieces of wire with the same corresponding number
belong to the same logo. Also one needs to check that a welded point
is not contained in more than one logo.
On the other hand each logo has three welded points, yet the whole
frame has only 20 welded points. Thus we cannot cut more than 20/3
logos and it follows that the maximum number of logos that can be
cut is 6.