10th VTRMC, 1988, Solutions
- Let ABDE be the parallelogram S and let the inscribed circle C
have center O. Thus
∠AED = θ.
Let S touch C at P, Q, R, T. It is well
known that S is a rhombus; to see this, note that EP = ET, AP = AQ, BQ = BR and DR = DT. In particular O is the intersection
of AD and EB,
∠EOD = π/2 and
∠OED = θ/2.
Let x = area of S. Then x = 4 times area of EOD. Since
ED = ET + TD = r cotθ/2 + r tanθ/2, we conclude that
x = 2r2(cotθ/2 + tanθ/2).
- Let the check be for x dollars and y cents, so the original check
is for 100x + y cents. Then
100y + x - 5 = 2(100x + y).
98y - 196x = 5 + 3x. Of course, x and y are integers,
0≤x, y≤99. Since 98 divides 5 + 3x, we see
that x = 31 and hence
y = 2x + (5 + 3x)/98 = 63. Thus the
original check was for $31.63.
- If we differentiate
y(x) + ∫1xy(t) dt = x2 with respect to
x, we obtain
y' + y = 2x. This is a first order linear
differential equations, and the general solution is
y = Ce-x + 2x - 2, where C is an arbitrary constant.
However when x = 1, y(1) = 1, so
1 = C/e + 2 - 2 and hence C = e.
y = 2x - 2 + e1-x.
- If a = 1, then
a2 + b2 = 2 = ab + 1 for all n, and we see that
a2 + b2 is always divisible by ab + 1. From now on, we assume
an+1 + 1 divides
a2 + a2n, where n is a positive
an+1 + 1 divides
an-1 - a2 and hence
an+1 + 1 divides a4 + 1. Thus in particular
n = 3, then
an+1 +1 = a4 + 1 divides
a2 + a2n = a2 + a6.
If n = 1, then a2 + 1 divides 2a2 implies a2 + 1 divides 2,
which is not possible. Finally if n = 2, we obtain
a3 + 1 divides a2 + a4, so a3 + 1 divides a2 - a which again
is not possible.
We conclude that if a > 1, then a2 + b2 is divisible by ab + 1
if and only if n = 3.
- Using Rolle's theorem, we see that f is either strictly monotonic
increasing or strictly monotonic decreasing; without loss of
generality assume that f is monotonic increasing. Then
f'(X)≥2 for all x, so
|α - x0| < .00005. Thus the smallest
upper bound is .00005.
f (x) = ax - bx3 has an extrema when f'(x) = 0, that is
a - 3bx2 = 0, so
x = ±a/(√3b). Then
f (x) = ±2a2(3√3b). Since f has 4 extrema on
[- 1, 1], two of the extrema must occur at
±1. Thus we have
| a - b| = 1. Thus a possible choice is a = .1 and b = 1.1.
|f ([0, 1])
||= [0, 1/3]∪[2/3, 1]
|f (f ([0, 1]))
||= [0, 1/9]∪[2/9, 3/9]∪[6/9, 7/9]∪[8/9, 1]
|f (f (f ([0, 1])))
||= [0, 1/27]∪[2/27, 3/27]∪[6/27, 7, 27]∪[8/27, 9/27]∪
||∪[20/27, 21/27]∪[24/27, 25/27]∪[26/27, 1]
T⊆R be a bounded set such that f (T) = T.
First note that T contains no negative numbers. Indeed if T
contains negative numbers, let
t0 = inft∈Tt and choose
t1 < t0/3. Then
there is no
f (t) = t1.
Therefore we may assume that T contains no negative numbers. Now
(x + 2)/3 = 1/2 implies x = - 1/6, we see
3/2∈T. Now let
t2 = supt∈Tt and choose
t3∈T such that
t3 > (t2 + 2)/3. Since f (T) = T, we see that
s∈T such that
f (s) > (t2 + 2)/3, which is not
We conclude that there is no bounded subset T such that f (T) = T
- If we have a triangle with integer sides a, b, c, then we obtain a
triangle with integer sides
a + 1, b + 1, c + 1. Conversely if we have a
triangle with integer sides a, b, c and
the perimeter a + b + c is even, then
none of a, b, c can be 1 (because in a triangle, the sum of the
lengths of any two sides is strictly greater than the length of the
third side). This means we can obtain a triangle with sides
a - 1, b - 1, c - 1. We conclude that
T(n) = T(n - 3) if n is even.