|
1.
Prove that (21n+4)/(14n+3) is irreducible for every natural number n.
|
|
|
|
Solution 3(14n+3)-2(21n+4)=1.
|
|
|
2.
For what real values of x is Ö(x + Ö(2x-1)) + Ö(x - Ö(2x-1)) = A given (a) A = Ö2, (b) A = 1, (c) A = 2,
where only non-negative real numbers are allowed in square roots and the root
always denotes the non-negative root?
|
|
|
|
Solution Care is needed to take the positive root. Taking one
term across to the other side before squaring is a bad idea, because the
product of the two main roots simplifies to |x-1|.
We find: (a) any x in the interval [1/2,1]; (b) no solutions (3/4 is
not a solution); (c) x=3/2.
|
|
|
3.
Let a, b, c be real numbers. Given the equation for cos x:
a cos2x + b cos x + c =
0,
form a quadratic equation in cos 2x whose roots are the same values of x.
Compare the equations in cos x and cos 2x for a=4, b=2, c=-1.
|
|
|
|
Solution You need that cos 2x = 2 cos2x
- 1. Some easy manipulation then gives:
a2cos22x + (2a2 + 4ac - 2b2)
cos 2x + (4c2 + 4ac - 2b2 + a2) = 0.
The equations are the same for the values of a, b, c given. The
angles are 2pi/5 (or 8pi/5) and 4pi/5 (or 6pi/5).
|
|
|
4.
Given c, construct a triangle ABC with angle ABC 90 degrees, AC=c, and the
median AM satisfying AM2 = AB.AC.
|
|
|
|
Solution Area =AB.BC/2 = AC2/8,
so B lies a distance AC/4 from AC. Take B as the intersection of a circle
diameter AC with a line parallel to AC distance AC/4.
|
|
|
5.
An arbitary point M is taken in the interior of the segment AB. Squares AMCD
and MBEF are constructed on the same side of AB. The circles circumscribed
about these squares, with centers P and Q, intersect at M and N.
(a) prove that AF and BC intersect at N;
(b) prove that the lines MN pass through a fixed point S (independent
of M);
(c) find the locus of the midpoints of the segments PQ as M varies.
|
|
|
|
Solution (a)Angle ANM = angle
BNM = 45 (half the angle at the center of the respective circle). So N lies
on the semicircle diameter AB. We can also easily check that angle ANM +
angle FNM = 180, so A, N, F are collinear. A similar idea works for B, C,
N.
(b) Let NM meet the circle diameter AB again at X. Angle ANS =
angle BNS implies AS = BS and hence S is a fixed point.
(c) Clearly the height of the midpoint above AB is AB/4. Since
it varies continuously with M, it must be the interval between the two
extreme positions, so the locus is a segment length AB/2 centered over AB.
|
|
|
6.
The planes P and Q are not parallel. The point A lies in P but not Q, and the
point C lies in Q but not P. Construct points B in P and D in Q such that the
quadrilateral ABCD satisfies the following conditions: it lies in a plane, AB
is parallel to CD, AD = BC, and a circle can be inscribed in ABCD touching
the sides.
|
|
|
Solution Let the planes meet in the line
L. Then AB and CD must be parallel to L. Let H be the foot of the
perpendicular from C to AB. The fact that a circle can be inscribed implies
AH = AD = BC. This gives enough to construct ABCD. Note that if CH>AH then
no construction is possible. If CH<AH, then there are two solutions, one
with AB>CD, the other with AB<CD.
|
网站首页 用户注册 用户登录 网站地图
数学频道首页