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1.
Solve the following equations for x, y and z:
x + y + z =
a; x2 + y2 + z2 = b2;
xy=z2.
What conditions must a and b satisfy for x, y and z to be distinct positive
numbers?
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Solution A routine slog gives z =
(a2 - b2)/2a, x and y = (a2 + b2)/4a
+/- Ö(10a2b2 - 3a4 - 3b4)/4a.
A little care is needed with the conditions. Clearly x, y, z positive
implies a > 0, and then z positive implies |b| < a. The expression
under the root must be positive. It helps if you notice that it factorizes
as (3a2 - b2)(3b2 - a2). The
second factor is positive because |b| < a, so the first factor must also
be positive and hence a < Ö3.|b|. These conditions are also
sufficient to ensure that x and y are distinct, but then z must also be
distinct because z2 = xy.
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2.
Let a, b, c be the sides of a triangle and A its area. Prove that:
a2 + b2
+ c2 >= 4Ö3 A
When do we have equality?
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Solution One approach is a routine slog from Heron‘s formula.
The inequality is quickly shown to be equivalent to a2b2
+ b2c2 + c2a2 <= a4
+ b4 + c4, which is true since a2b2
<= (a4 + b4)/2. We get equality iff the triangle
is equilateral.
Another approach is to take an altitude lying inside the triangle. If it
has length h and divides the base into lengths r and s, then we quickly
find that the inequality is equivalent to (h - (r + s)Ö3/2)2
+ (r - s)2 >= 0, which is true. We have equality iff r = s
and h = (r + s)Ö3/2, which means the triangle is equilateral.
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3.
Solve the equation cosnx - sinnx = 1, where n is a
natural number.
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Solution Since cos2x +
sin2x = 1, we cannot have solutions with n not 2 and 0 < |cos
x|, |sin x| < 1. Nor can we have solutions with n=2, because the sign is
wrong. So the only solutions have sin x = 0 or cos x = 0, and these are: x
= multiple of pi, and n even; x even multiple of pi and n odd; x = even
multiple of pi + 3pi/2 and n odd.
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4. P
is inside the triangle ABC. PA intersects BC in D, PB intersects AC in E, and
PC intersects AB in F. Prove that at least one of AP/PD, BP/PE, CP/PF does
not exceed 2, and at least one is not less than 2.
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Solution Take lines through the
centroid parallel to the sides of the triangle. The result is then obvious.
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5.
Construct the triangle ABC, given the lengths AC=b, AB=c and the acute angle
AMB = a, where M is the midpoint of BC. Prove that the construction is
possible if and only if
b tan(a/2) <= c
< b.
When does equality hold?
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Solution The key is to take N so
that A is the midpoint of NB, then angle NCB = a.
The construction is as follows: take BN length 2AB. Take circle through B
and N such that the angle BPN = a for points P on the arc BN. Take A as the
midpoint of BN and let the circle center A, radius AC cut the arc BN at C.
In general there are two possibilities for C.
Let X be the intersection of the arc BN and the perpendicular to the
segment BN through A. For the construction to be possible we require AX
>= AC > AB. But AB/AX = tan a/2, so we get the condition in the
question.
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6.
Given 3 non-collinear points A, B, C and a plane p not parallel to ABC and
such that A, B, C are all on the same side of p. Take three arbitary points
A’, B‘, C’ in p. Let A‘’, B‘’, C‘’ be the midpoints of AA‘, BB’, CC‘
respectively, and let O be the centroid of A’‘, B’‘, C’‘. What is the locus
of O as A’, B‘, C’ vary?
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Solution The key is to notice that O is
the midpoint of the segment joining the centroids of ABC and A‘B’C‘. The
centroid of ABC is fixed, so the locus is just the plane parallel to p and
midway between p and the centroid of ABC.
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摘自：数学人
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