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1.
Problems A, B and C were posed in a mathematical contest. 25 competitors
solved at least one of the three. Amongst those who did not solve A, twice as
many solved B as C. The number solving only A was one more than the number
solving A and at least one other. The number solving just A equalled the
number solving just B plus the number solving just C. How many solved just B?
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Solution Answer: 6.
Let a solve just A, b solve just B, c solve just C, and d solve B and C but
not A. Then 25 - a - b - c - d solve A and at least one of B or C. The
conditions give:
b + d = 2(c + d); a = 1 + 25 - a - b - c - d; a = b + c.
Eliminating a and d, we get: 4b + c = 26. But d = b - 2c >= 0, so b = 6,
c = 2.
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2.
Prove that if BC + AC = tan C/2 (BC tan A + AC tan B), then the triangle ABC
is isosceles.
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Solution A straight slog works. Multiply up to get (a + b)
cos A cos B cos C/2 = a sin A cos B sin C/2 + b cos A sin B sin C/2 (where
a = BC, b = AC, as usual). Now use cos(A + C/2) = cos A cos C/2 - sin A sin
C/2 and similar relation for cos (B + C/2) to get: a cos B cos(A + C/2) + b
cos A cos (B + C/2) = 0. Using C/2 = 90 - A/2 - B/2, we find that cos(A +
C/2) = - cos(B + C/2) (and = 0 only if A = B). Result follows.
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3.
Prove that a point in space has the smallest sum of the distances to the
vertices of a regular tetrahedron iff it is the center of the tetrahedron.
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Solution Let the tetrahedron be
ABCD and let P be a general point. Let X be the midpoint of CD. Let P‘ be
the foot of the perpendicular from P to the plane ABX. We show that if P
does not coincide with P’, then PA + PB + PC + PD > P‘A + P’B + P‘C +
P’D.
PA > P‘A (because angle PP’A = 90) and PB > P‘B. P’CD is isosceles
and PCD is not but P is the same perpendicular distance from the line CD as
P‘. It follows that PC + PD > P’C + P‘D. The easiest way to see this is
to reflect C and D in the line PP’ to give C‘ and D’. Then PC = PC‘, and
PC’ + PD > C‘D = P’C‘ + P’D = P‘C + P’D.
So if P has the smallest sum, it must lie in the plane ABX and similarly in
the plane CDY, where Y is the midpoint of AB, and hence on the line XY.
Similarly, it must lie on the line joining the midpoints of another pair of
opposite sides and hence must be the center.
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4.
Prove that 1/sin 2x + 1/sin 4x + ... + 1/sin 2nx = cot x - cot 2nx
for any natural number n and any real x (with sin 2nx non-zero).
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Solution cot y - cot 2y = cos
y/sin y - (2 cos2y - 1)/(2 sin y cos y) = 1/(2 sin y cos y) =
1/sin 2y. The result is now easy. Use induction. True for n = 1 (just take
y = x). Suppose true for n, then taking y = 2nx, we have 1/sin 2n+1x
= cot 2nx - cot 2n+1x and result follows for n + 1.
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5.
Solve the equations:
|ai - a1| x1 + |ai
- a2| x2 + |ai - a3| x3
+ |ai - a4| x4 = 1, i = 1, 2, 3, 4, where ai
are distinct reals.
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Solution Take a1 >
a2 > a3 > a4. Subtracting the
equation for i=2 from that for i=1 and dividing by (a1 - a2)
we get:
- x1 + x2 + x3
+ x4 = 0.
Subtracting the equation for i=4 from that for i=3 and dividing by (a3
- a4) we get:
- x1 - x2 - x3
+ x4 = 0.
Hence x1 = x4. Subtracting the equation for i=3 from
that for i=2 and dividing by (a2 - a3) we get:
- x1 - x2 + x3
+ x4 = 0.
Hence x2 = x3 = 0, and x1 = x4
= 1/(a1 - a4).
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6.
Take any points K, L, M on the sides BC, CA, AB of the triangle ABC. Prove
that at least one of the triangles AML, BKM, CLK has area <= 1/4 area ABC.
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Solution If not, then considering ALM
we have 4.AL.AM.sin A > AB.AC.sin A, so 4.AL.AM > AB.AC = (AM +
BM)(AL + CL), so 3.AL.AM > AM.CL + BM.AL + BM.CL. Set k = BK/CK, l
= CL/AL, m = AM/BM, and this inequality becomes:
3 > l + 1/m + l/m.
Similarly, considering the other two triangles we get: 3 > k + 1/l
+ k/l, and 3 > m + 1/k + m/k.
Adding gives: 9 > k + l + m + 1/k + 1/l
+ 1/m + k/l + l/m + m/k,
which is false by the arithmetic/geometric mean inequality.
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