1. One face of a pyramid with square base and all edges 2 is glued to a face of a regular tetrahedron with edge length 2 to form a polyhedron. What is the total edge length of the polyhedron?
2. P is a point on the circumcircle of the square ABCD between C and D. Show that PA2 - PB2 = PB·PD - PA·PC.
3. E is the midpoint of the arc BC of the circumcircle of the triangle ABC (on the opposite side of the line BC to A). DE is a diameter. Show that ∠DEA is half the difference between the ∠B and ∠C.
4. The triangle ABC (with sidelengths a, b, c as usual) satisfies log(a2) = log(b2) + log(c2) - log(2bc cos A). What can we say about the triangle?
5. Let X = {1, 2, 3, 4, 5, 6, 7}. How many 7-tuples (X1, X2, X3, X4, X5, X6, X7) are there such that each Xi is a different subset of X with three elements and the union of the Xi is X?
6. Each member of the sequence a1, a2, ... , an belongs to the set {1, 2, ... , n-1} and a1 + a2 + ... + an < 2n. Show that we can find a subsequence with sum n.
7. Put f(x) = x3 - x. Show that the set of positive real A such that for some real x we have f(x + A) = f(x) is the interval (0, 2].
8. The sequence of nonzero reals x1, x2, x3, ... satisfies xn = xn-2xn-1/(2xn-2 - xn-1) for all n > 2. For which (x1, x2) does the sequence contain infinitely many integral terms?
9. The year 1978 had the property that 19 + 78 = 97. In other words the sum of the number formed by the first two digits and the number formed by the last two digits equals the number formed by the middle two digits. Find the closest years either side of 1978 with the same property.
10. Show that (1 + x)n ≥ (1 - x)n + 2nx(1 - x2)(n-1)/2 for all 0 ≤ x ≤ 1 and all positive integers n.
11. Given a positive real k, for which real x0 does the sequence x0, x1, x2, ... defined by xn+1 = xn(2 - k xn) converge to 1/k?
12. Show that for n a positive integer we have cos4k + cos42k + ... cos4nk = 3n/8 - 5/16 where k = π/(2n+1).
13. ABC is a triangle with AB = 2·AC and E is the midpoint of AB. The point F lies on the line EC and the point G lies on the line BC such that A, F, G are collinear and FG = AC. Show that AG3 = AB·CE2.
14. a1, a2, ... , an are integers and m < n is a positive integer. Put Si = ai + ai+1 + ... + ai+m, and Ti = am+i + am+i+1 + ... + an-1+i, for where we use the usual cyclic subscript convention, whereby subscripts are reduced to the range 1, 2, ... , n by subtracting multiples of n as necessary. Let m(h, k) be the number of elements i in {1, 2, ... , n} for which Si = h mod 4 and Ti = b mod 4. Show that m(1, 3) = m(3, 1) mod 4 iff m(2, 2) is even.
15. X is a finite set. X1, X2, ... , Xn are distinct subsets of X (n > 1), each with 11 elements, such that the intersection of any two subsets has just one element and given any two elements of X, there is an Xi containing them both. Find n.
Solutions
Problem 1
One face of a pyramid with square base and all edges 2 is glued to a face of a regular tetrahedron with edge length 2 to form a polyhedron. What is the total edge length of the polyhedron?
Answer
18 (not 22).
Solution
The polyhedron has two less edges than you might expect. We start with 8 + 6 = 14 edges. We lose 3 when the two faces are glued. But we lose another 2 because in two cases adjacent faces are parallel and merge to form a rhombus. (That is easily seen if you note that the midpoints of the edges of a tetrahedron form the vertices of a regular octahedron, which can be regarded as two square pyramids joined at the base. ) Thus the total edge length is 9 x 2 = 18.
The polyhedron is a triangular prism with one square side face and two rhombic side faces (a total of 5 faces). It has 6 vertices, and 9 edges (note V + F = E + 2).
Problem 2
P is a point on the circumcircle of the square ABCD between C and D. Show that PA2 - PB2 = PB·PD - PA·PC.
Solution
Let O be the center of the circumcircle and put ∠DOP = 2x. Take the circumradius to be R. Then PD = 2R sin x, PC = 2R sin(45o - x), PA = 2R sin(45o + x), PB = 2R sin(90o - x). So the required result is equivalent to sin2(45o + x) - sin2(90o - x) = sin x sin(90o - x) - sin(45o + x) sin(45o - x). But 2 lhs = (1 - 2 sin2(90o - x)) - (1 - 2 sin2(45o + x) = cos(180o - 2x) - cos(90o + 2x) = - cos 2x + sin 2x, and 2 rhs = (cos(90o - 2x) - cos 90o) - (cos 2x - cos 90o) = sin 2x - cos 2x.
Problem 3
E is the midpoint of the arc BC of the circumcircle of the triangle ABC (on the opposite side of the line BC to A). DE is a diameter. Show that angle DEA is half the difference between the angles B and C.
Solution
Taking A between D and C as shown, we have ∠B = ∠AEC (cyclic points) = ∠OEC - ∠DEA, and ∠C = ∠AEB = ∠OEB + ∠DEA. But ∠OEC = ∠OEB, so result follows immediately. An exactly similar argument works if A is on the other side of D.
Problem 4
The triangle ABC (with sidelengths a, b, c as usual) satisfies log(a2) = log(b2) + log(c2) - log(2bc cos A). What can we say about the triangle?
Answer
Isosceles (specifically, BC equals one of the other two sides).
Solution
We have log(2bc cos A) = log(b2) + log(c2) - log(a2). Taking exponentials gives 2bc cos A = b2c2/a2. By the cosine formula, 2bc cos A = b2 + c2 - a2. Hence b2 + c2 - a2 = b2c2/a2. Multiplying by a2 and factorising gives (a2 - b2) (a2 - c2) = 0. Hence a = b or a = c.
Conversely it is easy to see that if a = b or c, then b2 + c2 - a2 = b2c2/a2 and hence 2bc cos A = b2c2/a2. Taking logs gives the relation given.
Problem 5
Let X = {1, 2, 3, 4, 5, 6, 7}. How many 7-tuples (X1, X2, X3, X4, X5, X6, X7) are there such that each Xi is a different subset of X with three elements and the union of the Xi is X?
Answer
7! x 6184400
Solution
Denote the binomial coefficient by mCn. There are 7C3 = 35 subsets of X with 3 elements. Suppose X1 = {x, y, z} and X - X1 = {a, b, c, d}. There are N = 34·33·32·31·30·29 possible choices for the remaining six subsets such that they are all different, but some of these choices will not cover all of {a, b, c, d}. Since 4C3 = 4 < 7, they must cover at least two of them. There are 5C3 = 10 sets not covering a or b. So N2 = 9·8·7·6·5·4 choices for (X2, X3, ... , X7) which do not cover a or b. Similarly, since 6C3 = 20, there are N1 = 19·18·17·16·15·14 choices which do not cover b (some of these may not cover a either).
Thus by the principle of inclusion and exclusion, there are 35(N - 4N1 + 6N2) = 7! x 6184400 possible arrangements.
This seems kludgy. Does anyone have a neater solution?
Problem 6
Each member of the sequence a1, a2, ... , an belongs to the set {1, 2, ... , n-1} and a1 + a2 + ... + an < 2n. Show that we can find a subsequence with sum n.
Solution
Ignore the restrictions for the moment. We show that given any n numbers a1, a2, ... , an, we can find a non-empty subset whose sum is a multiple of n. Consider, the n sums a1, a1 + a2, ... , a1 + ... + an. If any of them = 0 mod n, then we are home. If not, then two must be equal mod n, and then their difference is 0 mod n.
So returning to the problem given, we have a non-empty subsequence which is a multiple of n. It cannot be 0, because all the ai are positive. It cannot be 2n or more because the sum of all the ai is < 2n.
Problem 7
Put f(x) = x3 - x. Show that the set of positive real A such that for some real x we have f(x + A) = f(x) is the interval (0, 2].
Solution
It is clear from the graph (and the continuity of the function) that if we take x = 0 and A = 1 we get f(x + A) = f(x), and that as we increase x until we reach the minimum at x = 1/√3 (adjusting A so that f(x) remains equal to f(x + A), we reduce A to 0. That gives the interval [0, 1]. Equally as we reduce x from 0 to -1, A changes continuously from 1 to 2, so we get the interval [1, 2] (and possibly some values outside it).
It remains to show that we cannot get A > 2.
It is easy to check that if A = 2, then 6x2 + 12x + 6 = 0, so x = -1. In other words, there is a unique value of x for which A = 2.
Clearly if f(x + A) = f(x) (with A a positive real), then x must lie between -2/√3 (which gives the same image 1/(3√3) as the minimum at 1/√3) and 1/√3. But if x = -2/√3, then A = √3 < 2 and at x = -1/√3, A is again √3. Hence if A got above 2 between these two values of x, then there would have to be at least two values of x for which A = 2, which we have just shown to be false. So A ≤ 2 for x < -1/√3. Similarly A < 2 for x > 1/√3. But it is obvious that A < 2 for -1/√3 ≤ x ≤ 1/√3.
Problem 8
The sequence of nonzero reals x1, x2, x3, ... satisfies xn = xn-2xn-1/(2xn-2 - xn-1) for all n > 2. For which (x1, x2) does the sequence contain infinitely many integral terms?
Answer
x1 = x2 = non-zero integer
Solution
Put an = 1/xn. Then an = 2an-1 - an-2, so an - an-1 = an-1 - an-2. Iterating, an - an-1 = a2 - a1. Iterating, an = a1 + (n-1)(a2-a1).
Now if a2 ≠ a1 (or equivalently x1 ≠ x2), then for all but finitely many terms we have |an| > 1 and hence |xn| < 1, and hence (since xn is non-zero) xn not integral. If x1 = x2, then all xn = x1.
Thanks to Suat Namli
Problem 9
The year 1978 had the property that 19 + 78 = 97. In other words the sum of the number formed by the first two digits and the number formed by the last two digits equals the number formed by the middle two digits. Find the closest years either side of 1978 with the same property.
Answer
1868, 2307.
Solution
Suppose the number is abcd. So we have 10a + b + 10c + d = 10b + c, or 9(b - c) = 10a + d. So 10a + d is a multiple of 9. So it must be 18, 27, 36, 45, 54, 63, 72, 81, 90 or 99. Thus if a = 1, then d must be 8 and b = c + 2. Thus the closest such number below 1978 is 1868. The closest number above must have a = 2, d = 7, and b = c + 3. We want b as small as possible, so we take b = 3, c = 0, giving 2307.
Problem 11
Given a positive real k, for which real x0 does the sequence x0, x1, x2, ... defined by xn+1 = xn(2 - k xn) converge to 1/k?
Answer
0 < x0 < 2/k.
Solution
If xn ≤ 0, then xn+1 ≤ 0, so if x0 ≤ 0, the all terms are ≤ 0 and so the sequence cannot converge to the positive 1/k. If x0 ≥ 2/k, then x1 ≤ 0, so all subsequent terms are non-positive and the sequence cannot converge to 1/k.
If x0 = 1/k, then all xn = 1/k, so the sequence trivially converges to 1/k. So assume x0 ≠ 1/k. We have k(xn - 1/k)2 > 0, so kxn2 - 2xn + 1/k > 0, so xn+1 = 2xn - kxn2 < 1/k.
Suppose 0 < x0 < 2/k. Then 0 < x1. We have just shown that x1 < 1/k, so 0 < x1 < 1/k. Now if 0 < xn < 1/k, then xn+1 = xn + xn(1 - kxn) > xn. So for all n > 1 we have xn < xn+1 < 1/k. Hence xn must converge (it is a general result that an increasing sequence which is bounded above must converge). Suppose it converges to L. Then L = L(2 -kL), so L = 0 or 1/k. But L cannot be 0 because all terms are > x1 (except possibly x0). So the sequence converges to 1/k.
Thanks to Suat Namli
Problem 12
Show that for n a positive integer we have cos4k + cos42k + ... cos4nk = 3n/8 - 5/16 where k = π/(2n+1).
Solution
Multiplying by sin k and using sin(A+B) - sin(A-B) = 2 sin A cos B, we get 2 sin k (cos 2k + cos 4k + ... + cos 2nk = (sin 3k - sin k) + (sin 5k - sin 3k) + ... + (sin(2n+1)k - sin(2n-1)k) = sin(2n+1)k - sin k = -sin k. So ∑ cos 2rk = -1/2. Similarly, 2 sin 2k (cos 4k + cos 8k + ... + cos 4nk) = sin(2n+1)2k - sin 2k = -sin 2k. So ∑ cos 4rk = -1/2
We have cos4k = (cos 4k + 4 cos 2k + 3)/8. Hence 8(cos4k + cos42k + ... cos4nk) = ∑ cos 4rk + 4 ∑ cos 2rk + 3n = -5/2 + 6n. So ∑ cos4rk = 3n/8 - 5/16.
Labels:
Irish Mathematical Olympiad,
IrMO
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