Solutions

85.

Find all pairs (a, b) of positive integers with a ¹ b for which the system

cosax + cosbx = 0

asinax + bsinbx = 0 .

Suppose that a = 2^u p and b = 2^v q with u and v unequal integers and p and q odd. Then x = 2^-wp. where w is the minimum of u and v satisfies the system of equations.

Solution 2. First, observe that x ¹ 0 for any solution. If the system is satisfied, then

We now show that, for any pair a, b satisfying this condition, there is a solution. Wolog, let a = 2uw and v = (2v+1)w, where the greatest common divisor of 2u and 2v+1 is 1, and w is an arbitrary positive integer. Suppose that x = p/w. Then

Solution 3. Since cos² ax = cos² bx and a² sin² ax + b² sin² bx, then

Comment. Note that there are two parts to the solution of this problem, and your write-up should make sure that these are carefully delineated. First, assuming that there is a solution, you derive necessary conditions on a and b that the two equations are consistent. Then, you assume these conditions on a and b, and then display a solution to the two equations. A complete solution requires noting that suitable numbers a and b actually do lead to a solution.

86.

Let ABCD be a convex quadrilateral with AB = AD and CB = CD. Prove that

(a) it is possible to inscribe a circle in it;

(b) it is possible to circumscribe a circle about it if and only if AB ^BC;

(c) if AB ^BC and R and r are the respective radii of the circumscribed and inscribed circles, then the distance between the centres of the two circles is equal to the square root of R² + r² - rÖ[(r² + 4R²)].

Solution 1. (a) Triangles ABC and ADC are congruent (SSS) with the congruence implemented by a reflection in AC. Hence AC bisects angles DAB and DCB. The angle bisectors of ÐADB and ÐABC are reflected images and intersect in I, a point on AC. Since I is equidistant from the four sides of the kite ABCD, it is the centre of its incircle.

(b) If AB ^BC, then the circle with diameter AC passes through B. By symmetry about AC, it must pass through D as well. Conversely, let \frak C be the circumcircle of ABCD. The circle goes to itself under reflection in AC, so AC must be a diameter of \frak C. Hence ÐABC = ÐADC = 90^°.

(c) Let I be the incentre and O the circumcentre of ABCD; both lie on AC. Suppose that J and K are the respective feet of the perpendiculars to AB and BC from I, and P and Q the respective feet of the perpendiculars to AB and BC from O. Let x = |AB | and y = |BC |. Then

Solution 2. (a) Since triangles ADB and CDB are isosceles, the angle bisectors of A and C right bisect the base BD and so they coincide. The line AC is an axis of reflective symmetry that interchanges B and D, and also interchanges the angle bisectors of B and D. The point P where one of the bisectors intersects the axis AC is fixed by the reflection and so lies on the other bisector. Hence, P is common to all four angle bisectors, and so is equidistant from the four sides of the quadrilateral. Thus, we can inscribe a circle inside ABCD with centre P.

(b) Since AC is a line of symmetry, ÐB = ÐD. Note that, ABCD has a circumcircle Û pairs of opposite angles sum to 180^° ÛÐB + ÐD = 180^° Û ÐB = ÐD = 90^°. This establishes the result.

(c) [R. Barrington Leigh] Let a, b and c be the respective lengths of the segments BC, AC and AB. Let O and I be, respectively, the circumcentre and the incentre for the quadrilateral. Note that both points lie on the diagonal AC. Wolog, we may take a ³ c.

We observe that ÐABI = 45^° and that BI is the hypotenuse of an isosceles right triangle with arms of length r. We have, by the Law of Cosines,

By looking at the area of DABC in two ways, we have that ac = r(a+c). Now

87.

Prove that, if the real numbers a, b, c, satisfy the equation

ëna û+ ënb û = ënc û

for each positive integer n, then at least one of a and b is an integer.

Let a = ëa û+ a, b = ëb û+ b and c = ëc û+g. From the condition for n = 1, we have ëa û+ ëb û = ëc û. Then, ëna û = ënëa û+ naû = nëa û+ ënaû with similar equations for b and c. Putting this together gives that ënaû+ ënbû = ëngû, for all n ³ 1. As in the first part of the solution, we have that a+ b = g, from which { na} +{ nb} = { ng} for all n ³ 1, where { x } denotes the fractional part x - ëx û of x.

We first show that a, b and g cannot all be positive and rational. For, if they were rational, then for some positive integers i, j, k with k ³ 2, we would have a = i/k, b = j/k and g = (i+j)/k. Then ëkaû = i = ë(k-1)aû+ 1, with similar relations for b and g. Thus,

Wolog, suppose that all of a, b, g are positive with a irrational. Let p be a nonnegative integer for which a+ pb < 1 £ a+ (p+1)b and suppose that e = 1 - (a+ pb). Since a+ b = g < 1, it follows that p ³ 1.

We show that there is a positive integer m ³ 2 for which a+ pb < { ma}. Let t = é1/eù and consider the intervals [i/t, (i+1)/t) where 0 £ i £ t-1. By the Pigeonhole Principle, one of these intervals must contain two of the numbers { 2a},{ 4a}, ¼, { 2(t+1)a}, say { qa} and { ra} with q ³ r + 2. Thus, |{ qa} - { ra} | < 1/t £ e. Since

In the first case, we can find a positive integer s for which 1 > s { (q - r) a} > 1 - e. Since

Now, { ma} > a,

The only remaining possibility is that either a or b vanishes, i.e., that a or b is an integer. This possibility is feasible when the other two variables have the same fractional part.

88.

Let I be a real interval of length 1/n. Prove that I contains no more than ¹/₂(n+1) irreducible fractions of the form p/q with p and q positive integers, 1 £ q £ n and the greatest common divisor of p and q equal to 1.

Solution 1. We first establish a lemma: Let 1 £ q £ n. Then there exists a positive integer m for which ¹/₂(n+1) £ mq £ n. For, let m = ën/q û. If q > n/2, then m = 1 and clearly ¹/₂(n+1) £ mq = q £ n. If q £ n/2, then (n/q) - 1 < m £ (n/q), so that

Let p/q and p¢/q¢ be two irreducible fractions in I with m and m¢ corresponding integers as determined by the lemma. Suppose, if possible, that mq = m¢q¢. Then

It follows that the mapping p/q ® mq from the set of irreducible fractions in I into the set of integers in the interval [(n+1)/2, n] is one-one. But the latter set has at most n - ((n+1)/2) + 1 = (n+1)/2 elements, and the result follows.

Solution 2. [M. Zaharia] For 1 £ i £ ¹/₂(n+1), define

89.

Prove that there is only one triple of positive integers, each exceeding 1, for which the product of any two of the numbers plus one is divisible by the third.

Since a, b, c are distinct and so 2 £ a < b < c, we must have a ³ 2 and b ³ 3. Suppose, if possible that b ³ 4, so that c ³ 5. Then abc ³ 40 and

Solution 2. As in Solution 1, we show that abc £ ab + bc + ca + 1, so that

Solution 3. [P. Du] As in Solution 1, we show that a, b, c are pairwise coprime and that ab + bc + ca + 1 is a multiple of abc. Assume 2 £ a < b < c. Then abc > ac and bc(a - 1) ³ bc > ac ³ a(b + 1) > ab + 1, whence, adding these, we get 2abc - bc > ac + ab + 1, so that 2abc > ab + bc + ca + 1. Hence,

Solution 4. [O. Ivrii] As before, we show that a, b and c are pairwise coprime, and take 2 £ a < b < c. Then bc |ab + ac + 1. Since bc > ac + 1 and bc > ab + 1, we have that 2bc > ab + ac + 1. Hence bc = ab + ac + 1, so that (bc + 1) + ab = 2(ab + 1) + ac. Since a divides each of bc + 1, ab and ac, but is coprime with ab + 1, it follows that a divides 2. Hence a = 2 and

Solution 5. As above, we can take 2 £ a < b < c. Since

Suppose, if possible, that (a/c) + (a/b) + (1/(bc)) = 2. Then ab + ac + 1 = 2bc, whence b(c - a) + c(b - a) = 1, an impossibility. Hence a(b + c) + 1 = bc, so that