Solutions.

297.

The point P lies on the side BC of triangle ABC so that PC = 2BP, ÐABC = 45^° and ÐAPC = 60^°. Determine ÐACB.

Thus, AD bisects ÐEDP, where E lies on BD produced. Thus

Solution 2. [Y. Zhao] Let Q be the midpoint of PC and R the intersection of AP and the right bisector of PQ, so that PR = QR and BR = CR. Then ÐRPQ = ÐRQP = 60^° and triangle PQR is equilateral. Hence PB = PQ = PR = RQ = QC and ÐPBR = ÐPRB = ÐQRC = ÐQCR = 30^°.

Also, ÐRBA = 15^° = ÐPAB = ÐRAB, so AR = BR = CR. Thus, ÐRAC = ÐRCA. Now ÐARC = 180^° - ÐPRQ - ÐQRC = 90^°, so that ÐRCA = 45^° and ÐACB = 75^°.

Solution 3. [R. Shapiro] Let H be the foot of the perpendicular from C to AP. Then CPH is a 30-60-90 triangle, so that BP = ¹/₂PC = PH and ÐPBH = ÐPHB = 30^° = ÐPCH. Hence, BH = HC. As

Solution 4. From the equation expressing tan30^° in terms of tan15^°, we find that tan15^° = 2 - Ö3 and sin15^° = [(Ö3 - 1)/(2Ö2)]. Let ÐACP = q, so that

298.

Let O be a point in the interior of a quadrilateral of area S, and suppose that

2S = |OA |2 + |OB |2 + |OC |2+ |OD |2 .

Prove that ABCD is a square with centre O.

299.

Let s(r) denote the sum of all the divisors of r, including r and 1. Prove that there are infinitely many natural numbers n for which

s(n) n >  s(k) k

whenever 1 £ k < n.

Let r be a given positive integer, and select s £ 2^r such that u_s ³ u_k for 1 £ k £ 2^r (i.e., u_s is the largest value of u_k for k up to and including 2^r). Then, as u_2s > u_s, it must happen that 2^r £ 2s £ 2^r+1 and u_2s ³ u_k for 1 £ k £ 2^r.

Suppose that n is the smallest positive integer t for which 2^r £ t and u_k £ u_t for 1 £ k £ 2^r. Then 2^r £ n £ 2s £ 2^r+1. Suppose that 1 £ k £ n. If 1 £ k £ 2^r, then u_k £ u_n from the definition of n. If 2^r < k < n, then there must be some number k¢ not exceeding 2^r for which u_k < u_k¢ £ u_n. Thus, n has the desired property and 2^r £ n £ 2^r+1. Since such n can be found for each positive exponent r, the result follows.

Comment. The sequence selected in this way starts off: { 1, 2, 4, 6, 12, ¼}.

Solution 2. [P. Shi] Define u_m as in Solution 1. Suppose, if possible, that there are only finitely many numbers n satisfying the condition of the problem. Let N be the largest of these, and let u_s be the largest value of u_m for 1 £ m £ N. We prove by induction that u_n £ u_s for every positive integer n. This holds for n £ N. Suppose that n > N. Then, there exists an integer r < n for which u_r > u_n. By the induction hypothesis, u_r £ u_s, so that u_n < u_s. But this contradicts the fact (as established in Solution 1) that u_2s > u_s.

300.

Suppose that ABC is a right triangle with ÐB < ÐC < ÐA = 90^°, and let K be its circumcircle. Suppose that the tangent to K at A meets BC produced at D and that E is the reflection of A in the axis BC. Let X be the foot of the perpendicular from A to BE and Y the midpoint of AX. Suppose that BY meets K again in Z. Prove that BD is tangent to the circumcircle of triangle ADZ.

Solution 2. [Y. Zhao] As in Solution 1, we see that there is a circle through the vertices of AZPY and that BD is tangent to this circle. Let O be the centre of the circle K. The triangles OPA and OAD are similar, whereupon OP ·OD = OA². The inversion in the circle K interchanges P and D, carries the line BD to itself and takes the circumcircle of triangle AZP to the circumcircle of triangle AZD. As the inversion preserves tangency of circles and lines, the desired result follows.

301.

Let d = 1, 2, 3. Suppose that M_d consists of the positive integers that cannot be expressed as the sum of two or more consecutive terms of an arithmetic progression consisting of positive integers with common difference d. Prove that, if c Î M₃, then there exist integers a Î M₁ and b Î M₂ for which c = ab.

Since k + (k+1) = 2k+1, every odd integer exceeding 1 is the sum of two consecutive terms. Indeed, for each positive integers m and r,

Since 2n = (n + 1) + (n - 1), M₂ excludes all even numbers exceeding 2. Let k ³ 2 and m ³ 1. Then

We now examine M₃. The result will be established if we show that M₃ does not contain any number of the form 2^r u v where r is a nonnegative integer and u, v are odd integers with u ³ v > 1. Suppose first that r ³ 1 and let a = 2^ru - ³/₂(v-1). Then

Since m + (m+3) = 2m + 3, we see that M₃ excludes all odd numbers exceeding 3, and hence all odd composite numbers. Hence, every number in M₃ must be the product of a power of 2 and an odd prime or 1.

Comment. The solution provides more than necessary. It suffices to show only that M₁ contains all powers of 2, M₂ contains all primes and M₃ excludes all numbers with a composite odd divisor.

302.

In the following, ABCD is an arbitrary convex quadrilateral. The notation [ ¼] refers to the area.

(a) Prove that ABCD is a trapezoid if and only if

[ABC] ·[ACD] = [ABD] ·[BCD] .

(b) Suppose that F is an interior point of the quadrilateral ABCD such that ABCF is a parallelogram. Prove that

[ABC] ·[ACD] + [AFD] ·[FCD] = [ABD] ·[BCD] .

(b) Let the quadrilateral be in the horizontal plane of three-dimensional space and let F be at the origin of vectors. Suppose that u = [( ®) || FA], v = [( ®) || FC], and -pu - qv = [( ®) || FD], where p and q are nonnegative scalars. We have that [( ®) || FB] = u + v. Then

Solution 2. [Y. Zhao] Observe that, since (A + C) + (B + D) = 360^°,

(a) Hence

(b) From (a), we have that

303.

Solve the equation

tan2 2x = 2 tan2x tan3x + 1 .

Solution 2.

Solution 3. Let t = tanx, so that tan2x = 2t(1 - t²)^-1 and tan3x = (3t - t³)(1 - 3t²)^-1. Substituting for t in the equation and clearing fractions leads to

Solution 4. The equation is undefined if 2x or 3x is an odd multiple of p/2. We exclude this case. Then the equation is equivalent to

Solution 5. [Y. Zhao] Observe that, when tan(A - B) ¹ 0,

Solution 6. For a solution, neither 2x nor 3x can be a multiple of p/2, so we exclude these cases. Since