1. Let ABC be a triangle with sides a, b, c and corresponding angles α, β, γ. Prove that the equality α = 3β implies the inequality (a2− b2)(a− b) = bc2, and determine whether the converse also holds.
Solution: By the extended law of sines, a = 2R sin α, b = 2R sin β, c = 2R sin γ, where R is the circumradius of ABC. Thus,
(a2− b2)(a− b) = 8R3(sin2α− sin2β)(sin α− sin β)
= 8R3(sin23β− sin2β)(sin 3β− sin β)
= 8R3(sin 3β− sin β)2(sin 3β + sin β)
= 8R3(8 cos22β sin2β sin2β cos β)
= 8R3(sin2(180◦− 4β))(sin β)
= 8R3(sin2γ)(sin β)
= bc2.
The converse is false in general; we can also have α = 3β− 360◦, e.g.
for α = 15◦, β = 125◦, γ = 40◦.
2. Each side and diagonal of a regular n-gon (n ≥ 3) is colored red or blue. One may choose a vertex and change the color of all of the segments emanating from that vertex, from red to blue and vice versa. Prove that no matter how the edges were colored initially, it is possible to make the number of blue segments at each vertex even.
Prove also that the resulting coloring is uniquely determined by the initial coloring.
Solution: All congruences are taken modulo 2.
First, changing the order in which we choose the vertices does not affect the end coloring. Also, choosing a vertex twice has no net effect on the coloring. Then choosing one set of vertices has the same effect as choosing its “complement”: the latter procedure is equivalent to choosing the first set, then choosing all the vertices.
(Here, in a procedure’s complement, vertices originally chosen an odd number of times are instead chosen an even number of times, and vice versa.)
Label the vertices 1, . . . , 2n + 1. Let ai be the number of blue seg-ments at each vertex, bibe the number of times the vertex is chosen, and B be the sum of all bi. When vertex k is chosen, ak becomes 2n− ak ≡ ak; on the other hand, the segment from vertex k to each other vertex changes color, so the other ai change parity.
Summing the ai gives twice the total number of blue segments; so, there are an even number of vertices with odd ai— say, 2x vertices.
Choose these vertices. The parity of these aialternates 2x− 1 times to become even. The parity of the other ai alternates 2x times to remain even. Thus, all the vertices end up with an even number of blue segments. We now prove the end coloring is unique.
Consider a procedure with the desired results. At the end, ai be-comes ai+ B− bi (mod 2). All the ai equal each other at the end, so bj ≡ bk if and only if aj ≡ ak originally. Thus, either bi ≡ 1 if and only if ai ≡ 1 — the presented procedure — or bi ≡ 1 if and only if ai ≡ 0 – resuluting in an equivalent coloring from the first pargraph’s conclusions. Thus, the resulting coloring is unique.
This completes the proof.
Note: For a regular 2n-gon, n ≥ 2, choosing a vertex reverses the parities of all of the ai, so it is impossible to have all even ai unless the ai have equal parities to start with. And even if it is possible to have all even ai, the resulting coloring is not unique.
3. The tetrahedron ABCD is divided into five convex polyhedra so that each face of ABCD is a face of one of the polyhedra (no faces are di-vided), and the intersection of any two of the five polyhedra is either a common vertex, a common edge, or a common face. What is the smallest possible sum of the number of faces of the five polyhedra?
Solution: The smallest sum is 22. No polyhedron shares two faces with ABCD; otherwise, its convexity would imply that it is ABCD.
Then exactly one polyhedron P must not share a face with ABCD, and has its faces in ABCD’s interior. Each of P ’s faces must then be shared with another polyhedron, implying that P shares at least 3 vertices with each of the other polyhedra. Also, any polyhedron face not shared with ABCD must be shared with another polyhedron.
This implies that the sum of the number of faces is even. Each poly-hedron must have at least four faces for a sum of at least 20. Assume
this is the sum. Then each polyhedron is a four-vertex tetrahedron, and P shares at most 2 vertices with ABCD. Even if it did share 2 vertices with ABCD, say A and B, it would then share at most 2 vertices with the tetrahedron containing ACD, a contradiction.
Therefore, the sum of the faces must be at least 22. This sum can indeed be obtained. Let P and Q be very close to A and B, respec-tively; then the five polyhedra AP CD, P QCD, BQCD, ABDP Q, and ABCP Q satisfy the requirements.
4. Show that there exists an increasing sequence {an}∞n=1 of natural numbers such that for any k ≥ 0, the sequence {k + an} contains only finitely many primes.
Solution: Let pk be the k-th prime number, k ≥ 1. Set a1 = 2.
For n≥ 1, let an+1be the least integer greater than an that is con-gruent to−k modulo pk+1 for all k≤ n. Such an integer exists by the Chinese Remainder Theorem. Thus, for all k ≥ 0, k + an ≡ 0 (mod pk+1) for n≥ k + 1. Then at most k + 1 values in the sequence {k +an} can be prime; from the k +2-th term onward, the values are nontrivial multiples of pk+1and must be composite. This completes the proof.
5. For each natural number n≥ 2, determine the largest possible value of the expression
Vn= sin x1cos x2+ sin x2cos x3+· · · + sin xncos x1, where x1, x2, . . . , xn are arbitrary real numbers.
Solution: By the inequality 2ab≤ a2+ b2, we get Vn≤ sin2x1+ cos2x2
2 +· · · + sin2xn+ cos2x1
2 =n
2, with equality for x1=· · · = xn= π/4.
6. A parallelogram ABCD is given such that triangle ABD is acute and ∠BAD = π/4. In the interior of the sides of the parallelogram, points K on AB, L on BC, M on CD, N on DA can be chosen in various ways so that KLM N is a cyclic quadrilateral whose cir-cumradius equals those of the triangles AN K and CLM . Find the
locus of the intersection of the diagonals of all such quadrilaterals KLM N .
Solution: Since the arcs subtended by the angles ∠KLN , ∠KM N ,
∠LKM , ∠LN M on the circumcircle of KLM N and the arcs sub-tended by ∠KAN and ∠LCM on the circumcircles of triangles AKN and CLM , respectively, are all congruent, these angles must all be equal to each other, and hence have measure 45◦. The trian-gles SKL and SM N , where S is the intersection of KM and N L, are thus right isosceles triangles homothetic through S. Under the homothety taking K to M and L to N , AB is sent to CD and BC to DA, so S must lie on the segment BD.