37.

Let ABC be a triangle with sides a, b, c, inradius r and circumradius R (using the conventional notation). Prove that

Solution.

Comment. The identity in the solution can be obtained as follows. Let 2s = a + b + c. Then

38.

Let us say that a set S of nonnegative real numbers if hunky-dory if and only if, for all x and y in S, either x + y or |x - y | is in S. For instance, if r is positive and n is a natural number, then S(n, r) = { 0, r, 2r, ¼, nr } is hunky-dory. Show that every hunky-dory set is { 0 }, is of the form S(n, r) or has exactly four elements.

Suppose, if possible, that S contains an element that is not a positive integer multiple of a. Let b be the least nonmultiple of a. Then 0 < b - a < b. Since b - a cannot be a multiple of a (why?), we must have b - a \not Î S and b + a Î S. Since z is the largest element of S, z - a and z - b belong to S. However, (z - a) - (z - b) = b - a does not belong to S, so 2z - (a + b) = (z - a) + (z - b) Î S. Therefore, 2z - (a + b) £ z, whence z £ a + b, so that z = a + b. Thus, S contains { 0, a, b, a+b }, with a + b the largest element. This subset is already hunky-dory. But suppose, if possible, S contains more elements. Let c be the smallest such element. Then 0 < (a + b) - c Î S Þa £ (a + b) - c Þ c < b Þc = ma for some positive integer m ³ 2. Since b + ma > b + a, b - ma must belong to S, and so be a multiple of a. This yields a contradiction. Hence, S must be equal to { 0, a, b, a+b }.

The only remaining case is that S consists solely of nonnegative multiples of some element a. Let na be the largest such multiple. If n = 2, then S = S(2, a). Suppose that n > 2. Then (n - 1)a Î S, so S contains { 0, a, (n-1)a, na }, which is hunky-dory.

Suppose S contains a further multiple ma with 2 £ m £ n-2. Since a Î S and na + a > na, (n-1)a Î S, so that n - (m + 1)a = (n-1)a - ma Î S Þ(m+1)a = n - [n - (m+1)a] Î S. By induction, it can be shown that ka Î S for m £ k £ n. In particular, (n - 2)a Î S so that 2a = na - (n-2)a Î S. But then 3a, 4a, ¼, na are in S and so S = S(n, a). The desired result follows,

Solution 2. [S. Niu] Let S = { a₀, a₁, ¼, a_n }, with a₀ < a₁ < ¼ < a_n. The elements a_n - a_n, a_n - a_n-1, ¼, a_n - a₀ are n+1 distinct elements of S listed in increasing order, and so a₀ = 0, and for each i with 0 £ i £ n, we must have that a_n - a_i = a_n-i. Let i £ ⁿ/₂. Then i £ n - i and so a_i £ a_n-i; thus, a_i £ (a_n)/2 £ a_n-i. Thus, if j > k ³ n/2, a_j - a_k Î S.

Since 0 < a_n-1 - a_n-2 < a_n - a_n-2 = a₂, it follows that a_n-1 -a_n-2 = a₁. Also, 0 < a_n-1 - a_n-2 = a₁ < a_n-1 - a_n-3 < a_n - a_n-3 = a₃, so that a_n-1 - a_n-3 = a₂. Continuing on in this way, we find that, for i ³ n/2,

Now 0 < a_n-2 - a_n-3 < a_n-1 - a_n-3 = a₂ so a_n-2 - a_n-3 = a₁. We can proceed in this fashion to obtain that, for j ³ n/2, a_j+1 - a_j = a₁. Hence, for i £ (n/2) - 1, a_i+1 - a_i = (a_n - a_n-i-1) - (a_n- a_n-i) = a_n-i - a_n-i-1 = a₁.

Let n = 2m. Then a_i = ia₁ and a_n-i = a_n - ia₁ for 1 £ i £ m, so that a_m = ma₁ and a_n = a_m + ma₁ = na₁. It follows that a_k = ka₁ for 1 £ k £ n and S = S(n, a₁).

Let n = 2m+ 1. If m = 0, then S = { 0, a₁ } = S(1, a₁). If m = 1, then S = { 0, a₁, a₃ - a₁, a₃ } = { 0, a₁, a₂, a₁ + a₂ } is a 4-element hunky-dory set. Let m ³ 2. Then, for 1 £ i £ m, a_i = ia₁ and a_n-i = a_n - ia_i. Now a_m+1 = a_n - a_m > a_n-1 - a_m > ¼ ³ a_m+2 - a_m > a_m+1 - a_m ³ a₁. Since { a_n - a_m = a_{m + (m+1)} - a_m, a_n-1 - a_m, ¼, a_m+2 - a_m , a_m+1 - a_m } contains m+1 elements, we must have a_m+j - a_m = a_j for 1 £ j £ n - m = m + 1. Therefore, a_i = ia_i for 1 £ i £ n. (Why does this last statement fail to follow when m = 1?)

39.

(a) ABCDEF is a convex hexagon, each of whose diagonals AD, BE and CF pass through a common point. Must each of these diagonals bisect the area?

Solution 1. (a) No, they need not bisect the area. Let the vertices of the hexagon have coordinates (-1, 0), (-1, -1), (1, -1), (1, 0), (-t, t), (t, -t) with t > 0 but t ¹ 1. The diagonals with equations y = 0, y = x and y = -x intersect in the origin but do not bisect the area of the hexagon.

(b) Let the hexagon be ABCDEF and suppose that the intersection of the diagonals AD and BE is on the same side of CF as the side AB. Thus, AB, CD and EF border on triangles whose third vertices form a triangle at the centre of the hexagon (we will show this triangle to be degenerate). Let a, b, c, d, e, f be the lengths of the rays from the respective vertices A, B, C, D, E, F to the vertices of the central triangle, whose sides are x, y, z so that the lengths of AD, BE and CF are respectively a + x + d, b + y + e, c + z + f. All lower-case variables represent nonnegative real numbers.

Let the areas of the bordering on FA, AB, BC, CD, DE, EF be respectively a, b, g, d, e, f, and let the area of the central triangle be l. Then, since each diagonal bisects the area of the hexagon, we have that

Using the fact that the area of a trangle is half the product of adjacent sides and the sine of the angle between them, and the equality of opposite angles, we find that

Solution 2. (a) No. Let ABCDEF be a regular hexagon. The diagonals AD, BE, CF intersect and each diagonal does bisect the area. Let X be any point other than F on the diagonal CF for which ABCDEX is still a convex hexagon. The diagonals of this hexagon are the same as those of the regular hexagon, and so have a common point of intersection. However, the diagonals AD and BE no longer intersect the area of the hexagon.

(b) [X. Li] Let ABCDEF be a given convex hexagon, each of whose diagonals bisect its area. Suppose that the diagonals AD and CF intersect at G. As in Solution 1, we can determine that the areas of triangles AGF and DGC are equal, whence AG ·GF = CG ·GD, or AG/GD = CG/GF. Therefore, DAGC ~ DDGF (SAS). It follows that AC/DF = AG/GD = CG/GF, ÐCAG = ÐFDG, and so AC || DF. In a similar way, we find that BF || CE and AE || BD, so that DACE ~ DDFB and AC/DF = CE/FB = EA/BD.

Suppose diagonals AD and BE intersect at H. Then, as above, we find that AG/GD = AC/DF = EA/BD = EH/HB = AH/HD, so that H = G. Hence, the three diagonals have the point G in common.

40.

Determine all solutions in integer pairs (x, y) to the diophantine equation x² = 1 + 4y³(y + 2).

Suppose that y £ -3. We first observe that for a given product p of two positive integers, the sum of these positive integers has a minium value of 2Öp (why?) and a maximum value of 1 + p. This follows from the fact that, for integers u with 1 £ u £ p, 

Solution 2. Since x must be odd, we can let x = 2z + 1 for some integer z, so that the equation becomes z(z+1) = y⁴ + 2y³. We can deal with the cases that y = 0, -1, -2 directly to obtain the solutions (x, y, z) = (1, 0, 0), (-1, 0, -1), (1, -2, 0), (-1, -2, -1). Henceforth, suppose that y ³ 1 or y £ -3, so that y⁴ + 2y³ is positive. Let f(t) = t(t + 1). Then f(t) is increasing for t ³ 0 and f(-t) = f(t - 1) for every integer t; thus, we need check only that y⁴ + 2y³ does not coincide with a value taken by f(t) for nonnegative values of t.

Now

41.

Determine the least positive number p for which there exists a positive number q such that

Comments. Recall the binomial expansion

Solution 1. Observe that, for 0 £ x £ 1,

Suppose, for some constants p and c with 0 < p < 2 and c > 0,

Now we look at the situation when p = 2 and q > 0. For 0 £ x £ 1,

Solution 2. [R. Furmaniak] The given inequality is equivalent to

42.

G is a connected graph; that is, it consists of a number of vertices, some pairs of which are joined by edges, and, for any two vertices, one can travel from one to another along a chain of edges. We call two vertices adjacent if and only if they are endpoints of the same edge. Suppose there is associated with each vertex v a nonnegative integer f(v) such that all of the following hold:

(2) If f(v) > 0, then v is adjacent to at least one vertex w such that f(w) < f(v).

(3) There is exactly one vertex u such that f(u) = 0.

Prove that f(v) is the number of edges in the chain with the fewest edges connecting u and v.

Solution. We prove by induction that f(x) = n if and only if the shortest chain from u to x has n members. This is true for n = 0 (and for n = 1). Suppose that this holds for 0 £ n £ k.

Let f(x) = k + 1. There exists a vertex y adjacent to x for which h = f(y) < k + 1. By the induction hupothesis, y can be connected to u by a chain of h edges, so x can be connected to u by a chain of h + 1 edges. Hence, h + 1 ³ k + 1. From these two inequalities, we must have h = k, so x can be connected to u by a chain of k + 1 edges. There cannot be a shorter chain, as, by the induction hypothesis, this would mean that f(x) would have to be less than k + 1.

Let the shortest chain connecting x to u have k + 1 edges. Following along this chain, we can find an element z adjacent to x connected to u by k edges. This must be one of the shortest chains between u and z, so that f(z) = k. By hypothesis (1), f(x) must take one of the values k - 1 and k + 1. The first is not admissible, since there is no chain with k - 1 edges connecting u and x. Hence f(x) = k + 1.