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37.
Let ABC be a triangle with sides a, b, c, inradius r and circumradius R (using the conventional notation). Prove that
r
2R
£ abc
 ______________________Ö2(a2 + b2)(b2 + c2)(c2 + a2)
.
When does equality hold?

Solution.

 a2
 - (b2 + c2)(1 - cosA) = b2 + c2 - 2bc cosA - (b2 + c2) + (b2 + c2)cosA
 ³ (b - c)2 cosA ³ 0
 Þ a2 ³ (b2 + c2)(1 - cosA) = 2(b2 + c2)sin2 (A/2)  .
With similar inequalities for b and c, we find that
 a2 b2 c2 ³ 8(a2 + b2)(b2 + c2)(c2 + a2)sin2 (A/2) sin2 (B/2) sin2 (C/2) .
Since r = 4Rsin(A/2) sin(B/2) sin(C/2), the desired result follows. Equality holds if and only if the triangle is equilateral.

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

r
s - a
= tan A
2
=
 sin A2

 cos A2
while
 a = 2R sinA = 4R sin A2 cos A2 .
Hence
 ars - a = 4R sin2 A2 .
Using similar identities for the other sides, we find that
 abcr3(s - a)(s - b)(s - c) = 64R3 sin2 A2 sin2 B2 sin2 C2 .      (*)
Note that the area D of the triangle is given by
 D = rs = abc4R = ________________Ös(s - a)(s - b)(s - c) ,
so that the left side of (*) becomes 4RDr2(rs)D-2 = 4Rr2 . Substituting this in, dividing by 4R and taking the square root yields
 r = 4R sin A2 sin B2 sin C2 .

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.
Solution 1. { 0 } and sets of the form { 0, r } are clearly hunky-dory. Let S be a nontrivial hunky-dory set with largest postive element z. Then 2z \not Î S, so 0 = z - z Î S. Thus, every hunky-dory set contains 0. Suppose that S has at least three elements, with least positive element a.

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, an }, with a0 < a1 < ¼ < an. The elements an - an, an - an-1, ¼, an - 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 an - ai = an-i. Let i £ n/2. Then i £ n - i and so ai £ an-i; thus, ai £ (an)/2 £ an-i. Thus, if j > k ³ n/2, aj - ak Î S.

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

 0 < an-1 - an-2 < an-1 - an-3 < ¼ < an-1 - an - i < an - an - i = ai,
whence an-1 - an-j = aj-1 for 1 £ j £ (n/2).

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

Let n = 2m. Then ai = ia1 and an-i = an - ia1 for 1 £ i £ m, so that am = ma1 and an = am + ma1 = na1. It follows that ak = 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, ai = ia1 and an-i = an - iai. Now am+1 = an - am > an-1 - am > ¼ ³ am+2 - am > am+1 - am ³ a1. Since { an - am = am + (m+1) - am, an-1 - am, ¼, am+2 - am , am+1 - am } contains m+1 elements, we must have am+j - am = aj for 1 £ j £ n - m = m + 1. Therefore, ai = iai 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?
(b) ABCDEF is a convex hexagon, each of whose diagonals AD, BE and CF bisects the area (so that half the area of the hexagon lies on either side of the diagonal). Must the three diagonals pass through a common point?

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

 a+ b+ g+ l = d+ e+ f
 e+ f+ a+ l = b+ g+ d
 g+ d+ e+ l = f+ a+ b .
>From the first two equations, we find that d = a+ l. Similarly, f = g+ l and b = e+ l.

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

 1 = a+ ld = (a + x)(f + z)cd
 1 = g+ lf = (b + y)(c + z)ef
 1 = e+ lb = (d + x)(e + y)ab .
Multiplying these three equations together yields that
 abcdef = (a + x)(b + y)(c + z)(d + x)(e + y)c + z) ,
whence x = y = z = 0. Thus, the central triangle degenerates and the three diagonals intersect in a common point.

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).
Solution 1. Clearly, (x, y) = (±1, 0), (±1, -2) are solutions. When y = -1, the right side is negative and there is no solution. Suppose that y ³ 1; then
 (2y2 + 2y)2 = 4y4 + 8y3 + 4y2 > 4y4 + 8y3 + 1
and
 (2y2 + 2y - 1)2 = 4y4 + 8y3 - 4y + 1 < 4y4 + 8y3 + 1
so that the right side is between two consecutive squares, and hence itself cannot be square.

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,

 (1 + p) - (u + p/u) = (u - 1)[(p/u) - 1] ³ 0 .
We have that
 [(2y2 + 2y - 1) + x]
 [(2y2 + 2y - 1) - x] = (2y2 + 2y - 1)2 - x2
 = (4y4 + 8y3 - 4y + 1) - (4y4 + 8y3 + 1)
 = -4y .
Since 2y2 + 2y - 1 = y2 + (y + 1)2 - 2 is positive, at least one of the factors on the left is positive. Since the product is positive, both factors are positive. By our observation on the sum of the factors, we find that
 4y2 + 4y - 2 £ 1 - 4y ,
which is equivalent to
 4(y - 1)2 £ 7 .
However, this does not hold when y £ -3. Therefore, the only solutions are the four that we identified at the outset.

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

 f(y2 + y) = y4 + 2y3 + 2y2 + y > y4 + 2y3  ;
 f(y2 + y - 1) = y4 + 2y3 - y ¹ y4 + 2y3  ;
 f(y2 + y - 2)
 = y4 + 2y3 - 2y2 - 3y + 2
 = y4 + 2y3 - (2y + 1)(y + 1) + 3 < y4 + 2y3  .
It follows that f(t) can never assume the value y4 + 2y3 for any positive t, and hence for any t. Thus, the solutions already listed comprise the complete solution set.

41.
Determine the least positive number p for which there exists a positive number q such that
 ____Ö1 + x + ____Ö1 - x £ 2 - xpq
for 0 £ x £ 1. For this least value of p, what is the smallest value of q for which the inequality is satisfied for 0 £ x £ 1?

 (1 + x)n = 1 + nx + n(n-1)2! x2 + n(n-1)(n-2)3! x3 + ¼+ n(n-1)¼(n-r+1)r! xr + ¼ .
When n is not a nonnegative integer, this is an infinite series that converges when 0 £ |x | < 1 to (1 + x)n. The partial sums constitute a close approximation. When n = 1/2, we have that
 (1 ±x)1/2 = 1 ± 12 x - 18 x2 ± 116 x3 - 5128 x4±¼
so that
 (1 + x)1/2 + (1 - x)1/2 ~ 2 - x24 - x48 £ 2 - x44 .
This suggests that we are looking for (p, q) = (2, 4). However, the approximation approach is not sufficiently rigorous, and we need to find an argument in finite terms that will work.

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

 Ö 1 ±x £ 1 ± 12 x - 18 x2± 116 x3
 Û1 ±x £ 1 ±x + 564 x4 -± 164 x5+ 1256 x6
 Û 0 £ 5 -±x + 4x2 .
The last inequality clearly holds, so the first must as well. Hence
 ____Ö1 + x + ____Ö1 - x £ 2(1 - 18 x2) = 2 - 14 x2
so the pair (p, q) = (2, 4) works for all x Î [0, 1].

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

 ____Ö1 + x + ____Ö1 - x £ 2 - 2cxp
for 0 £ x £ 1. For this range of x, this is equivalent to
 2 + 2 _____Ö1 - x2 £ 4 - 8cxp + 4c2 x2p
 Û _____Ö1 - x2 £ 1 - 4cxp + 2c2 x2p
 Û 1 - x2 £ 1 - 8cxp + 20c2 x2p - 16c3 x3p + 4c4 x4p
 8c £ x2-p + 4c2 xp(5 - 4cx2p + c3 x3p) .
However, for x sufficiently small, the right side can be made less than 8c, yielding a contradiction. Hence, when 0 < p < 2, there is no value that yields the desired inequality.

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

 ____Ö1 + x + ____Ö1 - x £ 2 - x2q
 Û 2(1 + _____Ö1 - x2 ) £ 4 - 4x2q + x4q2
 Û _____Ö1 - x2 £ 1 - 2x2q + x42q2
 Û 1 - x2 £ 1 - 4x2q + 5x4q2 - 2x6q3 + x84q4
 Û 0 £ x2 éê ë æç è 1 - 4q ö÷ ø + x2q2 æç è 5 - 2x2q + x44q2 ö÷ ø ùú û .
If q < 4, then the quantity in square brackets is negative for small values of x. Hence, for the inequality to hold for all x in the interval [0, 1], we must have q ³ 4. Hence, p must be at least 2, and for p = 2, q must be at least 4.

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

 q
³ xp
 2 - ____Ö1 + x - ____Ö1 - x
=
 xp(2 + ____Ö1 + x + ____Ö1 - x )

 4 - (2 + 2 _____Ö1 - x2 )
=
 xp(2 + ____Ö1 + x + ____Ö1 - x )

 2(1 - _____Ö1 - x2 )
=
 xp (2 + ____Ö1 + x + ____Ö1 - x )(1 + _____Ö1 - x2 )

2x2
.
If p < 2, then the right side becomes arbitrarily large as x gets close to zero, so the inequality becomes unsustainable for any real q. Hence, for the inequality to be viable, we require p ³ 2. When p = 2, we can cancel x2 and see by taking x = 0 that q ³ 4. It remains to verify the inequality when (p, q) = (2, 4). We have the following chain of logically equivalent statements, where y = Ö[(1 - x2)] (note that 0 £ x £ 1):
 ____Ö1 + x + ____Ö1 - x £ 2 - x24
 Û 2 + 2 _____Ö1 - x2 £ 4 - x2 + x416
 Û 32 _____Ö1 - x2 £ x4 - 16x2 + 32
 Û 32y £ 1 - 2y2 + y4 - 16 + 16y2 + 32
 Û
 0 £ y4 + 14y2 - 32y + 17 = (y - 1)2(y2 + 2y + 17) = (y - 1)2[(y + 1)2 + 16] .
Since the last inequality is clearly true, the first holds and the result follows.

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:
(1) If v and w are adjacent, then |f(v) - f(w) | £ 1.

(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.

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