Solutions

283.

(a) Determine all quadruples (a, b, c, d) of positive integers for which the greatest common divisor of its elements is 1,

a b =  c d

and a + b + c = d.

(b) Of those quadruples found in (a), which also satisfy

1 b +  1 c +  1 d =  1 a  ?

(c) For quadruples (a, b, c, d) of positive integers, do the conditions a + b + c = d and (1/b) + (1/c) + (1/d) = (1/a) together imply that a/b = c/d?

Suppose, if possible, b and c were both odd. Then a and d would be odd as well. If b º c (mod 4), then bc º 1 and b + c º 2 (mod 4), whence ad º a(a + 2) º 3 \not º bc (mod 4). If b º c + 2 (mod 4), it can similarly be shown that ad \not º bc (mod 4), In either case, we get an untenable conclusion. Hence, exactly one of b and c is even and the other is odd.

Without loss of generality, we may suppose that a and b have opposite parity. Let g be the greatest common divisor of a and b, so that a = gu and b = gv for some coprime pair (u, v) of positive integers with opposite parity. Since d > c, it follows that b > a and v > u. Let w = v - u.

Since

(b)

(c) Note that the conditions imply that d - a and b + c are nonzero. The conditions yield that d - a = b + c and (1/a) - (1/d) = (1/b) + (1/c). The second of these can be rewritten

Solution 2. (a) [M. Lipnowski] Let a/b = c/d = r/s where the greatest common divisor of r and s is equal to 1. Then a = hr, b = hs, c = kr, d = ks. Since the greatest common divisor of a, b, c, d equals 1, the greatest common divisor of h and k is 1. From a + b + c = d, we have that (h + k)r = (k - h)s. Observe that gcd(h + k, k - h) = 1 when h and k have opposite parity and gcd(h + k, k - h) = 2 when h and k are both odd. (Why?)

Thus, when h and k have oppposite parity, r = k - h, s = k + h and

(b) Since a/b = c/(a + b + c), c = a(a + b)/(b - a) and d = (a + b) + [a(a+b)/(b-a)] = b(a + b)/(b - a). Hence

(c) [M. Lipnowski]

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284.

Suppose that ABCDEF is a convex hexagon for which ÐA + ÐC + ÐE = 360^° and

AB BC ·  CD DE ·  EF FA = 1 .

Prove that

AB BF ·  FD DE ·  EC CA = 1 .

Also,

Solution 2. [T. Yin] Lemma. Let ABCD be a convex quadrilateral with a, b, c, d, p, q the respective lengths of AB, BC, CD, DA, AC and BD. Then

Proof of Lemma. Locate E within the quadrilateral so that ÐEDC = ÐADB and ÐECD = ÐABD. Then DABD ~ DECD whence ac = qx where x is the length of EC. Now ÐADE = ÐBDC and AD:DE = BD:CD whence DADE ~ DBDC and bd = qy with y the length of AE.

Hence abcd = q²xy and ac + bd = q(x + y). Therefore,

Since ÐDEC = ÐBAD and ÐAED = ÐBCD,

Note that, when ÐA + ÐC = 180^°, then we get Ptolemy's Theorem. Consider the hexagon of the problem with |AB | = a, |BC | = b, |CD | = c, |DE | = d, |EF | = e, |FA | = f, |BF | = g, |CA | = h, |CF | = m, |DF | = u and |CE | = v. We are given that ace = bdf and need to prove that auv = dgh.

From the lemma applied to ABDF, we obtain that

Solution 3. [Y. Zhao] The proof uses inversion in a circle and directed angles. Recall that, if O is the centre of a circle of radius r, then inversion is that involution X« X¢ for which X¢ is on the ray from O through X and OX ·OX¢ = r². It is not too hard to check using similar triangles that ÐOPQ = ÐOQ¢P¢ and using the law of cosines that P¢Q¢ = PQ ·(r²/(OP ·OQ)). For this problem, we make F the centre of the inversion. Then

In the following, we suppress the factor r². We obtain that

Therefore

Solution 4. [M. Abdeh-Kolahchi] Let A, B, C, D, E, F be points in the complex plane with

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285.

(a) Solve the following system of equations:

(1 + 42x - y)(51 - 2x + y) = 1 + 22x - y + 1 ;

y2 + 4x = log2 (y2 + 2x + 1) .

(b) Solve for real values of x:

3x ·8x/(x+2) = 6 .

Express your answers in a simple form.

Substituting for x leads to

(b) The equation can be rewritten

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286.

Construct inside a triangle ABC a point P such that, if X, Y, Z are the respective feet of the perpendiculars from P to BC, CA, AB, then P is the centroid (intersection of the medians) of triangle XYZ.

Let I, J, K be the respective feet of the perpendiculars from the centroid G to the sides BC, AC and AB. The quadrilateral PYAZ is the image of the quadrilateral GJAK under a reflection in the angle bisector of A followed by a dilation with centre A and factor AP/AG. Hence PY:PZ = GK:GI. Since triangles AGB and AGC have the same area,

Solution 2. [M. Lipnowski] Erect squares ARSB, BTUC, CVWA externally on the edges of the triangle. Suppose that RS and VW intersect at A¢, RS and TU at B¢ and TU and UW at C¢.

We establish that AA¢, BB¢ and CC¢ are concurrent. They are cevians in the triangle A¢B¢C¢. We have that

To prove that this works, we first show that PX:PY:PZ = a:b:c, and then that [XPY] = [YPZ] = [ZPX]. Observe that, since DPZA ~ DARA¢ and DPYA ~ DAWA¢,

Comment. A. Zhang and Y. Zhao gave the same construction. Zhang first gave an argument that P, being the centroid of triangle XYZ is characterized by PX:PY:PZ = a:b:c. This is a result of the characterization [XPY] = [YPZ] = [ZPX] and the law of sines, with the argument similar to Lipnowski's. Zhao used the fact that PX:PY:PZ = BC:CA:AB and that the vectors [( ®) || PX], [( ®) || PY], [( ®) || PZ] were dilated versions of [( ®) || BC], [( ®) || CA], [( ®) || AB] after a 90^° rotation, so that [( ®) || PX] +[( ®) || PY] +[( ®) || PZ] = [( ®) || O].

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287.

Let M and N be the respective midpoints of the sides BC and AC of the triangle ABC. Prove that the centroid of the triangle ABC lies on the circumscribed circle of the triangle CMN if and only if

4 ·|AM |·|BN | = 3 ·|AC |·|BC | .

Solution 2. [A. Zhang] Since M and N are respective midpoints of BC and AC, [ABC] = 4[NMC], so that

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288.

Suppose that a₁ < a₂ < ¼ < a_n. Prove that

a1 a24 + a2 a34 + ¼+ an a14 ³ a2 a14 + a3 a24 + ¼+ a1 an4 .

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289.

Let n(r) be the number of points with integer coordinates on the circumference of a circle of radius r > 1 in the cartesian plane. Prove that

n(r) < 6 3 Ö   pr2    .

Let P₁P₂ ¼P_k be a regular k-gon inscribed in the circle. Locate the vertices so that none have integer coordinates. (How?) Identify P_k+1 = P₁ and P_k+2 = P₂, and let v_i = [( ®) || ( P_iP_i+1)] for 1 £ i £ k. Observe that v_i has length less than 2pr/k = (2/k)(pA)^1/2. Then, for each i, the area of triangle P_i P_i+1 P_i+2 is equal to