location:

# 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?
Solution 1. (a) Suppose that the conditions on a, b, c, d are satisfied. Note that b and c have symmetric roles. Since ad = bc, if b and c were both even, then either a or d would be even, whence both would be even (since a + b + c = d), contradicting the fact that the greatest common divisor of a, b, c, d is equal to 1. Hence, at most one of b and c is even.

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 a = a + b + c c = a + b c + 1 ,
it follows that
 b-a a(b + a) = 1 c
whence
 c = gu(u + v) w and            d = gv(u + v) w .
Since the greatest common divisor of u and v is 1, w has no positive divisor in common with either u or v, save 1. Any common divisor of w and u + v must divide 2u = (u + v) -(v - u) and 2v = (u + v) + (v - u); such a common divisor equals 1. Since u and v have opposite parity and so w is odd, w must divide g. Since the greatest common divisor of a, b, c, d is equal to 1, we must have that g = w. Hence
 (a, b, c, d) = (u(v - u), v(v - u), u(v + u), v(v + u))
where u and v are coprime with opposite parity. Interchanging, the roles of b and c leads also to
 (a, b, c, d) = (u(v - u), u(v + u), v(v - u), v(v + u))
with u, v coprime of opposite parity. On the other hand, any quadruples of this type satisfy the condition.

(b)

 1 b + 1 c + 1 d
 = 1 v(v - u) + 1 u(v + u) + 1 v(v + u)
 = 1 v(v - u) + 1 uv = u + (v - u) uv(v - u) = 1 v - u = 1 a .

(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

 ad d - a = bc b + c
so that ad = bc. Thus, all quadruples imply the required condition.

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

 (a, b, c, d) = (h(k-h), h(k+h), k(k-h), k(k+h))
and, when h and k are both odd, then r = 1/2(k - h), s = 1/2(k + h) and
 (a, b, c, d) = ((1/2)h(k-h), (1/2)h(k+h), (1/2)k(k-h),(1/2)k(k+h)) .
It can be checked that these always work. (Collate these with the result given in Solution 1.)

(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

 1 b + 1 c + 1 d
 = 1 b + b - a a + b æè 1 a + 1 b öø
 = 1 b + b-a ab = 1 a .

(c) [M. Lipnowski]

 1 b + 1 c + 1 a + b + c = 1 a
is equivalent to
 0
 = bc(a + b + c) - a(b + c)(a + b + c) - abc
 = (b + c)(bc - a2 - ab - ac) ,
which in turn is equivalent to
 0 = bc - a2 - ab - ac Û bc = a(a + b + c) = ad .

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 .
Solution 1. [A. Zhang] Since the hexagon is convex, all its angles are less than 180°. A dilation of factor |CD |/|DE | followed by a rotation, both with centre D, takes E to C and F to a point G so that DDCG ~ DDEF, ÐDEF = ÐDCG and DE:EF:FD = DC:CG:GD. Since DE:DC = FD:GD and ÐEDC = ÐFDG, DEDC ~ DFDC and DE:DC:CE = FD:DG:GF. Now
 ÐDCG + ÐBCD = ÐDEF + ÐBCD = 360°- ÐFAB > 180°
so that C lies within the triangle BDG and ÐBCG = 360° - (ÐDCG + ÐBCD) = ÐFAB.

Also,

 CG CD = EF DE = AF AB · BC CD
so that CG:BC = AF:AB, with the result that DBCG ~ DBAF, AB:BF:FA = CB:BG:GC and ÐFBG = ÐABC. From the equality of these angles and AB:CB = BF:BG, we have that DABC ~ DFBG and AB:BC:CA = FB:BG:GF. Hence
 AB BF · FD DE · EC CA = CA GF · GF CE · CE CA = 1
as desired.

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

 p2 q2 = (ac + bd)2 - 4abcd cos2 q
where 2q = ÐA + ÐC.

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,

 a2c2 + b2d2 + 2abcd = q2(x2 + 2xy + y2) = q2 (x2 + y2) + 2abcd
which reduces to a2 c2 + b2 d2 = q2(x2 + y2).

Since ÐDEC = ÐBAD and ÐAED = ÐBCD,

 ÐAEC = ÐAED + ÐDEC = ÐC + ÐA = 2q .
By the law of cosines,
 p2 = x2 + y2 - 2xy cos2q = x2 + y2 - 2xy(2 cos2 q- 1) Þ

 a2 c2 + b2 d2
 = p2 q2 + 4q2xy cos2 q- 2q2 xy
 = p2 q2 + 4abcd cos2 q- 2abcd
so that the desired result follows. ª

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

 g2 h2 = a2 m2 + 2abfm + b2 f2 - 4abfmcos2 a
where 2a = ÐBAC + ÐBCF. Applying the lemma to CDEF yields that
 u2 v2 = d2 m2 + 2cdem + c2 e2 - 4cdem cos2 b
where 2b = ÐFCD + ÐDEF. Since ÐA+ ÐC + ÐE = 360°, a+ b = 180° and cos2 a = cos2 b. Finally,
 d2g2h2 - a2u2v2
 = (a2d2m2 + 2abd2fm + b2d2f2 - 4abd2fmcos2a)
 - (a2d2m2 + 2a2cdem + a2c2e2 - 4a2cdemcos2b)
 = 2adm(bdf - ace) + (b2d2f2 - a2c2e2)- 4adm(bdf - ace)cos2a = 0 ,
whence auv = dgh as required.

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

 360°
 = ÐFAB + ÐBCD + ÐDEF = ÐFAB + ÐBCF + ÐFCD + ÐDEF
 = ÐA¢B¢F + ÐFB¢C¢+ ÐC¢D¢F + ÐFD¢E¢ = ÐA¢B¢C¢+ ÐC¢D¢E¢
whence ÐC¢B¢A¢ = ÐC¢D¢E¢.

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

 A¢B¢ B¢C¢ · C¢D¢ D¢E¢
 = æè AB FA ·FB · FB ·FC BC öø · æè CD FC ·FD · FD ·FE DE öø
 = AB FA · CD BC · EF DE = 1
so that A¢B¢: B¢C¢ = D¢E¢:C¢D¢. This, along with ÐC¢B¢A¢ = ÐC¢D¢E¢ implies that DC¢B¢A¢ ~ DC¢D¢E¢, so that A¢B¢:A¢C¢ = D¢E¢:E¢C¢ or A¢B¢·E¢C¢ = A¢C¢·E¢D¢.

Therefore

 AB BF · FD DE · EC CA
 = æè A¢B¢ FA¢·FB¢ ·B¢F öø · æè 1 F¢D¢ · FD¢·FE¢ D¢E¢ öø · æè E¢C¢ FE¢·FC¢ · FC¢·FA¢ C¢A¢ öø
 = A¢B¢ A¢C¢ · E¢C¢ E¢D¢ = 1 ,
as desired.

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

 B - A = a = |a |(cosa+ i sina)

 C - B = b = |b |(cosb+ i sinb)

 D - C = c = |c |(cosg+ i sing)

 E - D = d = |d |(cosd+ i sind)

 F - E = e = |e |(cose+ i sine)

 A - F = f = |f |(cosf+ i sinf) .
Modulo 360°, we have that
 ÐA = ÐFAB º 180° - (f- a)

 ÐC = ÐBCD º 180° - (d- b)

 ÐE = ÐDEF º 180° - (e- g) .
Also a + b + c + d + e + f = 0 and
 ace bdf
 = |a ||c ||e |(cosa+i sina)(cosg+ i sing)(cose+i sine) |b ||d ||f |(cosb+i sinb)(cosd+ i sind)(cosf+ i sinf)
 = 1 (cos(a- f+ d- b+ e- g))
 = cos(ÐA - 180° + ÐC - 180°+ ÐE - 180°) = cos(-180°) = -1 ,
whence ace + bdf = 0. Therefore,
 0 = ad(a + b + c + d + e + f) + (ace + bdf) = a(d + e)(c + d) + d(a + f)(a + b) ,
whence
 a(d+e)(c+d) d(a+f)(a+b) = -1Þ |a | |a + f | · |d + e | |d | · |c + d | |a + b | = 1

 Þ AB BF · FD DE · EC CA = 1 .

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 .
Solution. Let u = 2x - y. Then
 (1 + 4u)(51-u) = 1 + 2u+1
so that
 5u-1 = 1 + 22u 1 + 2u+1 = 2u-1 + 1 - 2u-1 1 + 2u+1 .
Thus,
 5u - 1 - 2u - 1 = 1 - 2u-1 1 + 2u+1 .
When u > 1, the left side of this equation is positive while the right is negative; when u < 1, the reverse is true. Hence, the only possible solution is u = 1, which checks out.

 y2 + 2y + 2 = log2 (y2 + y + 2) .
Since y2 + y + 2 = ( y + 1/2 )2 + 7/4 > 0, the right side is defined and is in fact positive. Let
 f(y) = y2 + 2y + 2 - log2 (y2 + y + 2) .
Then
 f¢(y) = 2y(y+1)2 + 4(y + 1) - (log2 e)(2y + 1) y2 + y + 2 .

 f¢(y) = 0 Û (y+1)2 = - æè (2 - log2 e) + 4 - log2 e 2y öø .
From the graphs of the two sides of the equation, we see that the left side and the right side have opposite signs when y > 0 and become equal for exactly one value of y. It follows that f¢(y) changes sign exactly once so that f(y) decreases and then increases. Thus, f(y) vanishes at most twice. Indeed, f(-2) = f(-1) = 0, and so (x, y) = (0, -1), (-1/2, -2) are the only solutions of the equation.

(b) The equation can be rewritten

 1 = 31-x 22(1-x)/(x+2)
whence
 0 = (1 - x)( log3 + (2/(x+2))log2)  .
Thus, either x = 1 or 0 = log2 3 + 2/(x+2). The latter leads to
 x = -2(1 + log3 2) = -2(log3 6) = -log3 36 .

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.
Solution 1. Let AU, BV, CW be the medians of triangle ABC and let AL, BM, CN be their respective images in the bisectors of angles A, B, C. Since AU, BV, CW intersect in a common point (the centroid of DABC). AL, BM, CN must intersect in a common point P. This follows from the sine version of Ceva's theorem and its converse. Let X, Y, Z be the respective feet of the perpendiculars from P to sides BC, AC, AB.

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,

 AB ·GK = AC ·GJÞ PY:PZ = AC:AB = b:c .
Applying a similar argument involving PX, we find that
 PX:PY:PZ = a:b:c .
Let PX = ae, PY = be, PZ = ce. Then, since ÐXPY + ÐACB = 180°,
 [PXY] = 1 2 abe2 sinÐXPY = e2 æè 1 2 absinC öø = e2[ABC] .
Similarly, [PYZ] = [PZX] = e2[ABC] = [PXY], whence P must be the centroid of triangle XYZ.

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

 sinÐRA¢A sinÐWA¢A
 · sinÐVC¢C sinÐUC¢C · sinÐTB¢B sinÐSB¢B
 = (AR/AA¢) (AW/AA¢) · (VC/CC¢) (UC/CC¢) · (TB/BB¢) (BS/BB¢)
 = AR AW · VC UC · TB BC = c b · b a · a c = 1 .
Hence AA¢, BB¢, CC¢ intersect in a point P by the converse to Ceva's Theorem. P is the desired point.

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¢,

 PY PZ = PY(AA¢/PA) PZ(AA¢/PA) = AW AR = b c ,
and similarly that PX:PZ = a:c. Now
 ÐXPY = 360° - ÐPXC - ÐPYC - ÐXCY = 180° - ÐXCY = 180° - ÐACB ,
so that [XPY] = 1/2PX ·PY sinÐXPY = 1/2 PX ·PY sinÐACB. We find that
 [XPY]
 : [YPZ] : [ZPX] = 1 2 PX ·PY sinÐACB : 1 2 PY ·PZ sinÐACB : 1 2 PZ ·PX sinÐABC
 = 1 2 ab sinC : 1 2 bc sinA : 1 2 ca sinB = [ABC] : [ABC] : [ABC] = 1:1:1 .
Hence [XPY] = [YPZ] = [ZPX] = 1/3[XYZ], so that the altitudes of these triangle from P to the sides of triangle XYZ are each one-third of the corresponding altitudes for triangle XYZ. Hence P must be the centroid of triangle XYZ.

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].

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 1.
 4 |AM ||BN | = 3 |AC ||BC |Û12 |AM ||GN | = 12 |AN ||MC |Û |AM |: |MC | = |AN |: |GN |

 Û DAMC ~ DANGÛ ÐAMC = ÐANG
Û GMGN is concyclic.

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

 [ABMN] = 3 4 [ABC] = 3 8 |AC ||BC |sinÐACB .
However, [ABMN] = 1/2|AM ||BN |sinÐNGM (why?). Hence
 4 |AM ||BN |sinÐNGM = 3 |AC ||BC |sinÐACB .
Observe that G lies inside triangle ABC, and so lies within the circumcircle of this triangle. Hence ÐNGM = ÐAGB > ÐACB. We deduce that
 4 |AM ||BN | = 3 |AC ||BC |Û sinÐNGM = sinÐACBÛ ÐNGM + ÐACB = 180°
Û CMGN is concyclic.

288.
Suppose that a1 < a2 < ¼ < an. Prove that
 a1 a24 + a2 a34 + ¼+ an a14 ³ a2 a14 + a3 a24 + ¼+ a1 an4 .
Solution. The result is trivial for n = 2. To deal with the n = 3 case, observe that, when x < y < z,
 (xy4 + yz4 + zx4)-(yx4 + zy4 + xz4) = (1/2)(z - x)(y - x)(z - y)[(x + y)2 + (x + z)2 + (y + z)2] ³ 0 .
As an induction hypothesis, assume that the result holds for the index n ³ 3. Then
 (a1 a24 + a2 a34 + ¼+ an an+14+ an+1 a14)
 - (a2 a14 + a3 a24 + ¼+an+1 an4 + a1 an+14)
 = (a1 a24 + a2 a34 + ¼+ an a14)- (a2 a14 + a3 a24 + ¼+ a1 an4)
 + (a1 an4 + an an+14 + an+1 a14)- (an a14 + an+1 an4 + a1 an+14) ³ 0 ,
as desired.

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 .
Solution. Let A = pr2 be the area of the circle, so that the right side of the inequality is 6A1/3. We observe that A > 3, p2 < (22/7)2 < 10 < (2.2)3.
 6A1/3 - 2p2/3A1/3
 = (6 - 2p2/3)A1/3 > (6 - 2 ×101/3)A1/3
 > (6 - 4.4) ×31/3 > 1.6 ×1.25 = 2 ,
so that there is an even integer k for which
 6 = 2 ×32/3 ×31/3 < 2 p2/3 A1/3 < k < 6A1/3 .
In particular, 8p2 A < k3.

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

 1 2 |vi ×vi+1 | = 1 2 |vi ||vi+1 |sin(2p/k) < 1 2 × 4 k2 ×pA × 2p k = 1 2 × 8p2 k3 ×A < 1 2 .
Suppose, if possible, that the arc joining Pi and Pi+2 (through Pi+1) contains points U, V, W, each with integer coordinates. Then, if u, v, w are the corresponding vectors for these points, then |(v - u) ×(w - u) | must be a positive integer, and so the area of triangle UVW must be at least 1/2. But each of the sides of triangle UVW has length less than the length of PiPi+2 and the shortest altitude of triangle UVW is less than the altitude of triangle Pi Pi+1Pi+2 from Pi+1 to side Pi Pi+2. Thus,
 1 2 £ [UVW] £ [Pi Pi+1 Pi+2] < 1 2 ,
a contradiction. Hence, each arc Pi Pi+2 has at most two points with integer coordinates. The whole circumference of the circle is the union of k/2 nonoverlapping such arcs, so that there must be at most k points with integer coordinates. The result follows.

 top of page | contact us | privacy | site map |