location:

# Solutions

Notes. A sequence x1, x2, ¼, xk is in arithmetic progression iff xi+1 - xi is constant for 1 £ i £ k-1. A triangular number is a positive integer of the form

 T(x) º 1 2 x(x+1) = 1 + 2 + ¼+ x ,
where x is a positive integer.

[ ¼] refers to the area of the figure between brackets.

332.
What is the minimum number of points that can be found (a) in the plane, (b) in space, such that each point in, respectively, (a) the plane, (b) space, must be at an irrational distance from at least one of them?
Solution 1. We solve the problem in space, as the planar problem is subsumed in the spatial problem. Two points will never do, as we can select on the right bisector of the segment joining them a point that is the same rational distance from both of them (why?).

However, we can find three points that will serve. Select three collinear points A, B, C such that |AB | = |BC | = u where u2 is not rational. Let P be any point in space. If P, A, B, C are collinear and |PA | = a, |PC | = c, then |PB | is equal to either a - u or c - u. If a and c are rational, then both a - u and c - u are nonrational. Hence at least one of the three distances is rational.

If P, A, C are not collinear, then PB is a median of triangle PAC. Let b = |PB |. Then a2 + c2 = 2b2 + 2u2 (why?). Since u2 is non rational, at least one of a, b, c is nonrational.

Comment. You can check that a2 + c2 = 2(b2 + u2) holds for the collinear case as well.

Solution 2. [F. Barekat] As in the foregoing, the number has to be at least three. Consider the points (0, 0, 0), (u, 0, 0) and (v, 0, 0), where v is irrational and u and v2 are rational. Let P ~ (x, y, z). Then the distances from P to the three points are the respective square roots of x2 + y2 + z2, x2 - 2ux + u2 + y2 + z2 and x2 - 2vx + v2 + y2 + z2. If the first of these is irrational, then we have one irrational distance. Suppose that x2 + y2 + z2 is rational. If x is irrational, then

 x2 - 2ux + u2 + y2 + z2 = (x2 + y2 + z2 + u2) - 2ux
is irrational. If x is rational, then
 x2 - 2vx + v2 + y2 + z2 = (x2 + y2 + z2 + v2) - 2vx
is irrational. Hence not all the three distances can be rational.

333.
Suppose that a, b, c are the sides of triangle ABC and that a2, b2, c2 are in arithmetic progression.
(a) Prove that cotA, cotB, cotC are also in arithmetic progression.
(b) Find an example of such a triangle where a, b, c are integers.
Solution 1. [F. Barekat] (a) Suppose, without loss of generality that a £ b £ c. Let AH be an altitude of the triangle. Then
 cotB = |BH | |AH | and      cotC = |CH | |AH | .
Therefore,
 2[ABC] (cotB + cotC) = a |AH | æè |BH |+ |CH | |AH | öø = a2 .
Similar equalities hold for b2 and c2. Therefore
 2b2 = a2 + c2 Û2(cotA + cotC) = (cotB + cotC) + (cotB + cotA)Û cotA + cotC = 2cotB .
The result follows from this.

(b) Observe that a2 + c2 = 2b2 if and only if (c - a)2 + (c + a)2 = (2b)2. So, if (x, y, z) is a Pythagorean triple with z even and x and y of the same parity, then

 (a, b, c) = æè y-x 2 , z 2 , y+x 2 öø .
Let (x, y, z) = (m2 - n2, 2mn, m2 + n2) where m and n have the same parity and m > n. Then
 (a, b, c) = æè n2 + 2mn - m2 2 , m2 + n2 2 , m2 + 2mn - n2 2 öø .
To ensure that these are sides of a triangle, we need to impose the additional conditions that
 n2 + 2mn - m2 > 0 Û 2n2 > (m - n)2Û m < (Ö2 + 1)n
and
 m2 + 2mn - n2 < (m2 + n2) + (n2 + 2mn - m2)Û m < nÖ3 .
Thus, we can achieve our goal as long as n2 < m2 < 3n2 and m º n (mod 2).

For example, (m, n) = (5, 3) yields (a, b, c) = (7, 17, 23).

Comment. Take (x, y, z) = (2mn, m2 - n2, m2 + n2) to give the solution

 (a, b, c) = æè m2 - 2mn - n2 2 , m2 + n2 2 , m2 + 2mn - n2 2 öø .
For example, (m, n) = (5, 1) yields (a, b, c) = (7, 13, 17). Here are some further numerical examples; note how they come in chains with the first of each triple equal to the last of the preceding one:
 (a, b, c) = [(1, 5, 7),] (7, 13, 17), (17, 25, 31), (31, 41, 49), ¼

 (a, b, c) = (7, 17, 23), (23, 37, 47), (47, 65, 79), ¼

Solution 2. Since c2 = a2 + b2 - 2ab cosC, we have that

 cotC = a2 + b2 - c2 2absinC = a2 + b2 - c2 4[ABC]
with similar equations for cotA and cotB. Hence
 cotA + cotC - 2cotB = (2[ABC])-1(2b2 - a2 - c2) .
Thus, a2, b2, c2 are in arithmetic progression if and only if cotA, cotB, cotC are in arithmetic progression.

(b) We need to solve the Diophantine equation a2 + c2 = 2b2 subject to the condition that c < a + b. The inequality is equivalent to 2b2 - a2 < (a + b)2 which reduces to (b - a)2 < 3a2 or b < (1 + Ö3)a. To ensure the inequality, let us try b = 2a + k, so that b2 = 4a2 + 4ka + k2 and we have to solve 7a2 + 8ka + 2k2 = c2. Upon multiplication by 7 and shifting terms, the equation becomes

 (7a + 4k)2 - 7c2 = 2k2 .
(Note that b/a = 2 + (k/a) < 1 + Ö3 as long as a > 1/2(Ö3 + 1)k.)

We solve a Pell's Equation x2 - 7y2 = 2k2 with the condition that x º 4k (mod 7). There is a standard technique for solving such equations. We find the fundamental solution of x2 - 7y2 = 1; this is the solution with the smallest positive values of x and y, and in this case is (x, y) = (8, 3). We need a particular solution of x2 - 7y2 = 2k2; the solution (x, y) = (3k, k) will do. Then we get an infinite set of solutions (xn, yn) for x2 - 7y2 = 2k2 by defining (x0, y0) = (3k, k) and, for n ³ 1,

 xn + ynÖ7 = (8 + 3Ö7)(xn-1 + yn-1Ö7) .
(Note that the same equation holds when we replace the plus signs in the three terms by minus signs, so we can see that this works by multiplying this equation by its surd conjugate.) Separating out the terms, we get the recursion
 xn = 8xn-1 + 21yn-1

 yn = 3xn-1 + 8yn-1
for n ³ 1. Observe that xn º xn-1 (mod 7). Thus, to get the solution we want, we need to select k such that k º 0 (mod 7).

An infinite family of solutions of x2 - 7y2 = 2k2 starts with

 (x, y) = (3k, k), (45k, 17k), (717k, 271k), ¼ .
All but the first of these will yield a triangle, since we will have 7a = x - 4k ³ 41k whence a > 5k > 1/2(Ö3+ 1)k. Let k = 7. Then, we get the triangles (a, b, c) = (41, 89, 119), (713, 1433, 1897), ¼. We get only similar triangles to these from other multiples of 7 for k.

334.
The vertices of a tetrahedron lie on the surface of a sphere of radius 2. The length of five of the edges of the tetrahedron is 3. Determine the length of the sixth edge.
Solution 1. Let ABCD be the tetrahedron with the lengths of AB, AC, AD, BC and BD all equal to 3. The plane that contains the edge AB and passes through the centre of the sphere is a plane of symmetry for the tetrahedron and is thus orthogonal to CD. This plane meets CD in P, the midpoint of CD. Likewise, the plane orthogonal to AB passing through the midpoint M of AB is a plane of symmetry of the tetrahedron that passes through C, D and P, as well as the centre O of the sphere.

Consider the triangle OCM with altitude CP. Since OM is the altitude of the triangle OAB with |OA | = |OB | = 2 and |AB | = 3, |OM | = (Ö7)/2. Since MC is an altitude of the equilateral triangle ABC, |MC | = (3Ö3)/2. Since OC is a radius of the sphere, |OC | = 2.

Let q = ÐOCM. Then, by the Cosine Law, 7/4 = 4 + [ 27/4] - 2Ö{27}cosq, so that cosq = (Ö3)/2 and sinq = 1/2. Hence, the area [OCM] of triangle OCM is equal to 1/2|OC ||MC |sinq = (3Ö3)/4. But this area is also equal to 1/2 |CP ||OM | = ((Ö7)/4)|CP |. Therefore,

 |CD | = 2 |CP | = 2(3Ö3)/(Ö7) = (6 Ö3)/(Ö7) = (6 Ö 21 )/7 .

Comment. An alterntive way is to note that CMP is a right triangle with hypotenuse CM with O a point on PM. Let u = |CP |. We have that |CM | = (3Ö3)/2, |OM | = (Ö7)/2, |CO | = 2. so that |OP |2 = 4 - u2 and |MP | = [(Ö7)/2]+ Ö{4 - u2}. Hence, by Pythagoras' Theorem,

 27 4 = éë Ö7 2 + Ö 4 - u2 ùû 2 + u2

 Þ 27 4 = 7 4 + Ö7 Ö 4 - u2 + 4

 Þ u2 = 27/7 Þ u = (3Ö3)/Ö7 .
Hence |CD | = 2u = (6Ö3)/Ö7.

Solution 2. [F. Barekat] As in Solution 1, let CD be the odd side, and let O be the centre of the sphere. Let G be the centroid of the triangle ABC. Since the right bisecting plane of the three sides of triangle ABC each pass through the centroid G and the centre O, OG ^ABC. Observe that |CG | = Ö3 and |GM | = (Ö3)/2, where M is the midpoint of AB.

As CGO is a right triangle, |GO |2 = |CO |2 - |CG |2 = 4 - 3 = 1, so that |GO | = 1. Hence, |OM |2 = |OG |2+ |GM |2 = 1 + 3/4 = 7/4, so that |OM | = (Ö7)/2. Therefore,

 sinÐOMC = sinÐOMG = |OG |/|OM | = 2/(Ö7) .

With P the midpoint of CD, the line MP (being the intersection of the planes right bisecting AB and CD) passes through O. Since MC and MD are altitudes of equilateral triangles, |MC | = |MD | = 3(Ö3)/2, so that MCD is isosceles with MP bisecting the apex angle. Hence

 |CD | = 2|MC |sin( 1 2 ÐCMD) = 2 |MC |sinÐOMC = (6 Ö3)/(Ö7) .

Solution 3. Let A be at (-3/2, 0, 0) and B be at (3/2, 0, 0) in space. The locus of points equidistant from A and B is the plane x = 0. Let the centre of the sphere be at the point (0, u, 0), so that u2 + 9/4 = 4 and u = (Ö7)/2. Suppose that C is at the point (0, y, z) where 9/4 + y2 + z2 = 9. We have that (y - (Ö7)/2))2 + z2 = 4 whence

 (Ö7)y + (9/4) = y2 + z2 = 9 - (9/4)
and y = 9/(2Ö7). Therefore z = (3Ö3)/(Ö7).

We find that C ~ (0, 9/(2Ö7), (3Ö3)/(Ö7)) and D ~ (0, 9/(2Ö7), (-3Ö3)/(Ö7)), whence |CD | = (6Ö3)/(Ö7).

Solution 4. [Y. Zhao] Use the notation of Solution 1, and note that the right bisecting plane of AB and CD intersect in a diameter of the sphere. Let us first determine the volume of the tetrahedron ABDO. Consider the triangle ABD with centroid X. We have that |AB | = |BD | = |AD | = 3, |AX | = Ö3 and [ABD] = (9Ö3)/4. Since O is equidistant from the vertices of triangle ABD, OX ^ABD. By Pythagoras' Theorem applied to triangle AXO, |OX| = 1, so that the volume of tetrahedron ABDO is (1/3)|OX |[ABD] = (3Ö3)/4.

Triangle ABO has sides of lengths 2, 2, 3, and (by Heron's formula) area (3Ö7)/4. Since CD ^ABO, CD is the production of an altitude of tetrahedron ABDO; the altitude has length 1/2|CO |. Hence

 |CD | = 2[3Volume(ABDO)] [ABO] = 2(9Ö3)/4 (3Ö7)/4 = 6Ö3 Ö7 .

Solution 5. [A. Wice] Let the tetrahedron ABCD have its vertices on the surface of the sphere of equation x2 + y2 + z2 = 4 with A at (0, 0, 2). The remaining vertices have coordinates (u, v, w) and satisfy the brace of equations: u2 + v2 + (w - 2)2 = 9 and u2 + v2 + w2 = 4. Hence, w = -1/4. Thus, the points B, C, D lie on the circle of equations z = -1/4, x2 + y2 = 63/16. The radius R of this circle and the circumradius of triangle BCD is 3Ö7/4.

Triangle BCD has sides of length (b, c, d) = (b, 3, 3) and area bcd/4R = 1/2bÖ{9 - (b2/4)}. Hence

 3 Ö7 = 1 2 Ö 9 - (b2/4) Û 36 = 7(9 - (b2/4)) Û144 = 7(36 - b2) Û b = 6Ö3 Ö7 .
Thus, the length of the remaining side is (6Ö3)/Ö7.

335.
Does the equation
 1 a + 1 b + 1 c + 1 abc = 12 a + b + c
have infinitely many solutions in positive integers a, b, c?
Comment. The equation is equivalent to
 (a + b + c)(bc + ca + ab + 1) = 12abc .
This is quadratic in each variable, and for any integer solution, has integer coefficients. The general idea of the solution is to start with a particular solution ((a, b, c) = (1, 1, 1) is an obvious one), fix two of the variables at these values and regard the equation as a quadratic in the third. Since the sum and the product of the roots are integers and one root is known, one can find another root and so bootstrap one's way to other solutions. Thus, we have the quadratic for a:
 (b + c)a2 + [(b + c)2 + (bc + 1) - 12bc]a + (b + c)(bc + 1) = 0 ,
so that if (a, b, c) satisfies the equation, then so also does (a¢, b, c) where
 a + a¢ = 9bc - b2 - c2 - 1 b + c and     aa¢ = bc + 1 .

We can start constructing solutions using these relations:

 (1, 1, 1), (1, 1, 2), (1, 2, 3), (2, 3, 7), (2, 5, 7),(3, 7, 11), (5, 7, 18), (5, 13, 18), ¼ .
However, some triples do not lead to a second solution in integers. For example, (b, c) = (2, 5) leades to (7, 2, 5) and (11/7, 2, 5), and (b, c) = (3, 11) leads to (7, 3, 11) and (34/7, 3, 11). So we have no guarantee that this process will not peter out.

Solution 1. [Y. Zhao] Yes, there are infinitely many solutions. Specialize to the case that c = a + b. Then the equation is equivalent to

 2(a + b)[(a + b)2 + ab + 1] = 12ab(a + b)Û a2 - 3ab + b2 + 1 = 0 .
This has at least one solution (a, b) = (1, 1). Suppose that (a, b) = (p, q) is a solution with p £ q. Then (a, b) = (q, 3q - p) is a solution. (To see this, note that x2 - 3qx + (q2 + 1) = 0 is a quadratic equation with one root x = p and root sum 3q.) Observe that q < 3q - p.

Define a sequence { xn } for n ³ 0 by x0 = x1 = 1 and xn = 3xn-1 - xn-2 for n ³ 2. Then (a, b) = (x0, x1) satisfies the equation, and by induction, so does (a, b) = (xn, xn+1) for n ³ 1. Since xn+1 - xn = 2xn - xn1, one sees by induction that { xn } is strictly increasing for n ³ 1. Hence, an infinite set of solutions for the given equation is given by

 (a, b, c) = (xn, xn+1, xn + xn+1)
for n ³ 0. Some examples are
 (a, b, c) = (1, 1, 2), (1, 2, 3), (2, 5, 7), (5, 13, 18), ¼

Comment. If { fn } is the Fibonacci sequence defined by f0 = 0, f1 = 1 and fn = fn-1 + fn-2 for n ³ 2, then the solutions are (a, b, c) = (f2k-1, f2k+1, f2k-1 +f2k+1).

Solution 2. Yes. Let u0 = u1 = 1, v0 = 1, v1 = 2 and

 un = 4un-1 - un-2

 vn = 4vn-1 - vn-2
for n ³ 2, so that { un } = { 1, 1, 3, 11, 41, ¼} and { vn } = { 1, 2, 7, 26, 97, ¼}. It can be proven by induction that both sequences are strictly increasing for n ³ 1. We prove that the equation of the problem is satisfied by
 (a, b, c) = (un, vn, un+1), (vn, un+1, vn+1)
for n ³ 0. In other words, the equation holds if (a, b, c) consists of three consecutive terms of the sequence { 1, 1, 1, 2, 3, 7, 11, 26, 41, 97, ¼}.

Observe that, for n ³ 2,

 unvn - un+1vn-1
 = un(4vn-1 - vn-2) - (4un - un-1)vn-1
 = un-1vn-1 - unvn-2
so that, by induction, it can be established that un vv - un+1vn-1 = -1. Similarly,
 un vn+1 - un+1vn
 = un (4vn - vn-1) - (4un - un-1)vn
 = un-1 vn - un vn-1 = 1 .
It can be checked that (a, b, c) = (1, 1, 1) satisfies the equation. Suppose, as an induction hypothesis, that (a, b, c) = (un, vn, un+1) satisfies the equation. Then un is a root of the quadratic
 0
 = [x + vn + un+1][x(vn + un+1) +vn un+1 + 1] - 12vn un+1 x
 = (vn + un+1)x2 + (vn2 + un+12 +1 - 9vn un+1)x + (vn + un+1)(vnun+1 + 1) .
The second root is
 vn+1 = vn un+1 + 1 un
so (a, b, c) = (vn, un+1, vn+1) satisfies the equation. Similarly, vn is a root of a quadratic, the product of whose roots is un+1vn+1 + 1. The other root of this quadratic is
 un+2 = un+1vn+1 + 1 vn
and so, (a, b, c) = (un+1, vn+1, un+2) satisfies the equation.

Solution 3. [P. Shi] Yes. Let u0 = v0 = 1, v0 = 1, v1 = 2, and

 un = un-1vn-1 + 1 vn-2 and     vn = un vn-1 + 1 un-1
for n ³ 2. We establish that, for n ³ 1,
 un = 2vn-1 - un-1 ;
(1)

 vn = 3un - vn-1 ;
(2)

 3un2 + 2vn-12 + 1 = 6unvn-1 ;
(3)

 3un2 + 2vn2 + 1 = 6unvn .
(4)
Note that (3un2 + 2vn2 + 1 - 6unvn) -(3un2 + 2vn-12 + 1 - 6unvn-1) = 2(vn - vn-1)(vn + vn-1 - 3un), so that the truth of any two of (2), (3), (4) implies the truth of the third.

The proof is by induction. The result holds for n = 1. Suppose it holds for 1 £ n £ k-1. From (3),

 3uk-12 + 2vk-22 + 1 = 6uk-1vk-2 Þ

 (3uk-1 - vk-2)uk-1 + 1 = 2(3uk-1 - vk-2)vk-2- uk-1vk-2 .
Substituting in (2) gives
 uk = uk-1 vk-1 + 1 vk-2 = uk-1(3uk-1 - vk-2) + 1 vk-2 = 2vk-1 - uk-1
which establishes (1) for n = k. From (4),
 (2vk-1 - uk-1)vk-1 + 1 = [3(2vk-1 - uk-1) - vk-1]uk-1
whence
 vk = uk vk-1 + 1 uk-1 = (2vk-1 - uk-1)vk-1 + 1 uk-1 = 3uk- vk-1
which establishes (2) for n = k. From (4),
 3(2vk-1 - uk-1)2 + 2vk-12 + 1 = 6(2vk-1 - uk-1)vk-1

 Þ 3uk2 + 2vk-12 + 1 = 6ukvk .©

Let w2n = un and w2n+1 = vn for n ³ 0. Then

 wn = wn-1wn-2 + 1 wn-3
for n ³ 3. It can be checked that (a, b, c) = (1, 1, 1) = (w0, w1, w2) satisfies the equation. Suppose that the equation is satisfied by (a, b, c) = (wn-1, wn, wn+1). Thus
 (x + wn + ww+1)(x(wn + wn+1) + wnwn+1 + 1) = 12xwnwn+1 = 0
is a quadratic equation in x one of whose roots is x = wn-1. The quadratic can be rewritten
 x2 + [(wn + wn+1) - (11wnwn+1 - 1)/(wn + wn+1)]x+ (wnwn+1 + 1) = 0 .
Since the product of the roots is equal to wnwn+1 + 1, the second root is equal to
 wn wn+1 + 1 wn-1 = wn+2 .
The result follows.

Solution 4. Let c = a + b. Then the equation becomes 2c(ab + c2 + 1) = 12abc, whence ab = (c2 + 1)/5. Hence, a, b are roots of the quadratic equation

 0 = t2 - ct + æè c2 + 1 5 öø = 1 4 éë (2t - c)2 - æè 5c2 - 20 25 öø ùû .
For this to have an integer solution, it is necessary that 5c2 -20 = s2, the square of an integer s. Now s2 - 5c2 = -20 is a Pell's equation, three of whose solutions are (s, c) = (0, 2), (5, 3), (15, 7). Since x2 - 5y2 = 1 is satisfied by (x, y) = (9, 4), solutions (sn, cn) of s2 - 5c2 = -20 are given by the recursion
 sn+1 = 9sn + 20cn

 cn+1 = 4sn + 9cn
for n ³ 0. where (s0, c0) is a starter solution. Taking (s0, c0) = (0, 2), we get the solutions
 (s; a, b, c) = (0; 1, 1, 2), (40; 5, 13, 18),(720; 89, 233, 322), ¼ .
Taking (s0, c0) = (5, 3), we get the solutions
 (s; a, b, c) = (5; 1, 2, 3), (105; 13, 34, 47),(1885; 233, 610, 843), ¼ .
Taking (s0, c0) = (15, 7), we get the solutions
 (s; a, b, c) = (15; 2, 5, 7), (275; 34, 89, 123), ¼ .

Comment. Note that the values of c seem to differ from a perfect square by 2; is this a general phenomenon?

Solution 5. [Z. Guo] Let r1 = r2 = 1, and, for n ³ 1,

 r2n+1 = 2r2n - r2n-1

 r2n+2 = 3r2n+1 - r2n .
Thus, { rn } = { 1. 1, 1, 2, 3, 7, 11, 26, 41, 87, 133, ¼}. Observe that (a, b, c) = (rm, rm+1, rm+2) satisfies the equation for m = 1, 2.

For each positive integer n, let kn = r2n - r2n-1, so that

 r2n-1 = r2n - kn  ,

 r2n+1 = r2n + kn  ,

 r2n+2 = 2r2n + 3kn .

Suppose that (a, b, c) = (r2m-1, r2m, r2m+1) and (a, b, c) = (r2m, r2m+1, r2m+2) satisfy the equation. Substituting (a, b, c) = (r2m - km, r2m, r2m + km) into the equation and simplifying yields that r2m = Ö{3km2 + 1}. (This can be verified by substituting (a, b, c) = (r2m, r2m + km, 2r2m + 3k2m) into the equation.) In fact, this condition is equivalent to these values of (a, b, c) satisfying the equation.

We have that

 3km+12 + 1
 = 3(r2m+2 - r2m+1)2 + 1
 = 3(r2m + 2km)2 + 1
 = 3(r2m2 + 4km2 + 4r2mkm) + 1
 = 3(7km2 + 1) + 12r2mkm + 1 = 21km2 + 12rrm km + 4 .
Thus
 r2m+22
 = (2r2m + 3km)2 = 4r2m2 + 9km2 + 12r2mkm
 = 12km2 + 4 + 9km2 + 12r2mkm = 3km+12 + 1 .
This is the condition that
 (a, b, c) = (r2m+2 - km+1, r2m+2, r2m+2 + km+1) = (r2m+1, r2m+2, r2m+3)
and
 (a, b, c) = (r2m+2, r2m+2 + km+1, 2r2m+2 + 3km+1) = (r2m+2, r2m+3, r2m+4)
satisfy the equation. The result follows by induction.

Solution 6. [D. Rhee] Try for solutions of the form

 (a, b, c) = æè x, 1 2 (x + y), y öø
where x and y are postive integers of the same parity. Plugging this into the equation and simplifying yields the equivalent equation
 x2 - 4xy + y2 + 2 = 0 Û x2 - 4yx + (y2 + 2) = 0 .

Suppose that z1 = 1, z2 = 3 and zn+1 = 4zn - zn-1 for n ³ 1. Then, it can be shown by induction that zn+1 > zn and zn is odd for n ³ 1. We prove by induction that (x, y) = (zn, zn+1) is a solution of the quadratic equation in x and y.

This is true for n = 1. Suppose that it holds for n ³ 1. Then zn is a root of the quadratic equation x2 - 4zn+1x +(zn+12 + 2) = 0. Since the sum of the roots is the integer 4zn+1, the second root is 4zn+1 - zn = zn+2 and the desired result holds because of the symmetry of the equation in x and y.

Thus, we obtain solutions (a, b, c) = (1, 2, 3), (3, 7, 11),(11, 26, 41), (41, 97, 153), ¼ of the given equation.

Solution 7. [J. Park; A. Wice] As before, we note that if (a, b, c) = (u, v, w) satisfies the equation, then so also does (a, b, c) = (v, w, (vw + 1)/u). Define a sequence { xn } by x1 = x2 = x3 = 1 and

 xn+3 = xn+2xn+1 + 1 xn
for n ³ 1. We prove by induction, that for each n, the following properties hold:

(a) x1, ¼, xn+3 are integers; in particular, xn divides xn+2xn+1 + 1;

(b) xn+1 divides xn + xn+2;

(c) xn+2 divides xn xn+1 + 1;

(d) x1 = x2 = x3 < x4 < x5 < ¼ < xn < xn+1 < xn+2;

(e) (a, b, c) = (xn, xn+1, xn+2) satisfies the equation.

These hold for n = 1. Suppose they hold for n = k. Since

 xk+2xk+3 + 1 = xk+2(xk+1xk+2 + 1) xk + 1 = xk+22 xk+1 + (xk + xk+2) xk ,
from (b) we find that xk+1 divides the numerator. Since, by (a), xk and xk+1 are coprime, xk+1 must divide xk+2xk+3 + 1.

Now

 xk+1 + xk+3 = (xk xk+1 + 1) + xk+1xk+2 xk .
By (a), xk and xk+2 are coprime, and, by (c), xk+2 divides the numerator. Hence xk+2 divides xk+1 +xk+3.

Since xk = (xk+1xk+2 + 1)/xk+3 is an integer, xk+3 divides xk+1xk+2 + 1. Next,

 xk+3 = xk+1xk+2 + 1 xk > æè xk+1 xk öø xk+2 > xk+2 .
Finally, the theory of the quadratic delivers (e) for n = k + 1. The result follows.

336.
Let ABCD be a parallelogram with centre O. Points M and N are the respective midpoints of BO and CD. Prove that the triangles ABC and AMN are similar if and only if ABCD is a square.
Comment. Implicit in the problem, but what should have been stated, is that the similarity intended is in the order of the vertices as given, i.e., AB:BC:CA = AM:MN:NA. In the first solution, other possible orderings of the vertices in the similarity are considered (in which case, the result becomes false); in the remaining solutions, the restricted sense of the similarity is discussed.

Solution 1. Let P be the midpoint of CN. Observe that

 [AMD] = 3 4 [ABD] = 3 4 [ABC]
and
 [AMC] = 1 2 [ABC] .
As N is the midpoint of CD, the area of triangle AMN is the average of these two (why?), so that
 [AMN] = 5 8 [ABC] .
Thus, in any case, we have the ratio of the areas of the two triangles, so, if they are similar, we know exactly what the factor of similarity must be.

Let |AB | = |CD | = 2a, |AD | = |BC | = 2b, |AC | = 2c and |BD | = 2d. Recall, that if UVW is a triangle with sides 2u, 2v, 2w opposite the respective vertices U, V, W and m is the length of the median from U, then

 m2 = 2v2 + 2w2 - u2 .
Since AN is a median of triangle ACD, with sides 2a, 2b, 2c,
 |AN |2 = 2c2 + 2b2 - a2 .
Since AM is a median of triangle ABO, with sides 2a, c, d,
 |AM |2 = 2a2 + (c2/2) - (d2/4) .
Since CM is a median of triangle CBO, with sides 2b, c, d,
 |CM |2 = 2b2 + (c2/2) - (d2/4) .
Since MN is a median of triangle MCD with sides |CM |, 3d/2, 2a,
 |MN |2
 = |CM |2 2 + 1 2 æè 3d 2 öø 2 - a2
 = b2 + c2 4 - d2 8 + 9d2 8 - a2 = b2 + c2 4 + d2 - a2 .

Suppose that triangle ABC and AMN are similar (with any ordering of the vertices). Then

 8(|AM |2 + |AN |2 + |MN |2) = 5(|AB |2 + |AC |2 + |BC |2)

 Û 24b2 + 22c2 + 6d2 = 20a2 + 20b2 + 20c2

 Û 2b2 + c2 + 3d2 = 10a2   .
(1)
Also, for the parallelogram with sides 2a, 2b and diagonals 2c, 2d, we have that
 c2 + d2 = 2(a2 + b2)   .
(2)
Equations (1) and (2) together yield d2 = 4a2 - 2b2 and c2 = 4b2 - 2a2. We now have that
 |AN |2 |AC |2 = 2c2 + 2b2 - a2 4c2 = 4c2 + c2 8c2 = 5 8
which is the desired ratio. Thus, when triangles ABC and AMN are similar, the sides AN and AC are in the correct ratio, and so must correspond in the similarity. (There is a little more to this than meets the eye. This is obvious when triangle AMN is scalene; if triangle AMN were isosceles or equilateral, then it is self-congruent and the similarity can be set up to make AN and BC correspond.)

We also have that

 |MN |2 |AB |2 = 4b2 + c2 + 4d2 - 4a2 16a2 = 4b2 + 4b2 - 2a2 + 16a2 - 8b2 - 4a2 16a2 = 10a2 16a2 = 5 8
and
 |AM |2 |BC |2 = 8a2 + 2c2 - d2 16b2 = 8a2 + 8b2 - 4a2 - 4a2 + 2b2 16b2 = 10b2 16b2 = 5 8 .
Thus, if triangle ABC and AMN are similar, then
 AB:BC:AC = MN:AM:AN
and cosÐABC = cosÐADC = 3(a2 - b2)/(2ab). The condition that the cosine has absolute value not exceeding 1 yields that (Ö{10} - 1)b £ 3a £ (Ö{10} + 1)b.

Now we look at the converse. Suppose that ABCD is a parallelogram with sides 2a and 2b as indicated above and that cosÐABC = cosÐADC = 3(a2 - b2)/(2ab). Then, the lengths 2c and 2d of the diagonals are given by

 4c2 = |AC |2 = 4a2 + 4b2 - 12a2 + 12b2 = 16b2 - 8a2 = 8(2b2 - a2)
and
 4a2 = |BD |2 = 4a2 + 4b2 + 12a2 - 12b2 = 16a2 - 8b2 = 8(2a2 - b2)
so that |AC | = 2Ö{4b2 - 2a2} and |BD | = 2Ö{4a2 - 2b2}. Using the formula for the lengths of the medians, we have that
 |AN |2 = 2(4b2 - 2a2) + 2b2 - a2 = 5(2b2 - a2)

 |AM |2 = 2a2 + 4b2 - 2a2 2 - 4a2 - 2b2 4 = 5b2 2

 |MN |2 = b2 + æè 4b2 - 2a2 4 öø + (4a2 - 2b2) - a2 = 5a2 2 .
Thus
 |AN |2 |AC |2 = |AM |2 |BC |2 = |MN |2 |AB |2 = 5 8
and triangles ABC and AMN are similar with AB:BC:AC=MN:AM:AN.

If triangles ABC and AMN are similar with AB:BC:AC=AM:MN:AN, then we must have that AB = BC, i.e. a = b and so ÐABC = ÐADC = 90°, i.e., ABCD is a square.

Comment. A direct geometric argument that triangles ABC and AMN are similar when ABCD is a square can be executed as follows. By reflection about an axis through M parallel to BC, we see that MN = MC. By reflection about axis BD, we see that AM = CM and ÐBAM = ÐBCM. Hence AM = MN and ÐBAM = ÐBCM = ÐCMP = ÐNMP where P is the midpoint of CN.

Consider a rotation with centre A through an angle BAM followed by a dilation of factor |AM |/|AB |. This transformation takes A® A, B ®M and the line BC to a line through M making an angle ÐBAM with BC; the image line must be MN. Since AB = BC and AM = MN, C ® N. Thus, the image of triangle ABC is triangle AMN and the two triangles are similar.

An alternative argument uses the fact that triangles AOM and ADN are similar, so that ÐAMO = ÐAND and AMND is concyclic. From this follows ÐAMN = 180° - ÐADN = 90°.

Solution 2. [S. Eastwood; Y. Zhao] If ABCD is a square, we can take B at 1 and D at i, whereupon M is at (3 + i)/4 and N at (1 + 2i)/2. The vector [( ®) || MN] is represented by

 1 + 2i 2 - 3 + i 4 = i(3 + i) 4 .
Since [( ®) || AM] is represented by (3 + i)/4, MN is obtained from AM by a 90° rotation about M so that MN = AM and ÐAMN = 90°. Hence the triangles ABC and AMN are similar.

For the converse, let the parallelogram ABCD be represented in the complex plane with A at 0, B at z and D at w, Then C is at z + w, M is at (3z + w)/4 and N is at 1/2(z + w) + 1/2w = (z + 2w)/2. Suppose that triangles ABC and AMN are similar. Then, since ÐAMN = ÐABC and AM:AN = AB:AC, we must have that

 1 4 (3z + w)

 1 2 (z + 2w)
=  z

z + w
Û (3z + w)(z + w) = 2z(z + 2w)

 Û 3z2 + 4zw + w2 = 2z2 + 4zwÛ z2 + w2 = 0 Û z = ±iw .
Thus AD is obtained from AB by a 90° rotation and ABCD is a square.

Comment. Strictly speaking, the reasoning in the last paragraph is reversible, so we could use it for the proof in both directions. However, the particularization may aid in understanding what is going on.

Solution 3. [F. Barekat] Let ABCD be a square. Then DADC ~ DAOB so that AB:AC = OB:DC = MB:NC. Since also ÐABM = ÐACN = 45°, DAMB ~ DANC. Therefore AM:AB = AN:AC. Also

 ÐMAB = ÐNAC Þ ÐCAB = ÐCAM + ÐMAB = ÐCAM + ÐNAC = ÐNAM .
Therefore DAMN ~ DABC.

On the other hand, suppose that DAMN ~ DABC. Then AM:AN = AB:AC and

 ÐNAC = ÐNAM - ÐCAM = ÐCAB - ÐCAM = ÐMAB ,
whence DAMB = DANC.

Therefore, AB:AC = MB:NC = BO:DC. Since also ÐABO = ÐACD, DABO ~ DACD, so that ÐABO = ÐACD and ÐAOB = ÐADC. Because ÐABO = ÐACD, ABCD is a concyclic quadrilateral and ÐDAB + ÐDCB = 180°. Since ABCD is a parallelogram, ÐDAB = ÐDCB = 90° and ABCD is a rectangle. Thus ÐAOB = ÐADC = 90°, from which it can be deduced that ABCD is a square.

Solution 4. [B. Deng] Let ABCD be a square. Since MN = MC = MA, triangles MNC and AMN are isosceles. We have that

 AM2 + MN2 = 2AM2 = 2(AO2 + OM2) = (5/4)AB2
and
 AN2 = AD2 + DN2 = (5/4)AB2 = AM2 + MN2 ,
whence ÐAMN = 90° and DAMN ~ DABC.

Suppose on the other hand that DAMN ~ DABC. Then AN:AM = AC:AB and ÐNAM = ÐCAB together imply that

 ÐNAC = ÐMAB Þ DNAC ~ DMABÞ ÐNCA = ÐABM .
But ÐNCA = ÐOAB Þ ÐOAB = ÐOBAÞ ABCD is a rectangle.

Let H be the foot of the perpendicular from M to CN. Then ON, MH and BC are all parallel and M is the midpoint of OB. Hence H is the midpoint of CN and DHMN º DHMC. Therefore MC = MN. Now, from the median length formula,

 AM2 = 1 4 (2BA2 + 2AO2 - BO2) = 1 4 (2BA2 + AO2)
and
 CM2 = 1 4 (2BC2 + 2CO2 - BO2) = 1 4 (2BC2 + CO2)
whence
 (2BA2 + AO2):(2BC2 + CO2) = AM2:CM2 = AM2:MN2 = BA2:BC2
so that
 2BA2·BC2 + AO2·BC2 = 2BA2·BC2 + BA2·CO2Þ BC2 = BA2 Þ BC = BA
and ABCD is a square.

337.
Let a, b, c be three real numbers for which 0 £ c £ b £ a £ 1 and let w be a complex root of the polynomial z3 + az2 + bz + c. Must |w | £ 1?
Solution 1. [L. Fei] Let w = u + iv, [`w] = u - iv and r be the three roots. Then a = -2u - r, b = |w |2 + 2ur and c = -|w |2 r. Substituting for b, ac and c, we find that
 |w |6 - b|w |4 + ac |w |2 - c2 = 0
so that |w |2 is a nonnegative real root of the cubic polynomial q(t) = t3 - bt2 + act - c2 = (t - b)t2+ c(at - c). Suppose that t > 1, then t - b and at - c are both positive, so that q(t) > 0. Hence |w | £ 1.

Solution 2. [P. Shi; Y.Zhao]

 0
 = (1 - w)(w3 + aw2 + bw + c)
 = -w4 + (1 - a)w3 + (a - b)w2 + (b - c)w + c
 Þ w4 = (1 - a)w3 + (a - b)w2 + (b - c)w + c
 Þ |w |4 £ (1 - a)|w |3+ (a - b)|w |2 + (b - c)|w |+ c .
Suppose, if possible, that |w | > 1. Then
 |w |4 £ |w |3[(1 - a) + (a - b) + (b - c) + c] = |w |3
which implies that |w | £ 1 and yields a contradiction. Hence |w | £ 1.

Solution 3. There must be one real solution v. If v = 0, then the remaining roots w and [`w], the complex conjugate of w, must satisfy the quadratic equation z2 + az + b = 0. Therefore |w |2 = w[`w] = b £ 1 and the result follows. Henceforth, let v ¹ 0.

Observe that

 f(-1) = -1 + a - b + c = -(1 - a) - (b - c) £ 0
and that
 f(-c) = -c3 + ac2 - bc + c ³ -c3 + c3 - bc + c = c(1 - b) ³ 0 ,
so that -1 £ v £ -c. The polynomial can be factored as
 (z - v)(z2 + pz + q)
where c = -qv so that q = c/(-v) £ 1. But q = w[`w], and the result again follows.

338.
A triangular triple (a, b, c) is a set of three positive integers for which T(a) + T(b) = T(c). Determine the smallest triangular number of the form a + b + c where (a, b, c) is a triangular triple. (Optional investigations: Are there infinitely many such triangular numbers a + b + c? Is it possible for the three numbers of a triangular triple to each be triangular?)

Solution 1. [F. Barekat] For each nonnegative integer k, the triple
 (a, b, c) = (3k + 2, 4k + 2, 5k + 3)
satisfies T(a) + T(b) = T(c). Indeed,
 (3k+2)(3k+3) + (4k+2)(4k+3) = (9k2+15k+6)+(16k2+20k+6) = 25k2 + 35k + 12 = (5k + 3)(5k + 4) .
We have that a + b + c = 12k + 7, so we need to determine whether there are triangular numbers congruent to 7 modulo 12. Suppose that T(x) is such. Then x(x + 1) must be congruent to 14 modulo 24. Now, modulo 24,
 x2 + x - 14 º x2 + x - 110 = (x - 10)(x + 11) º (x - 10)(x - 13)
so T(x) leaves a remainder 7 upon division by 12 if and only if x = 10 + 24m or x = 13 + 24n for some nonnegative integers m and n. This yields
 k = 4 + 21m + 24m2
and
 k = 7 + 27n + 24n2
for nonnegative integers m and n. The smallest triples according to these formula are (a, b, c) = (14, 18, 23) and (a, b, c) = (23, 30, 38), with the respective values of a + b + c equal to 55 = T(10) and 91 = T(13). However, it may be that there are others that do not come under this set of formulae.

The equation a(a+1) + b(b+1) = c(c+1) is equivalent to (2a + 1)2 + (2b + 1)2 = (2c + 1)2 + 1. It is straightforward to check whether numbers of the form n2 + 1 with n odd is the sum of two odd squares. We get the following triples with sums not exceeding T(10) = 55:

 (2, 2, 3), (3, 5, 6), (5, 6, 8), (4, 9, 10), (6, 9, 11), (8, 10, 13),(5, 14, 15),

 (9, 13, 16), (11, 14, 18), (6, 20, 21), (12, 17, 21), (9, 21, 23), (11, 20, 23), (14, 18, 23) .
The entries of none except the last of these sum to a triangular number.

Solution 2. [Y. Zhao] Let (a, b, c) be a triangular triple and let n = a + b + c. Now

 2T(a) + 2T(b)
 - 2T(n-a-b) = a2 + a + b2 + b - (n - a - b)2 - (n - a - b)
 = (n+1)(n+2) - 2(n+1-a)(n+1-b) .
Thus, (a, b, n-a-b) is a triangular triple if and only if (n+1)(n+2) = 2(n+1-a)(n+1-b). In this case, neither n+1 nor n+2 can be prime, as each factor on the right side is strictly less than either of them. When 1 and 2 are added to each of the first nine triangular numbers, 1, 3, 6, 10, 15, 21, 28, 36, 45, we get at least one prime. Hence n ³ 55. It can be checked that (a, b, c) = (14, 18, 23) is a triangular triple.

We claim that, for every nonnegative integer k, T(24k + 10) = a + b + c for some triangular triple (a, b, c). Observe that. with n = T(24k + 10) = 288k2 + 252k + 55,

 (n+1)(n+2)
 = (T(24k + 10) + 1)(T(24k + 10) + 2)
 = (288k2 + 252k + 56)(288k2 + 252k + 57)
 = 12(72k2 + 63k + 14)(96k2 + 84k + 19)  .
Select a and b so that
 n+1-a = 3(72k2 + 63k + 14) = 216k2 + 189k + 42
and
 n+1-b = 2(96k2 + 84k + 19) = 192k2 + 168k + 38 .
Let c = n - a - b. Thus,
 (a, b, c) = (72k2 + 63k + 14, 96k2 + 84k + 18, 120k2 + 105k + 23) .
From the first part of the solution, we see that (a, b, c) is a triangular triple.

We observe that (a, b, c) = (T(59), T(77), T(83)) = (1770, 3003, 3486) is a triangular triple.

Check:

 1770 ×1771
 + 3003 ×3004 = (2 ×3 ×7 ×11)(295 ×23 + 13 ×1502)
 = (2 ×3 ×7 ×11)(26311) = 2 ×3 ×7 ×11 ×83 ×317
 = (2 ×3 ×7 ×83)(11 ×317) = 3486 ×3487 .
However, the sum of the numbers is 1770 + 3003 + 3486 = 8259, which exceeds 8256 = T(128) by only 3.

Comment. David Rhee observed by drawing diagrams that for any triangular triple (a, b, c), T(a + b - c) = (c - b)(c - a). This can be verified directly. Checking increasing values of a + b - c and factoring T(a + b - c) led to the smallest triangular triple.

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