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Set 4

19.
The following statics problems involving vectors was given and two separate solutions provided. Determine whether either of them is correct and explain the discrepancy between the answers.
Problem. A force R of magnitude 200 N (Newtons) is the resultant of two forces F and G for which 2 |F | = 3 |G | and the angle between the resultant and G is twice the angle between the resultant and F. Determine the magnitudes of F and G.
Both solutions use the parallelogram representation of the vectors as illustrated, where 3u = |F |, 2u = |G| and v = |R | = 200. From the Law of Sines, we have that
 sin2q3 = sinq2
whence cosq = 3/4, cos2q = 1/8 and cos3q = 4cos3 q- 3cosq = -9/16. >From here, there are two ways to proceed:

(i) From the Law of Cosines, we find that
 v2 = 4u2 + 9u2 - 12u2 cos(180° - 3q) = 13u2 + 12u2 cos3q = 25u24
so that 200 = 5/2u, u = 80, |F | = 240 and |G | = 160.

(ii) From the Law of Cosines, we find that
 4u2 = 9u2 + v2 - 6uv cosq = 9u2 + 2002 - 900u
so that
 0 = 5(u2 - 180u + 8000) = 5(u - 80)(u - 100) .
Hence, u = 80, |F | = 240, |G| = 160 or u = 100, |F | = 300, |G | = 200.

Why does method (i) lead to one solution while method (ii) yields two? QED
20.
Using a pair of compasses with a fixed radius exceeding half the length of a line segment AB, it is possible to determine a point C for which the triangle ABC is equilateral. Here is how it is done.
With centres A and B construct circles using the compasses and let P be one of the two points of intersection of these circles. Draw the circle with centre P, and let it intersect the circle of centre A in Q and the circle of centre B in R. There are different possible configurations, but we will select one so that QR is parallel to AB. Now construct, using the compasses again, circles of centres Q and R. These will intersect in P and a second point C. Prove that triangle ABC is equilateral.
View solution
21.
In the following problem, we will begin by making some empirical observations. Your task will be to formulate some general results exemplified by them and provide proofs. You may observe some other general results not really pointed to below; if so, formulate and justify these.
The numbers 1, 2, 3, ¼ are placed in a triangular array and certain observations concerning row sums are made as indicated below:

The odd row sums of this array are 1, 35, 189, 559, ¼ and their running totals are 1 = 12, 36 = 22 ·32 = (1+2+3)2, 225 = 32 ·52 = (1+2+3+4+5)2, 784 = 42 ·72 = (1+2+3+4+5+6+7)2, ¼, You will notice that the square roots are the sum of odd sums of consecutive integers. Even sums are not to be left out. The even row sums in the array are 9, 91, 341, ¼ and their running totals are 9 = 12 ·32 = (1+2)2, 100 = 22 ·52 = (1+2+3+4)2, 441 = 32 ·72 = (1+2+3+4+5+6)2, ¼.
We can continue in this vein. Writing the triangular array with 1, 4, 7, 10, ¼ numbers in the consecutive rows, we find the running totals of the odd sums to be 1, 64 = 82 = 22 ·42, 441 = 212 = 32 ·72 and so on. With 1, 5, 9, 13, ¼ numbers in the consecutive rows, the running totals of the odd sums are 1, 100 = 102 = 22 ·52, 729 = 272 = 32·72.

View solution
22.
The diagonals of a concyclic quadrilateral ABCD intersect in a point O. Establish the inequality

 ABCD + CDAB + BCAD + ADBC £ OAOC + OCOA + OBOD + ODOB .

View solution
23.
Let A1, A2, ¼, Ar be subsets of { 1, 2, ¼, n } such that no Ai contains another. Suppose that Ai has ai elements (1 £ i £ r). Prove that

 å æç è n ai ö÷ ø -1 £ 1 .

View solution
24.
Without recourse to a calculator or a computer, give an argument that 5123 + 6753 + 7203 is composite. View solution
Problem 19.

19.
First solution. Method (i) leads to the equation 0 = (5u - 2v)(5u + 2v) while (ii) leads to 4u2 = 9u2 + v2 - 9uv/2 or 0 = (5u - 2v)(2u - v). Let us examine closely the second answer that is provided by (ii). In this case, the triangle formed by the two vectors and their resultant is isosceles with sides of magnitudes 3u, 2u and 2u with the base angle equal to q and the apex angle equal to 180° - 2q. In this configuration, cosq is indeed 3/4, but the apex angle is not 2q as specified in the statement of the problem. Indeed, method (i) made use of the angle between R and G. However in method (ii) the result of the sine law remained valid with 180° - 2q in place of 2q but the cosine law did not make use of the angle between R and G. So it is not surprising that the second method leads to a spurious possibility.

Problem 20.

20.
There are various configurations, and we give the solution for one of these. The solutions for the other are similar.
20.
First solution. See Figure 20.1. We have that AP = AQ = QP = QC. Let ÐQAC = ÐQCA = q, so that ÐAQC = 180°- 2q. Then ÐPQC = ÐAQC - 60° = 120° - 2q, so that ÐQCP = ÐQPC = 30° + q. Thus, ÐACP = 30°. Since CP ^AB, ÐCAB = 60°. Similarly, ÐCBA = 60° and so DABC is equilateral.

20.
Second solution. See Figure 20.2.Consider the reflection in the right bisector of AB. It interchanges circles with centres A and B, and hence fixes their intersection points, in particular P. This reflection also carries the circle with centre P to itself. Hence the point C is fixed by the reflection. It follows that P and C are on the right bisector of AB.
Suppose that PC intersects AB in S. Let ÐSAP = q. Then ÐAPB = 180° - 2q. Also
 ÐQPC = 180° - ÐAPQ - ÐAPS = 180° - 60° - (90° - q) = 30° + q
whence
 ÐPQC = 180° - 2(30° + q) = 120°- 2q
and
 ÐAQC = 60° + ÐPQC = 180° - 2q .
Hence DAQC º DAPB (SAS) so that AC = AB and the result follows.

Third solution. See Figure 20.3. A 60° clockwise rotation about A followed by a 60° clockwise rotation about B takes A ® A ® D and Q ® P ®R, where D is the third vertex of an equilateral triangle with side AB. Hence AQ ® DR so that |DR | = |AQ |, the radius of the circle.
Similarly, the composition of two counterclockwise rotations with respective centres B and A takes B ® B ®D and R ® P ® Q, so that |DQ | = |BR |, the radius of the circle. Hence D is a point of intersection of the circles with centres Q and R. When the radius is not equal to |AB |, this will be distinct from P, so that D = C, and the result follows.

Problem 21.

21.
First solution. Consider the triangular array in which the consecutive positive integers are written in rows with only the number 1 in the top row and with k ³ 0 more elements in each row than the previous one. For r ³ 1, the rth row has (r-1)k + 1 elements beginning with ((r-1) || 2)k + r and ending with (r || 2)k + r. The sum of the numbers in the rth row is
 12 [(r-1)k + 1]
 éê ë æç è æç è r-1 2 ö÷ ø + æç è r 2 ö÷ ø ö÷ ø k + 2r ùú û
 = 12 [(r-1)k + 1][(r-1)2k + 2r]
 = 12 [ (r-1)3 k2 + (3r2 - 4r + 1)k + 2r]  .
When r = 2s - 1, this sum is
 4(s-1)3 k2 + (6s2 - 10s + 4)k + (2s - 1)
and the sum of the first m odd-numbered rows is the sum of these terms over 1 £ s £ m, namely
 [m(m-1)k]2 + 2m2(m-1)k + m2 = [m((m-1)k + 1)]2 .

When k = 0, the sum of the elements in the sth odd-numbered row is 2s - 1 and the sum of the elments in the first m odd-numbered rows is m2, the sum of the first m odd integers.
When k = 1, the sum of the elements in the sth odd-numbered row is
 4(s-1)3
 + (6s2 - 10s + 4) + (2s - 1) = 4s3 - 6s2 + 4s - 1
 = s4 - (s-1)4 = [s2 - (s-1)2][s2 + (s-1)2]
 = [(s-1) + s][(s-1)2 + s2]
and the sum of all the elements in the first m odd-numbered rows is m4.

When k = 2, the sum of the elements in the first m odd-numbered rows is
 [m(2m - 1)]2 = [(1/2)(2m)(2m-1)]2 = [1 + 2 + ¼+ 2m-1 ]2  .

When r = 2s, the sum of the elements in the rth row is
 12 [(2s - 1)3 k2 + (12s2 - 8s + 1)k + 4s] .
For k = 0, this sum is 2s. When k = 1, the sum is 4s3 + s = s[1 + (2s)2] and the sum of all the elements in the first m even-numbered rows is
 m2 (m+1)2
 + 12 m(m+1) = æç è m+1 2 ö÷ ø [m2 + (m+1)2]
 = [1 + 2 + ¼+ m][m2 + (m+1)2]  .
For k = 2, the sum of the numbers in the sth even-numbered row is
 2(2s - 1)3 + (12s2 - 8s + 1) + 2s = 16s3 - 12s2 + 6s - 1 = (2s - 1)3 + (2s)3
and the sum of all the elements in the first m even-numbered rows is
 m å s = 1 [(2s - 1)3 + (2s)3] = 2m å r = 1 r3 = [1 + 2 + ¼+ 2m]2 .

Problem 22.

22.
First solution. Let the points and lengths be as labelled in the diagram, and let q = ÐAOB, a = ÐBAD, b = ÐABC. Then ÐBCD = 180° - a and ÐCDA = 180° - b. Then, where [ ¼] denotes area,
 2[ABC] = ab sinb = (p + q)rsinq
 2[ACD] = cd sinb = (p + q)s sinq
so that r/s = (ab)/(cd). Similarly p/q = (ad)/(bc). Hence
 pq + rs = ac æç è db + bd ö÷ ø ³ 2 ac
 pq + sr = db æç è ac + ca ö÷ ø ³ 2 db
 qp + rs = bd æç è ca + ac ö÷ ø ³ 2 bd
 qp + sr = ca æç è bd + db ö÷ ø ³ 2 ca
and the result follows.

Second solution. From similar triangles, we find that a/c = r/q = p/s, b/d = r/p = q/s, so that pq = rs. Then
 æç è pq
 + qp + rs + sr ö÷ ø - æç è ac + ca + bd + db ö÷ ø
 = æç è pq + qp + rs + sr ö÷ ø - æç è rq + sp + qs + pr ö÷ ø
 = p2 + q2 - pr - qspq + r2 + s2 - qr - psrs .
Since pq = rs, this is a fraction over a common denominator with numerator
 p2 + q2 +
 r2 + s2 - pr - qs - rq - sp
 = 12 [ (p - r)2 + (q - s)2 + (q - r)2 +(s - p)2] ,
which is nonnegative. The result follows.

Third solution. Begin as in the second solution. Then
 ac + bd + ca + bb
 = rq + qs + sp + pr
 = pr + qspq + rq + sprs
 = pr + qs + rq + sppq
£
 ______________Öp2 + q2 + r2 + s2 ______________Ör2 + s2 + q2 + p2

pq
 = p2 + q2 + r2 + s2pq
 = p2 + q2pq + r2 + s2rs
 = pq + qp + rs + sr ,
from the Cauchy-Schwarz Inequality.

Problem 23.

23.
First solution. There are n! arrangements of the first n natural numbers. Suppose that ai = k; consider the arrangements { x1, x2, ¼, xn } where Ai = { x1, x2, ¼, xk } (i.e., the first ai numbers constitute the set Ai). There are k! possible ways of ordering the first k numbers and (n-k)! ways of ordering the remaining numbers so that there are ai! (n - ai)! arrangements of this type. Let i ¹ j. No arrangement { x1, x2, ¼, xn } can at the same time have its first ai elements coincide with Ai and its first aj elements coincide with aj, since neither of Ai and Aj is contained in the other. Hence ai! (n - ai)! arrangements corresponding to Ai are distinct from the aj! (n - aj)! arrangements corresponding to Aj. It follows that
 n å i = 1 ai! (n - ai)! £ n!
and the result obtains.

Problem 24.

24.
First solution. Observe that 512 = 29, 675 = 33·52 and 720 = 24 ·32 ·5, so that 2 ·7202 = 3 ·512 ·675. We now use the identity
 x3 + y3 + z3 = x3 + y3 - z3 + 3xyz = (x + y - z)(x2 + y2 + z2 - xy + xz + yz)
which is valid when 2z2 = 3xy to obtain that
 5123 + 6753 + 7203 = (512 + 675 - 720)(5122 + 6752 + 7202 - 512·675+ 512 ·720 + 675 ·720) .
Thus, 467 = 512 + 675 - 720 is a factor of the sum of cubes.

24.
Second solution. [P. LeVan]
 5123
 + 6753 + 7203 = 5123 + 453 [153 + 163]
 = 5123 + 453 [(16 - 1)3 + 163 ]
 = 5123 - 453 + 453 ·16 [2 ·162 - 3 ·16 + 3]
 = (512 - 45)(5122 + 512 ·45 + 452) + 453 ·16[512 - 3 ·15]
 = 467[5122 + 512 ·45 + 452] + 453 ·16 [467]
 = 467 [5122 + 512 ·45 + 452 + 453 ·16]
so that 467 is a factor of the sum of the cubes.

24.
Third solution. [D. Pritchard] We have the identity
 (a9)3 + (b3 c2)3 + (a4 b2 c3)3 = a27 + b9 c6 + a12 b6 c3
 = (a9 + b3 c2 - a4 b2 c)(a18 - a9 b3 c2 + b6 c4+ a13 b2 c + a4 b5 c3 + a8 b4 c2) + (2b - 3a)a12 b5 c3 .
Now take (a, b, c) = (2, 3, 5) to yield the desired result, one factor being 467.

24.
Fourth solution. [D. Arthur] Writing 512 = 1/2(x + y), 675 = 1/2(x + z) and 720 = 1/2(y + z) yields (x, y, z) = (467, 557, 883). Modulo 467, we find that
 5123 + 6753 + 7203
 º 8-1(2y3 + 2z3 + 3y2z + 3yz2)
 = 8-1(y + z)(2y2 + 2z2 + yz)  .
[Note that 8-1 represents a number a for which 8a º 1 (mod 467).] Now y º 90 and z º -51, so that
 2y2 + 2z2 + yz
 º 2 ×902+ 2 ×512 - 90 ×51
 = 9 [4 ×450 + 2 ×289 - 510]
 º 9 [- 4 ×17 + 578 - 510] = 9 [-68 + 68] = 0
and so the sum of the cubes is divisible by 467.

24.
Comment. In fact,
 5123 + 6753 + 7203 = 229 ×467 ×7621 .
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