location:

# Solutions.

297.
The point P lies on the side BC of triangle ABC so that PC = 2BP, ÐABC = 45° and ÐAPC = 60°. Determine ÐACB.
Solution 1. Let D be the image of C under a reflection with axis AP. Then ÐAPC = ÐAPD = ÐDPB = 60°, PD = PC = 2BP, so that ÐDBP = 90°. Hence AB bisects the angle DBP, and AP bisects the angle DPC, whence A is equidistant from BD, PC and PD.

Thus, AD bisects ÐEDP, where E lies on BD produced. Thus

 ÐACB
 = ÐADP = 1 2 ÐEDP
 = 1 2 (180° - ÐBDP) = 1 2 (180° - 30°) = 75° .

Solution 2. [Y. Zhao] Let Q be the midpoint of PC and R the intersection of AP and the right bisector of PQ, so that PR = QR and BR = CR. Then ÐRPQ = ÐRQP = 60° and triangle PQR is equilateral. Hence PB = PQ = PR = RQ = QC and ÐPBR = ÐPRB = ÐQRC = ÐQCR = 30°.

Also, ÐRBA = 15° = ÐPAB = ÐRAB, so AR = BR = CR. Thus, ÐRAC = ÐRCA. Now ÐARC = 180° - ÐPRQ - ÐQRC = 90°, so that ÐRCA = 45° and ÐACB = 75°.

Solution 3. [R. Shapiro] Let H be the foot of the perpendicular from C to AP. Then CPH is a 30-60-90 triangle, so that BP = 1/2PC = PH and ÐPBH = ÐPHB = 30° = ÐPCH. Hence, BH = HC. As

 ÐHAB = ÐPAB = 180° - 120° - 45° = 15° = ÐABP - ÐHBP = ÐABH ,
AH = BH = HC. Therefore, ÐHAC = ÐHCA = 45°. Thus, ÐACB = ÐHCA + ÐPCH = 75°.

Solution 4. From the equation expressing tan30° in terms of tan15°, we find that tan15° = 2 - Ö3 and sin15° = [(Ö3 - 1)/(2Ö2)]. Let ÐACP = q, so that

 ÐPAC = 180° - 60° - q = 120° - q .
Suppose, wolog, we set |BP | = 1, so that |PC | = 2. Then by the Law of Sines in triangle ABP,
 |AP | = sin45° sin15° = Ö3 + 1.
By the Law of Sines in triangle APC,
 sinq Ö3 + 1 = sin(120° - q) 2 = Ö3cosq 4 + sinq 4
whence (3 - Ö3)sinq = (3 + Ö3)cosq. Hence
 tanq = 2 + Ö3 = (2 - Ö3)-1 = (tan15°)-1 ,
so that q = 75°.

298.
Let O be a point in the interior of a quadrilateral of area S, and suppose that
 2S = |OA |2 + |OB |2 + |OC |2+ |OD |2 .
Prove that ABCD is a square with centre O.
Solution.
 |OA |2
 + |OB |2 + |OC |2+ |OD |2
 = 1 2 (|OA |2 + |OB |2) + 1 2 (|OB |2 + |OC |2) + 1 2 (|OC |2 + |OD |2) + 1 2 (|OD |2 + |OA |2)
 ³ |OA ||OB |+ |OB ||OC |+ |OC ||OD |+ |OD ||OA |
 ³ 2[AOB] + 2[BOC] + 2[COD] + 2[DOA] = 2S
with equality if and only if OA = OB = OC = OD and all the angles AOB, BOC, COD and DOA are right. The result follows.

299.
Let s(r) denote the sum of all the divisors of r, including r and 1. Prove that there are infinitely many natural numbers n for which
 s(n) n > s(k) k
whenever 1 £ k < n.
Solution 1. Let um = s(m)/m for each positive integer m. Since d « 2d is a one-one correspondence between the divisors of m and some even divisor of 2m, s(2m) ³ 2s(m) + 1, so that
 u2m = s(2m) 2m ³ 2s(m) + 1 2m > s(m) m = um
for each positive integer m.

Let r be a given positive integer, and select s £ 2r such that us ³ uk for 1 £ k £ 2r (i.e., us is the largest value of uk for k up to and including 2r). Then, as u2s > us, it must happen that 2r £ 2s £ 2r+1 and u2s ³ uk for 1 £ k £ 2r.

Suppose that n is the smallest positive integer t for which 2r £ t and uk £ ut for 1 £ k £ 2r. Then 2r £ n £ 2s £ 2r+1. Suppose that 1 £ k £ n. If 1 £ k £ 2r, then uk £ un from the definition of n. If 2r < k < n, then there must be some number k¢ not exceeding 2r for which uk < uk¢ £ un. Thus, n has the desired property and 2r £ n £ 2r+1. Since such n can be found for each positive exponent r, the result follows.

Comment. The sequence selected in this way starts off: { 1, 2, 4, 6, 12, ¼}.

Solution 2. [P. Shi] Define um as in Solution 1. Suppose, if possible, that there are only finitely many numbers n satisfying the condition of the problem. Let N be the largest of these, and let us be the largest value of um for 1 £ m £ N. We prove by induction that un £ us for every positive integer n. This holds for n £ N. Suppose that n > N. Then, there exists an integer r < n for which ur > un. By the induction hypothesis, ur £ us, so that un < us. But this contradicts the fact (as established in Solution 1) that u2s > us.

300.
Suppose that ABC is a right triangle with ÐB < ÐC < ÐA = 90°, and let K be its circumcircle. Suppose that the tangent to K at A meets BC produced at D and that E is the reflection of A in the axis BC. Let X be the foot of the perpendicular from A to BE and Y the midpoint of AX. Suppose that BY meets K again in Z. Prove that BD is tangent to the circumcircle of triangle ADZ.
Solution 1. Let AZ and BD intersect at M, and AE and BC intersect at P. Since PY joints the midpoints of two sides of triangle AEX, PY || EX. Since ÐAPY = ÐAEB = ÐAZB = ÐAZY, the quadrilateral AZPY is concyclic. Since ÐAYP = ÐAXE = 90°, AP is a diameter of the circumcircle of AZPY and BD is a tangent to this circle. Hence MP2 = MZ ·MA. Since
triangles PAD and XBA are similar. Since
it follows that
so that triangles PAM and XBY are similar. Thus
 PM AP = XY XB = XA 2XB = PD 2PA Þ PD = 2PMÞ MD = PM .
Hence MD2 = MP2 = MZ ·MA and the desired result follows.

Solution 2. [Y. Zhao] As in Solution 1, we see that there is a circle through the vertices of AZPY and that BD is tangent to this circle. Let O be the centre of the circle K. The triangles OPA and OAD are similar, whereupon OP ·OD = OA2. The inversion in the circle K interchanges P and D, carries the line BD to itself and takes the circumcircle of triangle AZP to the circumcircle of triangle AZD. As the inversion preserves tangency of circles and lines, the desired result follows.

301.
Let d = 1, 2, 3. Suppose that Md consists of the positive integers that cannot be expressed as the sum of two or more consecutive terms of an arithmetic progression consisting of positive integers with common difference d. Prove that, if c Î M3, then there exist integers a Î M1 and b Î M2 for which c = ab.
Solution. M1 consists of all the powers of 2, and M2 consists of 1 and all the primes. We prove these assertions.

Since k + (k+1) = 2k+1, every odd integer exceeding 1 is the sum of two consecutive terms. Indeed, for each positive integers m and r,

 (m-r) + (m-r+1) + ¼+ (m-1) + m + (m+1) + ¼+ (m+r-1)+ (m+r) = (2r + 1)m ,
and,
 m + (m + 1) + ¼+ (m + 2r - 1) = r[2(m+r) - 1] ,
so that it can be deduced that every positive integer with at least one odd positive divisor exceeding 1 is the sum of consecutives, and no power of 2 can be so expressed. (If m < r in the first sum, the negative terms in the sum are cancelled by positive ones.) Thus, M1 consists solely of all the powers of 2.

Since 2n = (n + 1) + (n - 1), M2 excludes all even numbers exceeding 2. Let k ³ 2 and m ³ 1. Then

 m + (m + 2) + ¼+ (m + 2(k-1)) = km + k(k-1) = k(m + k - 1)
so that M2 excludes all multiples of k from k2 on. Since all such numbers are composite, M2 must include all primes. Since each composite number is at least as large as the square of its smallest nontrivial divisor, each composite number must be excluded from M2.

We now examine M3. The result will be established if we show that M3 does not contain any number of the form 2r u v where r is a nonnegative integer and u, v are odd integers with u ³ v > 1. Suppose first that r ³ 1 and let a = 2ru - 3/2(v-1). Then

 a ³ 2u - 3 2 (v-1) ³ v 2 + 1 > 1
and
 a + (a + 3) + ¼+ [a + 3(v-1)] = v[a + (3/2)(v-1)] = 2r uv .

Since m + (m+3) = 2m + 3, we see that M3 excludes all odd numbers exceeding 3, and hence all odd composite numbers. Hence, every number in M3 must be the product of a power of 2 and an odd prime or 1.

Comment. The solution provides more than necessary. It suffices to show only that M1 contains all powers of 2, M2 contains all primes and M3 excludes all numbers with a composite odd divisor.

302.
In the following, ABCD is an arbitrary convex quadrilateral. The notation [ ¼] refers to the area.
(a) Prove that ABCD is a trapezoid if and only if
 [ABC] ·[ACD] = [ABD] ·[BCD] .
(b) Suppose that F is an interior point of the quadrilateral ABCD such that ABCF is a parallelogram. Prove that
 [ABC] ·[ACD] + [AFD] ·[FCD] = [ABD] ·[BCD] .
Solution 1. (a) Suppose that AB is not parallel to CD. Wolog, let these lines meet at E with A between E and B, and D between E and C. Let P, Q, R, S be the respective feet of the perpendiculars from A to CD, B to CD, C to AB, D to AB produced. Then
 [ABC]·[ACD] = [ABD][BCD] Û|AB ||CR ||CD ||AP | = |AB ||DS ||CD ||BQ |Û CR : DS = BQ : AP .
By similar triangles, we find that CE : DE = CR : DS = BQ : AP = BE : AE. The dilation with centre E and factor |AE |/|BE | takes B to A, C to D and so the segment BC to the parallel segment AD. Thus ABCD is a trapezoid.

(b) Let the quadrilateral be in the horizontal plane of three-dimensional space and let F be at the origin of vectors. Suppose that u = [( ®) || FA], v = [( ®) || FC], and -pu - qv = [( ®) || FD], where p and q are nonnegative scalars. We have that [( ®) || FB] = u + v. Then

 2[ABC] = |u ×v | ;

 2[ACD] =
 = |u ×v |+ |u ×(pu + qv) |+ |v ×(pu + qv) |
 = (1 + q + p) |u ×v | ;

 2[FCD] = p |u ×v | ;

 2[AFD] = q |u ×v | ;

 2[ABD] = |(pu + qv + u) ×v | = (1 + p) |u ×v | ;

 2[BCD] = |(pu + qv + v) ×u | = (1 + q) |u ×v || .
The result follows.

Solution 2. [Y. Zhao] Observe that, since (A + C) + (B + D) = 360°,

 sinA sinC - sinB sinD
 = 1 2 [ cos(A - C) - cos(A + C) - cos(B - D)+ cos(B + D)]
 = 1 2 [ cos(A - C) - cos(B - D) ] = 1 2 [ cos(B + A - B - C) - cos(B + A + B + C) ]
 = sin(B + A)sin(B + C) .

(a) Hence

 4 [ABD][BCD]
 - 4[ABC][ACD] = (AB ·DA sinA)(BC ·CD sinC) -(AB ·BC sinB)(CD ·DA sinD)
 = (AB ·BC ·CD ·DA) (sinA sinC - sinB sinD)
 = (AB ·BC ·CD ·DA) sin(B + A) sin(B + C) .
The left side vanishes if and only if A + B = C + D = 180° or B + C = A + D = 180°, i.e., AD || BC or AB || CD.

(b) From (a), we have that

 4[ABD][BCD]
 - 4[ABC][ACD] = (AB ·BC ·CD ·DA) sin(A + B)sin(B + C)
 = (AB ·BC ·CD ·DA) sin(A + B - 180°)sin(B + C - 180°)
 = (FC ·AF ·CD ·DA) (sin(ÐBAD - ÐBAF)sin(ÐBCD - ÐBCF))
 = [(DA ·AF)sinÐDAF][ (DC ·CF) sinÐDCF]
 = 4[AFD][FCD] ,
as desired.

303.
Solve the equation
 tan2 2x = 2 tan2x tan3x + 1 .
Solution 1. Let u = tanx and v = tan2x. Then
 v2 - 2v æè u + v 1 - uv öø - 1 = 0

 Û v2 - uv3 - 2uv - 2v2 - 1 + uv = 0

 Û0 = uv + 1 + v2 + uv3 = (uv + 1)(1 + v2)

 Û uv = -1 .
Now v = 2u(1 - u2)-1, so that 2u = v - u2v = u + v and u = v. But then u2 = -1 which is impossible. Hence the equation has no solution.

Solution 2.

 0
 = tan2 2x - 2 tan3x tan2x - 1
 = tan2 2x - 2 tan3x tan2x + tan2 3x - sec2 3x
 = (tan2x - tan3x)2 - sec2 3x
 = (tan2x - tan3x - sec3x)(tan2x - tan3x + sec3x) .
Hence, either tan2x = tan3x + sec3x or tan2x = tan3x - sec3x. Suppose that the former holds. Multiplying the equation by cos2x cos3x yields sin2x cos3x = sin3x cos2x + cos2x. Hence
 0
 = cos2x + (sin3x cos2x - sin2x cos3x)
 = 1 - 2sin2 x + sinx = (1 - sinx)(1 + 2sinx) ,
whence
 x º p 2 , - p 6 , 7p 6
modulo 2p. But tan3x is not defined at any of these angles, so the equation fails. Similarly, in the second case, we obtain 0 = (2 sinx - 1)(sinx + 1) so that
 x º -p 2 , p 6 , 5p 6
modulo 2p, and the equation again fails. Thus, there are no solutions.

Solution 3. Let t = tanx, so that tan2x = 2t(1 - t2)-1 and tan3x = (3t - t3)(1 - 3t2)-1. Substituting for t in the equation and clearing fractions leads to

 4t2(1 - 3t2) = 4t(3t - t3)(1 - t2) + (1 - t2)2(1 - 3t2)

 Û4t2 - 12t4 = (12t2 - 16t4 + 4t6) + (1 - 5t2 + 7t4 - 3t6)

 Û0 = t6 + 3t4 + 3t2 + 1 = (t2 + 1)3 .
There are no real solutions to the equation.

Solution 4. The equation is undefined if 2x or 3x is an odd multiple of p/2. We exclude this case. Then the equation is equivalent to

 sin2 2x - cos2 2x cos2 2x = 2 sin2x sin3x cos2x cos3x
or
 0
 = 2 sin2x sin3x cos2x cos3x + cos4x cos2 2x
 = sin4x sin3x + cos4x cos3x cos2 2x cos3x
 = cosx cos2 2x cos3x .
Since cosx vanishes only if x is an odd multiple of p, we see that the equation has no solution.

Solution 5. [Y. Zhao] Observe that, when tan(A - B) ¹ 0,

 1 + tanA tanB = tanA - tanB tan(A - B) .
In particular,
 1 + tanx tan2x = tan2x - tanx tanx and   1 + tan2x tan3x = tan3x - tan2x tanx .
There is no solution when x º 0 (mod p), so we exclude this possibility. Thus
 0
 = (1 + tan2x tan3x) + (tan2x tan3x - tan2 2x)
 = (tan3x - tan2x)(cotx + tan2x) = cotx (tan3x - tan2x)(1 + tanx tan2x)
 = cot2 x (tan3x - tan2x)(tan2x - tanx)
 = cot2 x æè sinx cos2x cos3x öø æè sinx cosx cos2x öø .
This has no solution.

Solution 6. For a solution, neither 2x nor 3x can be a multiple of p/2, so we exclude these cases. Since

 tan4x = 2 tan2x 1 - tan2 2x ,
we find that
 cot4x = 1 - tan2 2x 2 tan2x = - tan3x ,
whence 1 + tan3x tan4x = 0. Now
 tan4x - tan3x = (1 + tan3x tan4x)tanx = 0 ,
so that 4x º 3x (mod p). But we have excluded this. Hence there is no solution to the equation.
 top of page | contact us | privacy | site map |