mr 1 2014 solutions

Junior problems

J289. Let a be a real number such that 0 ≤ a < 1. Prove that⌊a

(1 +

⌊1

1− a

⌋)⌋+ 1 =

⌊1

1− a

⌋.

Proposed by Arkady Alt, San Jose, California, USA

Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, SpainSince 0 ≤ a < 1, then 0 < 1− a ≤ 1.

If k ∈ N such that1

1 + k< 1 − a ≤ 1

k, then k ≤ 1

1− a< k + 1, and

k − 1

k≤ a <

k

k + 1, so

⌊1

1− a

⌋= k.

On the other hand, for the left-hand side of the proposed identity we have⌊a

(1 +

⌊1

1− a

⌋)⌋+ 1 = ba (1 + k)c+ 1

= k − 1 + 1 = k.

Also solved by Archisman Gupta, RKMV, Agartala, Tripura, India; Joshua Benabou, Manhasset HighSchool, NY, USA; Daniel Lasaosa, Pamplona, Navarra, Spain; Arber Igrishta, Eqrem Qabej, Vushtrri, Ko-sovo; Mathematical Group Galaktika shqiptare, Albania; Paolo Perfetti, Università degli studi di Tor VergataRoma, Italy; Alessandro Ventullo, Milan, Italy; Polyahedra, Polk State College, FL, USA; AN-anduud Pro-blem Solving Group, Ulaanbaatar, Mongolia; Ioan Viorel Codreanu, Satulung, Maramures, Romania; VietQuoc Hoang, University of Auckland, New Zealand.

Mathematical Reflections 1 (2014) 1

J290. Let a, b, c be nonnegative real numbers such that a+ b+ c = 1. Prove that

3√

13a3 + 14b3 +3√

13b3 + 14c3 +3√

13c3 + 14a3 ≥ 3.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, UzbekistanFrom Hölder’s inequality we easily obtain (13a3 + 14b3)(13 + 14)(13 + 14) ≥ (13a+ 14b)3 =>

3√

13a3 + 14b3 ≥ 13a+ 14b

9(1)

Similarly, we have

3√

13b3 + 14c3 ≥ 13b+ 14c

9,

3√

13c3 + 14a3 ≥ 13c+ 14a

9(2)

From (1) and (2) we get

3√

13a3 + 14b3 +3√

13b3 + 14c3 +3√

13c3 + 14a3 ≥ 27(a+ b+ c)

9= 3

Also solved by Bodhisattwa Bhowmik, RKMV, Agartala, Tripura, India; Daniel Lasaosa, Pamplona,Navarra, Spain; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Arkady Alt, San Jose,California, USA; An Zhen-ping, Xianyang Normal University, China; Viet Quoc Hoang, University of Auc-kland, New Zealand; Ioan Viorel Codreanu, Satulung, Maramures, Romania; AN-anduud Problem SolvingGroup, Ulaanbaatar, Mongolia; Polyahedra, Polk State College, FL, USA; Paolo Perfetti, Università deglistudi di Tor Vergata Roma, Italy; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; SayakMukherjee, Kolkata, India; Sayan Das, Indian Statistical Institute, Kolkata, India; Neculai Stanciu, Buzău,Romania and Titu Zvonaru, Comănes,ti, Romania; Alessandro Ventullo, Milan, Italy; Georgios Batzolis,Mandoulides High School, Thessaloniki, Greece.


J291. Let ABC be a triangle such that ∠BCA = 2∠ABC and let P be a point in its interior such thatPA = AC and PB = PC. Evaluate the ratio of areas of triangles PAB and PAC.

Proposed by Panagiote Ligouras, Noci, Italy

Solution by Polyahedra, Polk State College, USA

∑

E

P

ADB

C

Let Σ be the circle with center A and radius AC. Suppose that the bisector of ∠ACB intersects AB at D andΣ at E. Then PD is the perpendicular bisector of BC. Since ∠AEC = ∠ACE = ∠BCE, EA ‖ BC. Thus∠BAE = ∠ABC, so PD is the perpendicular bisector of AE as well. Hence4APE is equilateral. Therefore,∠PCE = 1

2∠PAE = 30, ∠APC = 30+B, and ∠APB = 60+∠EPB = 60+∠APC = 90+B. Finally,let [·] denote area, then

[PAB]

[PAC]=

sin∠APBsin∠APC

=cosB

sin (30 +B)=

2

1 +√

3 tanB.

Also solved by Andrea Fanchini, Cantú, Italy; Daniel Lasaosa, Pamplona, Navarra, Spain; Arkady Alt,San Jose, California, USA.


J292. Find the least real number k such that for every positive real numbers x, y, z, the following inequalityholds: ∏

cyc

(2xy + yz + zx) ≤ k(x+ y + z)6.

Proposed by Dorin Andrica, Babes-Bolyai University, Romania

Solution by Albert Stadler, Herrliberg, Switzerland

We claim that k =64

729.

Let x = y = z = 1⇒ ∏cyc

(2xy + yz + zx)

(x+ y + z)6=

64

729≤ k

By the Cauchy-Schwarz inequality , xy + yz + zx ≤ x2 + y2 + z2.So 3(xy + yz + zx) ≤ 2xy + 2yz + 2zx+ x2 + y2 + z2 = (x+ y + z)2.By AM-GM:

∏cyc

(2xy + yz + zx) ≤

∑cyc

(2xy + yz + zx)

3

3

=

4∑cycxy

3

3

≤(

4(x+ y + z)2

9

)3

=64

729(x+ y + z)6,

Which proves that k ≤ 64

729and the conclusion follows.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Alessandro Ventullo, Milan, Italy; Titu Zvona-ru, Comănes,ti, Romania and Neculai Stanciu, Buzău, Romania; Shohruh Ibragimov, Lyceum Nr.2 under theSamIES, Samarkand, Uzbekistan; Sayan Das, Indian Statistical Institute, Kolkata, India; Sayak Mukherjee,Kolkata, India; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Sun Mengyue Lansheng,Fudan Middle School, Shanghai, China; Arkady Alt, San Jose, California, USA; Georgios Batzolis, Mandou-lides High School, Thessaloniki, Greece; Ioan Viorel Codreanu, Satulung, Maramures, Romania; AN-anduudProblem Solving Group, Ulaanbaatar, Mongolia; Polyahedra, Polk State College, FL, USA; Jan Jurka, Brno,Czech Republic.


J293. Find all positive integers x, y, z such that(x+ y2 + z2

)2 − 8xyz = 1.

Proposed by Aaron Doman, University of California, Berkeley , USA

Solution by Alessandro Ventullo, Milan, ItalyWe rewrite the equation as

x2 + 2x(y2 + z2 − 4yz) + (y2 + z2)2 − 1 = 0.

Since x must be a positive integer, the discriminant of this quadratic equation in x must be non-negative,i.e.

(y2 + z2 − 4yz)2 − (y2 + z2)2 + 1 ≥ 0,

which is equivalent to−8yz(y − z)2 + 1 ≥ 0,

which gives yz(y − z)2 ≤ 1/8. Since y and z are positive integers, it follows that

yz(y − z)2 = 0,

so y − z = 0, i.e. y = z. The given equation becomes (x− 2y2)2 = 1, which yields x = 2y2 ± 1. Therefore,all the positive integer solutions to the given equation are

(2n2 − 1, n, n), (2n2 + 1, n, n), n ∈ Z+.

Also solved by Albert Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Navarra, Spain; SimaSharifi, College at Brockport, SUNY, USA; Sayan Das, Indian Statistical Institute, Kolkata, India; GeorgiosBatzolis, Mandoulides High School, Thessaloniki, Greece; Viet Quoc Hoang, University of Auckland, NewZealand; Ioan Viorel Codreanu, Satulung, Maramures, Romania; Polyahedra, Polk State College, FL, USA.


J294. Let a, b, c be nonnegative real numbers such that a+ b+ c = 3. Prove that

1 ≤ (a2 − a+ 1)(b2 − b+ 1)(c2 − c+ 1) ≤ 7.

Proposed by An Zhen-ping, Xianyang Normal University, China

Solution by Daniel Lasaosa, Pamplona, Navarra, SpainDenote p = abc and s = ab + bc + ca, where clearly 0 ≤ p ≤ 1, with p = 0 iff at least one of a, b, c is zero,and p = 1 iff a = b = c = 1 by the AM-GM inequality, while 0 ≤ s ≤ 3, with s = 0 iff two out of a, b, c arezero, and s = 3 iff a = b = c = 1 by the scalar product inequality. Note that

(a2 − a+ 1)(b2 − b+ 1)(c2 − c+ 1) = p2 + s2 − ps− 4s− p+ 7.

The lower bound then rewrites as

(2 + p− s)2 + 2(1− p) ≥ p(3− s).

Now, 9p = 3abc(a+ b+ c) ≤ (ab+ bc+ ca)2 = s2, or 9(1− p) ≥ (3 + s)(3− s), or it suffices to show that

(2 + p− s)2 +6 + 2s− 9p

9(3− s) ≥ 0.

Assume that 9p > 6, or p > 23 , hence by the AM-GM inequality, s ≥ 3 3

√p2 ≥ 3

√12 > 2, or 6 + 2s > 10 > 9p.

It follows that both terms in the LHS are non-negative, being zero iff s = 3 and simultaneously s = p + 2,for p = 1.

On the other hand, the upper bound rewrites as

p(1− p) + s(3− s) + s+ ps ≥ 0,

clearly true because s, ps, p(1− p), s(3− s) are all non-negative. Note that equality requires s = 0, and sincea, b, c are non-negative, this requires at least two out of a, b, c to be zero, resulting in p = 0.

The conclusion follows, equality holds in the lower bound iff a = b = c = 1, and in the upper bound iff(a, b, c) is a permutation of (3, 0, 0).

Also solved by Polyahedra, Polk State College, FL, USA; AN-anduud Problem Solving Group, Ulaanbaa-tar, Mongolia; Viet Quoc Hoang, University of Auckland, New Zealand; Arkady Alt, San Jose, California,USA; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania; Paolo Perfetti, Università de-gli studi di Tor Vergata Roma, Italy; Shivang jindal, Jaipur, Rajasthan, India; Albert Stadler, Herrliberg,Switzerland.


Senior problems

S289. Let x, y, z be positive real numbers such that x ≤ 4, y ≤ 9 and x+ y + z = 49. Prove that

1√x

+1√y

+1√z≥ 1.

Proposed by Marius Stanean, Zalau, Romania

Solution by Li Zhou, Polk State College, FL, USAApplying Jensen’s inequality to the convex function f(t) = 1/

√t for t > 0, we get

1√x

+1√y

+1√z

=1

2f(x

4

)+

1

3f(y

9

)+

1

6f( z

36

)≥ f

(1

2· x

4+

1

3· y

9+

1

6· z

36

)=

√216√

27x+ 8y + z=

√216√

26x+ 7y + 49≥

√216√

26(4) + 7(9) + 49= 1.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Sima Sharifi, College at Brockport, SUNY,USA; Arkady Alt, San Jose, California, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;Ioan Viorel Codreanu, Satulung, Maramures, Romania; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES,Samarkand, Uzbekistan; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Sun Mengyue ,Lansheng Fudan Middle School, Shanghai, China.


S290. Prove that there is no integer n for which

1

22+

1

32+ · · ·+ 1

n2=

(4

5

)2

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by Alessandro Ventullo, Milan, ItalyLet p be the greatest prime number such that p ≤ n < 2p. Then the given equality can be written as

52k = (4 · n!)2,

where k =

n∑i=2

(n!)2

i2. Observe that k ≡ (n!)2

p26≡ 0 (mod p). Since p|52k and p does not divide k, it follows

that p|52, i.e. p = 5. So, n ∈ 5, 6, 7, 8, 9. An easy check shows that none of these values satisfies theequality.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Yassine Hamdi, Lycée du Parc, Lyon, France;Sima Sharifi, College at Brockport, SUNY, USA; Li Zhou, Polk State College, FL, USA; Arghya Datta,Hooghly Collegiate School, Kolkata, India; Albert Stadler, Herrliberg, Switzerland.


S291. Let a, b, c be nonnegative real numbers such that ab+ bc+ ca = 3. Prove that(2a2 − 3ab+ 2b2

) (2b2 − 3bc+ 2c2

) (2c2 − 3ca+ 2a2

)≥ 5

3

(a2 + b2 + c2

)− 4.

Proposed by Titu Andreescu, USA and Marius Stanean, Romania

Solution by Daniel Lasaosa, Pamplona, Navarra, SpainNote that

27(2a2 − 3ab+ 2b2

) (2b2 − 3bc+ 2c2

) (2c2 − 3ca+ 2a2

)+ 27 · 4− 45

(a2 + b2 + c2

)=

= 27(2a2 − 3ab+ 2b2

) (2b2 − 3bc+ 2c2

) (2c2 − 3ca+ 2a2

)+ 4 (ab+ bc+ ca)3−

−5(a2 + b2 + c2

)(ab+ bc+ ca)2 =

= 211(a4b2 + b4c2 + c4a2 + a2b4 + b2c4 + c2a4)− 320(a3b3 + b3c3 + c3a3)−

−334abc(a3 + b3 + c3) + 164abc(a2b+ b2c+ c2a+ ab2 + bc2 + ca2)− 288a2b2c2,

or it suffices to show that this last expression is non-negative. Note that it can be rearranged as∑cyc

(3c4 + 160a2b2 + 48c2(a+ b)2 − 164abc(a+ b)

)(a− b)2 =

=∑cyc

(3c4 + 5c2(a+ b)2 + 39(bc+ ca− 2ab)2 + 4(bc+ ca− ab)2

)(a− b)2,

clearly non-negative, and being zero iff a = b = c. The conclusion follows.

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy; Nicuşor Zlota, “TraianVuia” Technical College, Focşani, Romania; Arkady Alt, San Jose, California, USA.


S292. Given triangle ABC, prove that there exists X on the side BC such that the inradii of triangles AXBand AXC are equal and find a ruler and compass construction.

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by Li Zhou, Polk State College, USA

Y

X FE

KJ

D

I

A

B C

As usual, let a, b, c, s, and r be the sides BC, CA, AB, semiperimeter, and inradius of 4ABC. Let I, J , Kbe the incenters of triangles ABC, ABX, AXC, and D, E, F be the feet of perpendiculars from A, J , Konto BC. Let h = AD, r1 = JE, and r2 = KF . As X moves from B to C, r1 increases from 0 to r while r2decreases from r to 0. Hence, there exists X such that r1 = r2 = t. Now t

r = BEs−b , so BE = t(s−b)

r . Likewise,FC = t(s−c)

r . Let [·] denote area. Then

1

2ah = [ABC] = [ABX] + [AXC] = t(c+ EX) + t(b+XF )

= t(c+ b+ a−BE − FC) =t

r(2rs− ta) =

t

r(ah− at).

Hence, t2 − ht+ 12hr = 0, which yields t = 1

2

(h−

√h(h− 2r)

). This suggests an easy construction:

Construct the length 2r on AD. Then construct the length AY =√h(h− 2r). Taking away the length

AY from AD gives us h−√h(h− 2r). Halving this yields t, as in the figure.

Also solved by Titu Zvonaru, Comănes,ti, Romania, and Neculai Stanciu, ”George Emil Palade” School,Buzău, Romania; Arkady Alt, San Jose, California, USA.


S293. Let a, b, c be distinct real numbers and let n be a positive integer. Find all nonzero complex numbersz such that

azn + bz +c

z= bzn + cz +

a

z= czn + az +

b

z.


Solution by Daniel Lasaosa, Pamplona, Navarra, SpainSince z 6= 0, we may multiply by z all terms in the proposed equation, or

azn+1 + b|z|2 + c = bzn+1 + c|z|2 + a = czn+1 + a|z|2 + b,

yielding(a− b)zn+1 + (b− c)|z|2 + (c− a) = (b− c)zn+1 + (c− a)|z|2 + (a− b) = 0,

and eliminating the terms with zn+1, we obtain

|z|2 =(a− b)2 − (b− c)(c− a)

(b− c)2 − (a− b)(c− a)=a2 + b2 + c2 − ab− bc− caa2 + b2 + c2 − ab− bc− ca

= 1,

since a2 + b2 + c2 > ab + bc + ca because of the Cauchy-Schwarz Inequality, a, b, c being distinct. Insertingthis result in the original equation and rearranging terms, we obtain

a(zn+1 − 1

)= b

(zn+1 − 1

)= c

(zn+1 − 1

),

or z must be one of the n+ 1-th roots of unity. For any one of those n+ 1 roots, we have z = 1z = zn, and

all n+ 1-th roots of unity are clearly solutions of the proposed equation.

Also solved by Arkady Alt, San Jose, California, USA; Moubinool Omarjee Lycée Henri IV, Paris, France;Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Albert Stadler, Herrliberg, Switzerland; LiZhou, Polk State College, FL, USA.


S294. Let s(n) be the sum of digits of n2 + 1. Define the sequence (an)n≥0 by an+1 = s(an), with a0 anarbitrary positive integer. Prove that there is n0 such that an+3 = an for all n ≥ n0.

Proposed by Roberto Bosch Cabrera, Havana, Cuba

Solution by Alessandro Ventullo, Milan, ItalyWe have to prove that the given sequence is 3-periodic. Since f(5) = 8, f(8) = 11 and f(11) = 5, it sufficesto prove that for every positive integer a0 there exists some n ∈ N such that an ∈ 5, 8, 11. Let m be thenumber of digits of a0. We prove the statement by induction on m. For m ≤ 2 we proceed by a direct check.If a0 ∈ 5, 8, 11 there is nothing to prove. If a0 is a two-digit number, then a20 ≤ 10000, so a1 ≤ 37 and wereduce to analyze the cases for a0 ≤ 37.

(i) If a0 ∈ 2, 7, 20, then a1 = 5. If a0 ∈ 1, 10, 26, 28, then a1 ∈ 2, 20, so a2 = 5. Finally, ifa0 ∈ 3, 6, 9, 12, 15, 18, 27, 30, 33, then a3 = 5.

(ii) If a0 ∈ 4, 13, 23, 32, then a1 = 8.

(iii) If a0 ∈ 17, 19, 21, 35, 37, then a1 = 11. If a0 ∈ 14, 22, 24, 31, 36, then a1 ∈ 17, 19, so a2 = 11.Finally, if a0 ∈ 16, 25, 29, 34, then a3 = 11.

Thus, we have proved that if a0 is a one or two-digit number, then an ∈ 5, 8, 11 for some n ∈ N, i.e. thesequence is 3-periodic. Let m ≥ 2 and suppose that the statement is true for all k ≤ m. Let a0 be an(m+ 1)-digit number. Then, 10m ≤ a0 < 10m+1 which implies 102m ≤ a20 < 102(m+1). Hence,

a1 = f(a0) ≤ 9 · 2(m+ 1) + 1 < 10m,

and by the induction hypothesis, the sequence (an)n≥1 is 3-periodic, which implies that the sequence (an)n≥0is 3-periodic, as we wanted to prove.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Arghya Datta, Hooghly Collegiate School,Kolkata, India; Li Zhou, Polk State College, FL, USA; AN-anduud Problem Solving Group, Ulaanbaatar,Mongolia.


Undergraduate problems

U289. Let a ≥ 1 be such that (banc)1n ∈ Z for all sufficiently large integers n. Prove that a ∈ Z.

Proposed by Mihai Piticari, Campulung Moldovenesc, Romania

Solution by Li Zhou, Polk State College, FL, USAFor the purpose of contradiction, suppose that a = m+h for some positive integerm and some real h ∈ (0, 1).Then for all n ≥ 1

h ,mn + 1 ≤ mn + nh < an < (m+ 1)n,

and thus mn < banc < (m + 1)n. Hence, m < (banc)1n < m + 1 for all sufficiently large n, a desired

contradiction.

Also solved by Arkady Alt, San Jose, California, USA; Daniel Lasaosa, Pamplona, Navarra, Spain;Alessandro Ventullo, Milan, Italy; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia.


U290. Prove that there are infinitely many triples of primes (pn−1, pn, pn+1) such that 12(pn+1 + pn−1) ≤ pn.

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by G.R.A.20 Problem Solving Group, Roma, ItalyAssume for the sake of contradiction that there is an integer n0 such that for all n > n0,

pn−1 + pn+1 > 2pn

that is, dn > dn−1 ≥ 1 where dn = pn+1 − pn. Hence, for k ≥ 1,

dn0+k−1 ≥ dn0+k−2 + 1 ≥ · · · ≥ dn0 + k − 1 ≥ k

and

pn0+k = dn0+k−1 + dn0+k−2 + · · ·+ dn0 + pn0 > k + (k − 1) + · · ·+ 1 =k(k + 1)

2>k2

2.

On the other hand, by the Prime Number Theorem, π(n) ∼ n/ ln(n) and since x/ ln(x) is strictly increasingfor x ≥ e, it follows that

1← π(pn0+k) ln(pn0+k)

pn0+k=

(n0 + k) ln(pn0+k)

pn0+k<

(n0 + k) ln(k2/2)

k2/2→ 0

which yields a contradiction.

Also solved by Julien Portier, Francois 1er, France; Daniel Lasaosa, Pamplona, Navarra, Spain; ArpanSadhukhan, Indian Statistical Institute, Kolkata; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;Li Zhou, Polk State College, FL, USA.


U291. Let f : R→ R be a bounded function and let S be the set of all increasing maps ϕ : R→ R. Prove thatthere is a unique function g in S satisfying the conditions

a) f(x) ≤ g(x) for all x ∈ R.b) If h ∈ S and f(x) ≤ h(x) for all x ∈ R then g(x) ≤ h(x) for all x ∈ R.

Proposed by Marius Cavachi, Constanta, Romania

Solution by Arkady Alt, San Jose, California, USAa) Since f is bounded then for any x ∈ R set G (x) := f (t) | t ∈ R and t ≤ x is bounded. Therefore forany x ∈ R we can define g (x) := supG (x) and, obviously, that function g (x) defined by such way satisfy tocondition (a).

b) Let now h ∈ S and f(x) ≤ h(x) for all x ∈ R.Since f(t) ≤ h(t) for any t ≤ x then g (x) = sup

t≤xf(t) ≤ sup

t≤xh(t) = h (x) (since h is increasing in t ∈ (−∞, x] ).

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Italy.


U292. Let r be a positive real number. Evaluate∫ π2

0

1

1 + cotr xdx.

Proposed by Ángel Plaza, Universidad de Las Palmas de Gran Canaria, Spain

Solution by Robinson Higuita, Universidad de Antioquia , ColombiaWe denote by

I(r) :=

∫ π2

0

1

1 + cotr(x)dx =

∫ π2

0

sinr x

sinr x+ cosr xdx.

If we make the substitution x = π2 − y, we have

I(r) =

∫ 0

π2

sinr(π2 − y)

sinr(π2 − y) + cosr(π2 − y)(−dy) =

∫ π2

0

cosr y

cosr y + sinr y.

Therefore

2I(r) =

∫ π2

0

sinr x

sinr x+ cosr xdx+

∫ π2

0

cosr y

cosr y + sinr y

=

∫ π2

0

sinr y + cosr y

cosr y + sinr ydx =

∫ π2

0dx =

π

2.

ThusI(r) =

π

4.

We note that the hypothesis on r is not important.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Albert Stadler, Herrliberg, Switzerland; SayakMukherjee, Kolkata, India; Samin Riasat, Dhaka, Bangladesh; Paolo Perfetti, Università degli studi di TorVergata Roma, Italy; Antoine Barré and Dmitry Chernyak, Lycée Stanislas, Paris, France; Arkady Alt, SanJose, California, USA; Moubinool Omarjee Lycée Henri IV, Paris, France; Li Zhou, Polk State College, FL,USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia.


U293. Let f : (0,∞) → R be a bounded continuous function and let α ∈ [0, 1). Suppose there exist realnumbers a0, ..., ak, whith k ≥ 2, so that

∑kp=0 ap = 0 and

limx→∞

xα

∣∣∣∣∣∣k∑p=0

apf(x+ p)

∣∣∣∣∣∣ = a.

Prove that a = 0.

Proposed by Marcel Chirita, Bucharest, Romania

Solution by the authorLet us denote M = sup|f(x)| and g(x) =

∑kp=0 apf(x+ p).

g : (0,∞)→ R is a continuous bounded function.Assume that a > 0.

Then there is an N > 0, such that xα|g(x)| > a, for ∀x > N (1)

From (1) we conclude that function g has the same sign ∀x > N , since it is continuous, and therefore hasthe intermediate value property.First, suppose that function g is positive on the interval (N,∞).

From (1) we have |g(x)| > a

xα(2)

Denoting n1 = [N ] + 1 and x = n1, n1 + 1, n1 + 2, ... and summing up yields to∑n≥n1

|g(k)| > a∑n≥n1

1

nα.

Since the series∑n≥1

1

nαis divergent for α ∈ (0, 1) it follows that

∑n≥n1

1

nαis divergent as well

⇒∑n≥n1

|g(k)| =∞.

Now, let T = max|a0|, |a1|, |a2|, ..., |ak|. Taking into consideration positiveness of g on (N,∞) and thesame values of n we get

Sn =∑n≥n1

|g(k)| =∑n≥n1

|apf(n+ p)| = |∑n≥n1

apf(n+ p)| =

|akf(n+ k) + (ak + ak−1)f(n+ k − 1) + ...+ (ak + ak−1 + ...+ a1)f(n+ 1) + (ak−1 + ak−2 + ...+ a0)f(n1 +k − 1) + (ak−2 + ak−3 + ...+ a0)f(n1 + k − 2) + ...+ (a1 + a0)f(n1 + 1) + a0f(n1)| ≤

|ak|M + |ak + ak−1|M + ...+ |ak + ak−1 + ...+ a1|M+

|ak−1 + ak−2 + ...+ a0|M + |ak−2 + ak−3 + ...+ a0|M + ...+ |a1 + a0|M + |a0|M ≤MT + 2MT + ...+ kMT +kMT + ...+ 2MT +MT = k(k + 1)MT and the series converges.

Therefore, we have reached a contradiction ⇒ a = 0. Similarly to prove for g(x) ≤ 0.


U294. Let p1, p2, . . . , pn be pairwise distinct prime numbers. Prove that

Q (√p1,√p2, . . . ,

√pn) = Q (

√p1 +

√p2 + · · ·+√pn) .

Proposed by Marius Cavachi, Constanta, Romania

Solution by Alessandro Ventullo, Milan, ItalyClearly, Q(

√p1+√p2+. . .+

√pn) is a subfield of Q(

√p1,√p2, . . . ,

√pn). Observe that Q(

√p1,√p2, . . . ,

√pn)

is Galois over Q, since it is the splitting field of the polynomial (x2−√p1) · · · (x2−√pn). Every automorphism

σ is completely determined by its action on √p1, . . . ,√pn, which must be mapped to ±√p1, . . . ,±

√pn,

respectively. Therefore, Gal(Q(√p1,√p2, . . . ,

√pn)/Q) is the group generated by σ1, σ2, . . . , σn, where σi

is the automorphism defined by

σi(√pj) =

−√pj if i = j√pj if i 6= j.

Clearly, the only automorphism that fixes Q(√p1,√p2, . . . ,

√pn) is the identity. Moreover, it’s easy to see

that the only automorphism that fixes the element √p1 + . . . +√pn is the identity, which means that the

only automorphism that fixes Q(√p1 + . . . +

√pn) is the identity. Hence, by the Fundamental Theorem of

Galois Theory, Q(√p1 +

√p2 + . . .+

√pn) = Q(

√p1,√p2, . . . ,

√pn).

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain.


Olympiad problems

O289. Let a, b, x, y be positive real numbers such that x2 − x+ 1 = a2, y2 + y + 1 = b2, and(2x− 1)(2y + 1) = 2ab+ 3. Prove that x+ y = ab.


Solution by Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, UzbekistanFrom the conditions (2x− 1)2 = 4a2 − 3 and (2y + 1)2 = 4b2 − 3 and

(2ab+ 3)2 = (2x− 1)2(2y + 1)2 = (4a2 − 3)(4b2 − 3),

a2b2 − ab− a2 − b2 = 0 (1).

We have (2x− 1)2 + (2y + 1)2 = 4(a2 + b2)− 6, and since (2x− 1)(2y + 1) = 2ab+ 3, we get that

(2(x+ y))2 = 4(a2 + b2)− 6 + (4ab+ 6),

(x+ y)2 = a2 + b2 + ab (2).

Since (1), (2), we get that (x+ y)2 = a2b2 and

x+ y = ab.

and we are done.

Also solved by Daniel Lasaosa, Pamplona, Navarra, Spain; Moubinool Omarjee Lycée Henri IV, Paris,France; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Daniel Văcaru, Piteşti,Romania; Alessandro Ventullo, Milan, Italy; Li Zhou, Polk State College, FL, USA; AN-anduud ProblemSolving Group, Ulaanbaatar, Mongolia; Arkady Alt, San Jose, California, USA; Ayoub Hafid, Elaraki School,Morroco.


O290. Let Ω1 and Ω2 be the two circles in the plane of triangle ABC. Let α1, α2 be the circles through A thatare tangent to both Ω1 and Ω2. Similarly, define β1, β2 for B and γ1, γ2 for C. Let A1 be the secondintersection of circles α1 and α2. Similarly, define B1 and C1. Prove that the lines AA1, BB1, CC1 areconcurrent.


Solution by Daniel Lasaosa, Pamplona, Navarra, SpainNote that we must assume that Ω1,Ω2 have different radii. Otherwise, the figure clearly has symmetryaround the perpendicular bisector of the segment joining their centers, and lines AA1, BB1, CC1 would beparallel to the line joining their centers (which could be considered equivalent to AA1, BB1, CC1 meeting atinfinity). We will assume in the rest of the problem that Ω1,Ω2 have distinct radii, hence a homothety withcenter O and scaling factor ρ with |ρ| 6= 1 can be found, which transforms Ω1 into Ω2.

Claim 1: Let Ω1,Ω2 be two circles with different radii, and let ω be a circle simultaneously tangent toboth, respectively at points T1, T2. Then, T1, T2 and the center O of the homothety that transforms Ω1 intoΩ2 are collinear.

Proof 1: Let O1, O2 be the respective centers, and R1, R2 the radii, of Ω1,Ω2. Let O′, r be the centerand radius of ω. Consider triangle O1O2O

′, where T1, T2 are clearly inside segments O1O′ and O2O

′. Let Pbe the second point where T1T2 intersects O1O2, where by Menelaus’ theorem, we have

O1P

PO2=T2O

′

O2T2· T1O1

O′T1=

r

R2· R1

r=R1

R2,

or indeed P is a center of a homothety that transforms O1 into O2 with scaling factor with absolute valueR1R2

, hence P = O. The Claim 1 follows.

Claim 2: The power of O with respect to ω is invariant for all the possible circles ω which are simulta-neously tangent to Ω1,Ω2.

Proof 2: Since T1, T2 are points on ω which are collinear with O, the power of O with respect to ω isOT1 ·OT2. Let P1, P2 be the respective powers of O with respect to Ω1,Ω2, which clearly satisfy P2 = ρ2P1.Consider now points T ′1, resulting from applying the homothety to T1. Clearly, T ′1 is collinear withO, T1, henceon line OT2; at the same time, T1 ∈ Ω1, or T ′1 ∈ Ω2 by construction, hence T ′1, T2 are the two points where aline through O intersects T2, or OT ′1 ·OT2 = P2. But since OT ′1 = |ρ|OT1, we have OT1 ·OT2 = P2

|ρ| = |ρ|P1,independently on the choice of ω. The Claim 2 follows.

By the Claim 2, the power of O with respect to α1, α2 is the same, or since AA1 is their radical axis, thenA,A1, O are collinear, or line AA1 passes through O. Similarly, BB1, CC1 pass through O. The conclusionfollows.

Also solved by Saturnino Campo Ruiz, Salamanca, Spain; Shohruh Ibragimov, Lyceum Nr.2 under theSamIES, Samarkand, Uzbekistan; Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, Uzbekistan;Alessandro Pacanowski, PECI, Rio de Janeiro, Brazil.


O291. Let a, b, c be positive real numbers. Prove that

a2√4a2 + ab+ 4b2

+b2√

4b2 + bc+ 4c2+

a2√4c2 + ca+ 4a2

≥ a+ b+ c

3.


First solution by Marius StaneanFrom Hölder’s Inequality, we have(∑

cyc

a2√4a2 + ab+ 4b2

)2(∑cyc

a2(4a2 + ab+ 4b2)

)≥ (a2 + b2 + c2)3. (1)

But ∑cyc

a2(4a2 + ab+ 4b2) = 4(a4 + b4 + c4) + a3b+ b3c+ c3a+ 4(a2b2 + b2c2 + c2a2)

= 4(a2 + b2 + c2)2 + a3b+ b3c+ c3a− 4(a2b2 + b2c2 + c2a2)

≤ 13(a2 + b2 + c2)2 − 12(a2b2 + b2c2 + c2a2)

4,

where the last line follows from Cartoaje’s Inequality,

(a2 + b2 + c2)2 ≥ 3(a3b+ b3c+ c3a).

Hence, considering (1), it follows that it is sufficient to prove that

27(a2 + b2 + c2)3 ≥ [13(a2 + b2 + c2)2 − 12(a2b2 + b2c2 + c2a2)](a+ b+ c)2

9(a2 + b2 + c2)2[3(a2 + b2 + c2)− (a+ b+ c)2]− 4(a+ b+ c)2[(a2 + b2 + c2)2 − 3(a2b2 + b2c2 + c2a2)] ≥ 0

9(a2 + b2 + c2)2[(a− b)2 + (a− c)(b− c)]− 2(a+ b+ c)2[(a+ b)2(a− b)2 + (a+ c)(b+ c)(a− c)(b− c)] ≥ 0

[9(a2 + b2 + c2)2 − 2(a+ b+ c)2(a+ b)2](a− b)2 + [9(a2 + b2 + c2)2 − 2(a+ b+ c)2(a+ c)(b+ c)](a− c)(b− c) ≥ 0.

Without loss of generality, we may assume a ≤ b ≤ c. Then it suffices to show that

9(a2 + b2 + c2)2 ≥ 2(a+ b+ c)2(a+ c)(b+ c),

but this is true because3(a2 + b2 + c2) ≥ (a+ b+ c)2

by Cauchy-Schwarz and

3(a2 + b2 + c2)− 2(a+ c)(b+ c) ≥ 0

⇐⇒ c2 − 2(a+ b)c+ 3(a2 + b2)− 2ab ≥ 0

⇐⇒ (c− a− b)2 + 2(a− b)2 ≥ 0,

so we are done.


Second solution by the authorWe will begin by finding x, y such that

a2√4a2 + ab+ 4b2

≥ xa+ yb

for all a, b > 0. Letting a = b = 1, the ”best” such x, y will satisfy x + y = 13 . Note that the inequality is

homogeneous, so letting t = ab , we have

t2√4t2 + t+ 4

≥ xt+ y = x(t− 1) +1

3. (1)

The inequality clearly holds if the RHS is negative. Otherwise, squaring and multiplying both sides by 9yields

9t4

4t2 + t+ 4≥ [3x(t− 1) + 1]2

9t4 − (4t2 + t+ 4)[3x(t− 1) + 1]2 ≥ 0.

Considering the LHS as a function of t, say f(t), we want it to have a double root at t = 1. This means

f ′(1) = 27− 54x = 0,

so x = 12 and y = −1

6 . This implies

f(t) =1

4(15t− 4)(t− 1)2.

Clearly, f is positive for t > 415 . For t ≤

415 , (1) becomes

3t2√4t2 + 4t+ 4

≥ 3t− 1

2,

which must be true since the RHS is negative. Thus,

a2√4a2 + ab+ 4b2

≥ 3a− b6

for all a, b > 0. Adding up the similar inequalities with a and c and b and c yields∑cyc

a2√4a2 + ab+ 4b2

≥∑cyc

3a− b6

=a+ b+ c

3,

as desired.

Also solved by Arkady Alt, San Jose, California, USA; Paolo Perfetti, Università degli studi di TorVergata Roma, Italy; Batzolis, Mandoulides High School, Thessaloniki, Greece; Nicuşor Zlota, “Traian Vuia”Technical College, Focşani, Romania.


O292. For each positive integer k let

Tk =

k∑j=1

1

j2j.

Find all prime numbers p for which

p−2∑k=1

Tkk + 1

≡ 0 (mod p).

Proposed by Gabriel Dospinescu, Ecole Normale Superieure, Lyon

Solution by G.R.A.20 Problem Solving Group, Roma, ItalyIt is known that the following identity holds

∑1≤j≤k≤n

(n

k

)(−1)k(1− x)j

jk=

n∑k=1

xk

k2−

n∑k=1

1

k2.

Let p > 2 be a prime (for p = 2 the congruence trivially holds).Then,

(p−1k

)≡ (−1)k (mod p) and the above identity imply

p−2∑k=1

Tkk + 1

=

p−2∑j=1

1

j2j

p−2∑k=j

1

k + 1=

∑1≤j<k≤p−1

(1/2)j

jk=

∑1≤j≤k≤p−1

(1/2)j

jk−

p−1∑k=1

(1/2)k

k2

≡∑

1≤j≤k≤p−1

(p− 1

k

)(−1)k(1− 1/2)j

jk−

p−1∑k=1

(1/2)k

k2(mod p)

=

p−1∑k=1

(1/2)k

k2−

p−1∑k=1

1

k2−

p−1∑k=1

(1/2)k

k2= −

p−1∑k=1

1

k2≡

1 if p = 3,0 if p > 3. (mod p).


O293. Let x, y, z be positive real numbers and let t2 = xyzmax(x,y,z) . Prove that

4(x3 + y3 + z3 + xyz

)2 ≥ (x2 + y2 + z2 + t2)3.

Proposed by Nairi Sedrakyan, Yerevan, Armenia

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, ItalyThe inequality is implied by

4(x3 + y3 + z3 + xyz)2 ≥

(x2 + y2 + z2 +

(a+b+c)3

27a+b+c

3

)3

Now define

a+ b+ c = 3u, ab+ bc+ ca = 3v2, abc = w3

We have

a2 + b2 + c2 = 9u2 − 6v2, a3 + b3 + c3 = 27u3 − 27uv2 + 3w3

The inequality reads as

4(27u3 − 27uv2 + 4w3)2 ≥((9u2 − 6v2) + u2

)3that is

64w6 + 864w3(−uv2 + u3) + 1836u2v4 + 1916u6 − 4032u4v2 + 216v6 ≥ 0

This a convex parabola in w3 whose minimum has abscissa negative. Since w3 ≥ 0, itfollows that the parabola is nonnegative if and only it is nonnegative its value for w3 = 0.The theory states that, fixed the values of (u, v), the minimum of w3 occursfor a = 0 or a = b.

If a = 0 we have

479(b6 + c6) + 1156b3c3 ≥ 780(b4c2 + b2c4) + 150(c5b+ cb5)

which implied by

479(b6 + c6) + 1156b3c3 ≥ 930(c5b+ cb5)

This follows by AM-GM since

479

2b6 +

479

2c6 + 578b3c3 ≥ 3

3

√(479)2578

4b5c > 963b5c

and the same with (c, b) in place of (b, c).

If a = b we have

(254b4 + 1972b3c+ 525b2c2 + 658c3b+ 479c4)(b− c)2 ≥ 0

and the proof is complete.

Also solved by Ioan Viorel Codreanu, Satulung, Maramures, Romania; Daniel Lasaosa, Pamplona, Na-varra, Spain; Shohruh Ibragimov, Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Ángel Plaza,University of Las Palmas de Gran Canaria, Spain; Khakimboy Egamberganov, S.H.Sirojiddinov lyceum,Tashkent, Uzbekistan; Arkady Alt, San Jose, California, USA.


O294. Let ABC be a triangle with orthocenter H and let D,E, F be the feet of the altitudes from A, B andC. Let X,Y, Z be the reflections of D,E, F across EF , FD, and DE, respectively. Prove that thecircumcircles of triangles HAX,HBY,HCZ share a common point, other than H.


Solution by Daniel Lasaosa, Pamplona, Navarra, SpainClaim: Let ABC be any triangle, I its incenter, Ia its excenter opposite vertex A, and A′ the symmetric ofA with respect to side BC. Define similarly Ib, Ic and B′, C ′. Then, the circles through I, Ia, A′, throughI, Ib, B

′ and through I, Ic, C ′, pass through a second common point other than I.

Proof 1: In exact trilinear coordinates, I ≡ (r, r, r), Ia ≡ (−ra, ra, ra) and A′ ≡ (−ha, 2ha cosC, 2ha cosB),or using non-exact trilinear coordinates, we have

I ≡ (1, 1, 1), Ia ≡ (−1, 1, 1), A′ ≡ (−1, 2 cosC, 2 cosB).

The equation of a circle in trilinear coordinates is given by

(`α+mβ + nγ)(aα+ bβ + cγ) + k(aβγ + bγα+ cαβ) = 0,

where substitution of the coordinates of the three given points yields k = −(`+m+ n) when applied to theincenter, and consequently m+ n = 0 and k = −` when applied to the excenter, yielding finally

m = −n = `1− 2 cosA

2(cosC − cosB)

when applied to A′. Indeed, the circle equation(α+

(1− 2 cosA)(β − γ)

2(cosC − cosB)

)=aβγ + bγα+ cαβ

aα+ bβ + cγ

is easily checked to be satisfied by I, Ia, A′. Analogous equations may be found for the circles through I, Ib, B′

and I, Ic, C ′. The intersection of the circles through I, Ib, B′ and I, Ic, C ′ must clearly satisfy

β +(1− 2 cosB)(γ − α)

2(cosA− cosC)= γ +

(1− 2 cosC)(α− β)

2(cosB − cosA),

or1 + 2 cosA− 2 cosB − 2 cosC

(cosA− cosB)(cosA− cosC)α+

1 + 2 cosB − 2 cosC − 2 cosA

(cosB − cosA)(cosB − cosC))β+

+1 + 2 cosC − 2 cosB − 2 cosA

(cosC − cosA)(cosC − cosB)γ = 0.

This line equation represents the radical axis of these two circles, and is clearly satified by I, and since itis invariant under cyclic permutation of A,B,C and α, β, γ, it is also therefore the equation of the radicalaxes of the other two pairs of circles, which intersect the three circles at I, and at another point. The Claimfollows.

Consider triangle DEF . Clearly, X,Y, Z are the reflections of its vertices with respect to its sides, H isits incenter, and A,B,C its excenters. The conclusion follows by direct application of the Claim.

Also solved by Khakimboy Egamberganov, S.H.Sirojiddinov lyceum, Tashkent, Uzbekistan; SebastianoMosca, Pescara, Italy


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