Junior problems - AwesomeMath · J430. IntriangleABC,∠C>90 and3a+ 15ab+5b=7c. Provethat∠C≤120 . ProposedbyTituAndreescu,UniversityofTexasatDallas,USA SolutionbyPolyahedra,PolkStateCollege,USA

Junior problems

J427. Find all complex numbers x, y, z which satisfy simultaneously the equations:

x + y + z = 1, x3 + y3 + z3 = 1, x2 + 2yz = 4.

Proposed by Mircea Becheanu, University of Bucharest, Romania

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italyx3 + y3 + (1 − x − y)3 = 1 ⇐⇒ (x + y)(x + y − 1 − xy) = 0 (1)

If y = −x we have z = 1 and from the third equation we get

x2 − 2x − 4 = 0 ⇐⇒ x = 1 ± i√

5

so here the solutions are

(x, y, z) = (1 − i√

5,−1 + i√

5,1), (1 + i√

5,−1 − i√

5,1)

From (1) we have alsox + y + 1 − xy = 0 ⇐⇒ x(1 − y) = 1 − y (2)

If y = 1 we get x = −z and from the third equation we have

x2 − 2x = 4 ⇐⇒ x = 1 ± i√

5

so the solutions we get are

(x, y, z) = (1 + i√

5,1,−1 − i√

5), (1 − i√

5,1,−1 + i√

5)

If in (2) we take x = 1 we get y = −z and from the third equation

1 − 2y2 = 4 ⇐⇒ y = ±i√

3/2

so the solutions are

(x, y, z) =⎛⎝

1, i

√3

2,−i

√3

2

⎞⎠,

⎛⎝(1,−i

√3

2, i

√3

2

⎞⎠

Also solved by Daniel Lasaosa, Pamplona, Spain; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina;Arkady Alt, San Jose, CA, USA; Kelvin Kim, St. George’s School, RI, USA; Konstantinos Kritharidis,Evangelliki Model School of Smyrna, Athens, Greece; Nicus,or Zlota‚ Traian Vuia Technical College, Focs,ani,Romania; Oana Prajitura, University of Pittsburgh, PA, USA; Naïm Mégarbané, UPMC, Paris, France;Soohyun Ahn, Middlesex School, Concord, MA, USA; Akash Singha Roy, Kolkata, India; Joehyun Kim,Fort Lee High School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Polyahedra, Polk StateCollege, FL, USA; Joonsoo Lee, Dwight Englewood School, NJ, USA; Yejin Kim, The Taft School, Watertown,CT, USA; Adrienne Ko, Fieldston School, New York, NY, USA; Celine Lee, Chinese International School,Hong Kong; Bekhzod Kurbonboev, NUUz, Tashkent, Uzbekistan; Luca Ferrigno, Universitá degli studi di TorVergata, Roma, Italy; Nikos Kalapodis, Patras, Greece; Seo Yeong Kwag, Academy, Blairstown, NJ, USA;Pedro Acosta De Leon, Massachusetts Institute of Technology, Cambridge, MA, USA; P.V.Swaminathan,Smart Minds Academy, Chennai, India; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti,Romania.

Mathematical Reflections 6 (2017) 1

J428. Solve the equation2x[x] + 2{x} = 2017,

where [a] denotes the greatest integer not greater than a and {a} is the fractional part of a.

Proposed by Adrian Andreescu, Dallas, Texas

Solution by Oana Prajitura, University of Pittsburgh, PA, USA2x[x] + 2{x} = 2017 ⇐⇒ 2([x] + {x})[x] + 2{x} = 2017

⇐⇒ 2[x]2 + 2[x]{x} + 2{x} = 2017 ⇐⇒ 2{x}([x] + 1) = 2017 − 2[x]2

If [x] = −1 the equation becomes 0 = 2019, a contradiction. Thus [x] ≠ 1 and so

{x} = 2017 − 2[x]22([x] + 1)

which implies that

0 ≤ 2017 − 2[x]22([x] + 1) < 1.

If [x] ≥ 0,

0 ≤ 2017 − 2[x]22([x] + 1) ⇐⇒ 2017 − 2[x]2 ≥ 0 ⇐⇒ [x]2 ≤ 1008 ⇐⇒ 0 ≤ [x] ≤ 31.

and2017 − 2[x]22([x] + 1) < 1 ⇐⇒ 2017 − 2[x]2 < 2[x] + 2 ⇐⇒ 2[x]2 + 2[x] − 2015 > 0

⇐⇒ [x] > −1 +√

4031

2⇐⇒ [x] ≥ 32.

Thus, in this case there is no solution. If [x] ≤ −2,

0 ≤ 2017 − 2[x]22([x] + 1) ⇐⇒ 2017 − 2[x]2 ≤ 0 ⇐⇒ [x]2 ≥ 1009 ⇐⇒ [x] ≤ −32.

and2017 − 2[x]22([x] + 1) < 1 ⇐⇒ 2017 − 2[x]2 > 2[x] + 2 ⇐⇒ 2[x]2 + 2[x] − 2015 < 0

−1 −√

4031

2< x ≤ −2 ⇐⇒ −32 ≤ [x] ≤ −2

Thus [x] = −32 and{x} = .5,

which gives us x = −31.5.

Also solved by Daniel Lasaosa, Pamplona, Spain; Naïm Mégarbané, UPMC, Paris, France; SoohyunAhn, Middlesex School, Concord, MA, USA; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort LeeHigh School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Polyahedra, Polk State College,FL, USA; Joonsoo Lee, Dwight Englewood School, NJ, USA; Yejin Kim, The Taft School, Watertown, CT,USA; Adrienne Ko, Fieldston School, New York, NY, USA; Celine Lee, Chinese International School, HongKong; Bekhzod Kurbonboev, NUUz, Tashkent, Uzbekistan; Luca Ferrigno, Universitá degli studi di Tor Ver-gata, Roma, Italy; Seo Yeong Kwag, Academy, Blairstown, NJ, USA; Nermin Hodžic, Dobošnica, Bosnia andHerzegovina; Arkady Alt, San Jose, CA, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Ioan Viorel Codreanu, Satulung, Maramures,,Romania; Joel Schlosberg, Bayside, NY, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma,Rome, Italy; Paraskevi-Andrianna Maroutsou, Charters Sixth Form, Sunningdale, England, UK; Daniel Cor-tild; Chanyeol Paul Kim, Seoul International School, Seoul, South Korea; P.V.Swaminathan, Smart MindsAcademy, Chennai, India; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania.


J429. Let x, y be positive real numbers such that x + y ≤ 1. Prove that

(1 − 1

x3)(1 − 1

y3) ≥ 49.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution by AN-anduud Problem Solving Group, Ulaanbaatar, MongoliaUsing the given condition and AM-GM inequality we get

1 ≥ x + y ⇔ 1 ≥ (x + y)3 = x3 + y3 + 3xy(x + y) ⋅ 1≥ x3 + y3 + 3xy(x + y)4

≥ x3 + y3 + 3xy(2√xy)4

= x3 + y3 + 48x3y3,

1 ≥ x3 + y3 + 48x3y3 ⇔ (1 − x3)(1 − y3) ≥ 49x3y3.

and we are done. Given equation equality holds only when x = y = 12 .

Also solved by Daniel Lasaosa, Pamplona, Spain; Naïm Mégarbané, UPMC, Paris, France; Soohyun Ahn,Middlesex School, Concord, MA, USA; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort Lee HighSchool, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Polyahedra, Polk State College, FL,USA; Joonsoo Lee, Dwight Englewood School, NJ, USA; Yejin Kim, The Taft School, Watertown, CT, USA;Adrienne Ko, Fieldston School, New York, NY, USA; Celine Lee, Chinese International School, Hong Kong;Bekhzod Kurbonboev, NUUz, Tashkent, Uzbekistan; Nikos Kalapodis, Patras, Greece; Daniel Cortild; NerminHodžic, Dobošnica, Bosnia and Herzegovina; Arkady Alt, San Jose, CA, USA; Dionysios Adamopoulos, 3rdHigh School, Pyrgos, Greece; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; KonstantinosKritharidis, Evangelliki Model School of Smyrna, Athens, Greece; Kunihiko Chikaya, Tokyo, Japan; Nicus,orZlota‚ Traian Vuia Technical College, Focs,ani, Romania; Paolo Perfetti, Università degli studi di Tor VergataRoma, Rome, Italy; Fong Ho Leung, Hoi Ping Chamber of Commerce Secondary School, Hong Kong; MichaelTang, MN, USA; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania.


J430. In triangle ABC, ∠C > 90○ and 3a +√

15ab + 5b = 7c. Prove that ∠C ≤ 120○.

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

Solution by Polyahedra, Polk State College, USABy the Cauchy-Schwarz inequality,

a2 + b2 + ab = ( 9

49+ 15

49+ 25

49)(a2 + ab + b2) ≥ (3a

7+

√15ab

7+ 5b

7)2

= c2.

Thus cosC = a2+b2−c2

2ab ≥ −12 , so C ≤ 120○.

Also solved by Daniel Lasaosa, Pamplona, Spain; Naïm Mégarbané, UPMC, Paris, France; SoohyunAhn, Middlesex School, Concord, MA, USA; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort Lee HighSchool, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Yejin Kim, The Taft School, Watertown,CT, USA; Adrienne Ko, Fieldston School, New York, NY, USA; Celine Lee, Chinese International School,Hong Kong; Nikos Kalapodis, Patras, Greece; Seo Yeong Kwag, Academy, Blairstown, NJ, USA; NerminHodžic, Dobošnica, Bosnia and Herzegovina; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia;P.V.Swaminathan, Smart Minds Academy, Chennai, India; Albert Stadler, Herrliberg, Switzerland.


J431. Let a, b, c, d, e be real numbers in the interval [1,2]. Prove that

a2 + b2 + c2 + d2 + e2 − 3abcde ≤ 2.

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Albert Stadler, Herrliberg, SwitzerlandWe need to prove that

2 + 3(a + 1)(b + 1)(c + 1)(d + 1)(e + 1) − (a + 1)2 − (b + 1)2 − (c + 1)2 − (d + 1)2 − (e + 1)2 ≥ 0

if 0 ≤ a, b, c, d, e ≤ 1. Indeed,

2 + 3(a + 1)(b + 1)(c + 1)(d + 1)(e + 1) − (a + 1)2 − (b + 1)2 − (c + 1)2 − (d + 1)2 − (e + 1)2 ≥

≥ 3(a + b + c + d + e) − a2 − b2 − c2 − d2 − e2 − 2(a + b + c + d + e) =

= a(1 − a) + b(1 − b) + c(1 − c) + d(1 − d) + e(1 − e) ≥ 0.

Also solved by Daniel Lasaosa, Pamplona, Spain; Naïm Mégarbané, UPMC, Paris, France; Soohyun Ahn,Middlesex School, Concord, MA, USA; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort Lee HighSchool, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Polyahedra, Polk State College, USA;Yejin Kim, The Taft School, Watertown, CT, USA; Adrienne Ko, Fieldston School, New York, NY, USA;Nikos Kalapodis, Patras, Greece; Chanyeol Paul Kim, Seoul International School, Seoul, South Korea; FongHo Leung, Hoi Ping Chamber of Commerce Secondary School, Hong Kong; Michael Tang, MN, USA; NerminHodžic, Dobošnica, Bosnia and Herzegovina; Arkady Alt, San Jose, CA, USA; Paolo Perfetti, Università deglistudi di Tor Vergata Roma, Rome, Italy; Bekhzod Kurbonboev, NUUz, Tashkent, Uzbekistan; Joonsoo Lee,Dwight Englewood School, NJ, USA; Titu Zvonaru, Comănes,ti, Romania.


J432. Let m and n be integers greater than 1. Prove that

(m3 − 1) (n3 − 1) ≥ 3m2n2 + 1.


Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, SpainConsider the substitution m = 1 + x, and n = 1 + y. After some algebra the proposed inequality becomes

x3y3 + 3x3y2 + 3x3y + 3x2y3 + 6x2y2 + 3x2y − 3x2 + 3xy3 + 3xy2 − 3xy − 6x − 3y2 − 6y − 4 ≥ 0

which it is equivalent to

(xy − 1) (x2y2 + 3x2y + 3x2 + 3xy2 + 7xy + 6x + 3y2 + 6y + 4) ≥ 0

which it is true, since x, y > 0 are integers.

Also solved by Daniel Lasaosa, Pamplona, Spain; Naïm Mégarbané, UPMC, Paris, France; SoohyunAhn, Middlesex School, Concord, MA, USA; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort LeeHigh School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Polyahedra, Polk State College,FL, USA; Yejin Kim, The Taft School, Watertown, CT, USA; Adrienne Ko, Fieldston School, New York,NY, USA; Celine Lee, Chinese International School, Hong Kong; Seo Yeong Kwag, Academy, Blairstown, NJ,USA; Pedro Acosta De Leon, Massachusetts Institute of Technology, Cambridge, MA, USA; Chanyeol PaulKim, Seoul International School, Seoul, South Korea; Bekhzod Kurbonboev, NUUz, Tashkent, Uzbekistan;Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Dionysios Adamopoulos, 3rd High School, Pyrgos,Greece; Nikos Kalapodis, Patras, Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; IoanViorel Codreanu, Satulung, Maramures,, Romania; Duy Quan Tran, University of Medicine and Pharmacy,Ho Chi Minh, Vietnam; Joel Schlosberg, Bayside, NY, USA; Nguyen Ngoc Tu, Ha Giang, Vietnam; Nicus,orZlota‚ Traian Vuia Technical College, Focs,ani, Romania; Paolo Perfetti, Università degli studi di Tor VergataRoma, Rome, Italy; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania.


Senior problems

S427. Solve in complex numbers the system of equations:

z + 2017

w= 4 − i,

w + 2018

z= 4 + i.


Solution by Joseph Currier, SUNY Binghamton, NY, USASolving for z in the first equation yields z = 4− i − 2017

w and substituting that value into the second equationyields w + 2018

w(4−i)−2017 = 4 + i. Clearing fractions gives a quadratic equation in terms of w:

w2(4 − i) − 16w + 2017(4 + i) = 0

The solutions to the above equation are w = −9 + 44i and w = 21717 − 732

17 i. Hence the solutions to the systemof equations are

(w, z) = (−9 + 44i,13 + 43i)

and(w, z) = (217

17− 732

17i,−149

17− 749

17i)

Also solved by Albert Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Spain; Akash SinghaRoy, Kolkata, India; Joehyun Kim, Fort Lee High School, NJ, USA; Erica Choi, Blair Academy, Blairstown,NJ, USA; Yejin Kim, The Taft School, Watertown, CT, USA; Celine Lee, Chinese International School,Hong Kong; Seo Yeong Kwag, Academy, Blairstown, NJ, USA; Pedro Acosta De Leon, Massachusetts In-stitute of Technology, Cambridge, MA, USA; P.V.Swaminathan, Smart Minds Academy, Chennai, India;Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Ángel Plaza, University of Las Palmas de Gran Cana-ria, Spain; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Xingze Xu, HangzhouForeign Languages School A-level Centre, China; Arkady Alt, San Jose, CA, USA; Titu Zvonaru, Comănes,ti,Romania.


S428. Let a, b, c be nonnegative real numbers, not all zero, such that ab + bc + ca = a + b + c. Prove that

1

1 + a +1

1 + b +1

1 + c ≤5

3.

Proposed by An Zenping, Xianyang Normal University, China

Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia1

1 + a +1

1 + b +1

1 + c ≤5

3⇔ a + b + c + 5abc ≥ 4 (1)

Using Schur’s inequality, we get

a2 + b2 + c2 + 9abc

a + b + c ≥ 2(ab + bc + ca).

Deduce that(a + b + c)2 + 9abc

a + b + c ≥ 4(ab + bc + ca) (2)

using a + b + c = ab + bc + ca identities, we get

(2) ⇔ a + b + c + 9abc

(a + b + c)2 ≥ 4 (3)

Other hand, we have

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)≥ (ab + bc + ca) + 2(ab + bc + ca)= 3(ab + bc + ca).

and a + b + c = ab + bc + ca, hence we have a + b + c ≥ 3.Thus we get

(a + b + c)2 ≥ +9 ⇔ 9

(a + b + c)2 ≤ 1 (4)

From (3) and (4), we geta + b + c + abc ≥ 4.

Usinga + b + c + 5abc ≥ a + b + c + abc

we have a + b + c + 5abc ≥ 4. Equality holds only when, {a, b, c} = {2,2,0}.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim,Fort Lee High School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Yejin Kim, The TaftSchool, Watertown, CT, USA; Albert Stadler, Herrliberg, Switzerland; Nermin Hodžic, Dobošnica, Bosniaand Herzegovina; Arkady Alt, San Jose, CA, USA; Nicus,or Zlota‚ Traian Vuia Technical College, Focs,ani,Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Joonsoo Lee, DwightEnglewood School, NJ, USA; Titu Zvonaru, Comănes,ti, Romania.


S429. Let ABC be a triangle and let M be a point in its plane. Prove that for all positive real numbersx, y, z the following inequality holds

xMA2 + yMB2 + zMC2 > yz

2(y + z)a2 + zx

2(z + x)b2 + xy

2(x + y)c2


Solution by Nermin Hodžic, Dobošnica, Bosnia and HerzegovinaUsing Triangle inequality we have

MB +MC ≥ BC = a (1)

Using Cauchy-Shwarz inequality we have

(1

y+ 1

z) (yMB2 + zMC2) ≥ (MB +MC)2 ≥(1) a2 ⇒

yMB2 + zMC2 ≥ (MB +MC)2 ≥ yz

y + z a2

Similarly we obtain

zMC2 + xMA2 ≥ (MC +MA)2 ≥ zx

z + xb2

xMA2 + yMB2 ≥ (MA +MB)2 ≥ xy

x + y c2

Adding these three inequalities and dividing by 2 we obtain

xMA2 + yMB2 + zMC2 ≥ yz

2(y + z)a2 + zx

2(z + x)b2 + xy

2(x + y)c2

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim,Fort Lee High School, NJ, USA; Adrienne Ko, Fieldston School, New York, NY, USA; Celine Lee, ChineseInternational School, Hong Kong; Albert Stadler, Herrliberg, Switzerland; Kevin Soto Palacios, Huarmey,Perú; Titu Zvonaru, Comănes,ti, Romania.


S430. Prove thatsin

π

2n≥ 1

n,

for all positive integers n.

Proposed by Florin Rotaru, Focşani, Romania

Solution by Arkady Alt, San Jose, CA, USA

Note that f (x) ∶= sinx

xis decreasing function on (0, π/2].

Indeed, f ′ (x) = x cosx − sinx

x2< 0 for any x ∈ (0, π/2].Hence, f (x) ≥ f (π/2) =

sinπ

2π

2

⇐⇒

sinπ

2nπ

2n

≥ 1π

2

⇐⇒ sinπ

2n≥ π

2n⋅ 2

π= 1

n

for any x ∈ (0, π/2].

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim,Fort Lee High School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Yejin Kim, The TaftSchool, Watertown, CT, USA; Adrienne Ko, Fieldston School, New York, NY, USA; Bekhzod Kurbonboev,NUUz, Tashkent, Uzbekistan; Pedro Acosta De Leon, Massachusetts Institute of Technology, Cambridge, MA,USA; Chanyeol Paul Kim, Seoul International School, Seoul, South Korea; P.V.Swaminathan, Smart MindsAcademy, Chennai, India; Albert Stadler, Herrliberg, Switzerland; Nermin Hodžic, Dobošnica, Bosnia andHerzegovina; Dionysios Adamopoulos, 3rd High School, Pyrgos, Greece; AN-anduud Problem Solving Group,Ulaanbaatar, Mongolia; Joel Schlosberg, Bayside, NY, USA; José Hernández Santiago, México; Nicus,orZlota‚ Traian Vuia Technical College, Focs,ani, Romania; Paolo Perfetti, Università degli studi di Tor VergataRoma, Rome, Italy; Titu Zvonaru, Comănes,ti, Romania.


S431. Let a, b, c be positive numbers such that ab + bc + ca = 3. Prove that

1

(1 + a)2 +1

(1 + b)2 +1

(1 + c)2 ≥ 3

4.

Proposed by Konstantinos Metaxas, Athens, Greece

Solution by Daniel Lasaosa, Pamplona, SpainDefine s = a+b+c and p = abc, or after multiplying both sides of the proposed equation by 4(1+a)2(1+b)2(1+c)2and rearranging terms, results in the equivalent inequality

0 ≤ 5s2 + 16s − 14ps − 3p2 − 48p = (5s + p − 16)(s − 3p).

Now, by the AM-GM inequality we have ab + bc + ca ≥ 33√a2b2c2 and a + b + c ≥ 3

3√abc, and using that

ab + bc + ca = 3, we have3s = (ab + bc + ca)(a + b + c) ≥ 9abc,

or s ≥ 3p with equality iff a = b = c = 1. It then suffices to show that 5s + p ≥ 16, where by the scalar productinequality we have s2 − 6 = a2 + b2 + c2 ≥ 3, or s ≥ 3, and by previous results we have 3

√a2b2c2 ≤ 1, or abc ≤ 1.

Now, denoting s = 3 +∆ for some nonnegative real ∆, note that

a3 + b3 + c3 − 3abc = (a + b + c) (a2 + b2 + c2 − ab − bc − ca) =

= s (s2 − 3(ab + bc + ca)) = 18∆ + 9∆2 +∆3,

and by Schur’s inequality,

0 ≤ a(a − b)(a − c) + b(b − c)(b − a) + c(c − a)(c − b) =

= a3 + b3 + c3 + 6abc − (a + b + c)(ab + bc + ca),

or3p = a3 + b3 + c3 − 18∆ − 9∆2 −∆3 ≥ 3s − 6p − 18∆ − 9∆2 −∆3,

and finally

p ≥ 1 − 15

9∆ −∆2 − ∆3

9.

Now, if ∆ ≤ 1, then p ≥ 1−∆ (159 + 1 + 1

9) ≥ 1− 3∆, and 5s+ p ≥ 16+ 2∆ ≥ 16, with equality iff ∆ = 0, whereas

if ∆ > 1, then s > 4 and 5s + p ≥ 5s > 20 > 16, and the inequality holds strictly. The conclusion follows,equality holds iff ∆ = 0, or iff a = b = c = 1.

Also solved by Joehyun Kim, Fort Lee High School, NJ, USA; Akash Singha Roy, Kolkata, India; Ye-jin Kim, The Taft School, Watertown, CT, USA; Adrienne Ko, Fieldston School, New York, NY, USA;Celine Lee, Chinese International School, Hong Kong; Nikos Kalapodis, Patras, Greece; Seo Yeong Kwag,Academy, Blairstown, NJ, USA; Albert Stadler, Herrliberg, Switzerland; Nermin Hodžic, Dobošnica, Bo-snia and Herzegovina; Arkady Alt, San Jose, CA, USA; AN-anduud Problem Solving Group, Ulaanbaatar,Mongolia; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Ioan Viorel Codreanu, Satulung,Maramures,, Romania; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Nicus,or Zlota‚ TraianVuia Technical College, Focs,ani, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Ro-me, Italy; Paraskevi-Andrianna Maroutsou, Charters Sixth Form, Sunningdale, England, UK; Giurgi Vasile,C.N. Dragos-Voda, Sighetu Marmatiei, Romania; Titu Zvonaru, Comănes,ti, Romania.


S432. Let d be an open half-disk of diameter AB and h be the half-plane defined by the line AB and containingd. Let X be a point on d and let Y and Z be points in h on the semicircles of diameters AX and BX,respectively. Prove that

AY ⋅BZ +XY ⋅XZ ≤ AX2 −AX ⋅BX +BX2.


Solution by Arkady Alt, San Jose, CA, USALet a ∶= AX, b ∶= BX,y ∶= AY, z ∶= BZ,u ∶=XY, v ∶=XZ.

Then inequality becomesyz + uv ≤ a2 + b2 − ab.

Since y2 + u2 = a2 and z2 + v2 = b2 (because ∠AYX =∠BZX = 90○) then by Cauchy Inequality

ab =√y2 + u2 ⋅

√z2 + v2 ≥ yz + uv

and alsoa2 + b2 − ab ≥ ab.

Hence, a2 + b2 − ab ≥ ab ≥ yz + uv.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim,Fort Lee High School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Adrienne Ko, FieldstonSchool, New York, NY, USA; Celine Lee, Chinese International School, Hong Kong; Seo Yeong Kwag, Aca-demy, Blairstown, NJ, USA; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Titu Zvonaru, Comănes,ti,Romania.


Undergraduate problems

U427. Let f ∶ R2 → R be the function defined by

f(x, y) = 1(0, 1

y)(x) ⋅ 1(0,1)(y) ⋅ y,

where 1 is the characteristic function. Evaluate

∫R2f(x, y)dxdy.

Proposed by Alessandro Ventulo, Milan, Italy

Solution by Henry Ricardo, Westchester Area Math Circle, NY, USAWe have

f(x, y) = 1(0,1/y)(x) ⋅ 1(0,1)(y) ⋅ y = { 1 if y ∈ (0,1) and 0 < x < 1/y0 if y ∉ (0,1) or x ∉ (0,1/y) } ⋅ y.

Therefore

∫R2f(x, y) dx dy = ∫

1

0∫

1/y

0y dx dy = ∫

1

0y (∫

1/y

01 dx)dy = ∫

1

0y ⋅ 1

ydy = 1.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim, FortLee High School, NJ, USA; Yejin Kim, The Taft School, Watertown, CT, USA; Albert Stadler, Herrliberg,Switzerland; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Paolo Perfetti, Università degli studi diTor Vergata Roma, Rome, Italy; Naïm Mégarbané, UPMC, Paris, France.


U428. Let a, b, c positive real numbers such that a + b + c = 1. Prove that

(1 + a2b2)c (1 + b2c2)a (1 + c2a2)b ≥ 1 + 9a2b2c2.


Solution by Daniel Lasaosa, Pamplona, SpainTaking natural logarithms on both sides, the proposed inequality is equivalent to

a ln (1 + b2c2) + b ln (1 + c2a2) + c ln (a2 + b2) ≥ ln (1 + 9a2b2c2) .

Consider now function f(x) = ln (1 + x2), whose first and second derivatives are respectively

f ′(x) = 2x

1 + x2 , f ′′(x) =2 (1 − x2)

1 + x2 .

Now, since a+b+c = 1 and a, b, c are positive, we have a, b, c < 1, or ab, bc, ca < 1, and function f(x) is strictlyconvex in an interval which contains ab, bc, ca. It follows from Jensen’s inequality that

a ln (1 + b2c2) + b ln (1 + c2a2) + c ln (a2 + b2) = af(bc) + bf(ca) + cf(ab) ≥

≥ (a + b + c)f ( 3abc

a + b + c) = f(3abc) = ln (1 + 9a2b2c2) ,

where we have used that a+ b+ c = 1, and where equality holds iff ab = bc = ca, ie iff a = b = c. The conclusionfollows, equality holds iff a = b = c = 1

3 .

Also solved by Albert Stadler, Herrliberg, Switzerland; Yejin Kim, The Taft School, Watertown, CT,USA; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort Lee High School, NJ, USA; Nermin Hodžic,Dobošnica, Bosnia and Herzegovina; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; PaoloPerfetti, Università degli studi di Tor Vergata Roma, Rome, Italy.


U429. Let n ≥ 2 be an integer and let A be an n × n real matrix in which exactly (n − 1)2 entries are zero.Prove that if B is an n×n matrix with all entries nonzero numbers, then BA can not be a nonsingulardiagonal matrix.

Proposed by Alessandro Ventullo, Milan, Italy

Solution by Nermin Hodžic, Dobošnica, Bosnia and HerzegovinaLet A = (aij)n×n and B = (bij)n×n, the by the condition bij ≠ 0,∀i, j

Since there are n2 entries at A and (n − 1)2 of them are zero, there are

n2 − (n − 1)2 = 2n − 1

nonzero entries at A.If all the columns have at least two nonzero elements then the totalnumber of nonzero elements is at least 2n > 2n − 1 which is contradiction.Hence there is acolumn which contains at most one nonzero element, let it be Aj ,the j−th column of A.If all the entries of Aj are zero then all the entries of j−th column of the prodact BA arezeros, hence the matrix BA is singular.Suppose Aj has one nonzero entry, say aij .Then

the j−th column of the product BA is⎛⎜⎜⎜⎝

b1i ⋅ aijb2i ⋅ aij

⋮bni ⋅ aij

⎞⎟⎟⎟⎠.Since by the condition we have

n

∏k=1

bki ≠ 0

and the matrix is diagonal, so at least two of the b1i ⋅ aij , b2i ⋅ aij , ..., bni ⋅ aij are zero thenthe entry aij equals to zero, so rezulting product BA is a singular matrix.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Deep Ghoshal,Indian Statistical Institute, Kolkata, India.


U430. Let A and B be 3 × 3 matrices with complex numbers entries, such that

(AB −BA)2 = AB −BA.

Prove that AB = BA.

Proposed by Florin Stănescu, Găeşti, Romania

Solution by Albert Stadler, Herrliberg, SwitzerlandWe note that the trace of AB −BA equals tr(AB −BA) = tr(AB) − tr(BA) = 0.We will prove a slightly more general result: Let C be a 3× 3 matrix with complex number entries and trace0 such that C2 = C. Then C = 0.

If λ is an eigenvalue of C and x an associated eigenvector then C2x = λ2x = Cx = λx. So λ ∈ {0,1}.tr(C) = 0 equals the sum of the eigenvalues of C. Therefore λ = 0 is the only eigenvalue of C. C is similarto an upper triangular matrix whose diagonal entries are the eigenvalues of C. Hence there are complexnumbers r, s, t such that C is similar to

C ∼⎛⎜⎝

0 r t0 0 s0 0 0

⎞⎟⎠

Then

C2 ∼⎛⎜⎝

0 0 rs0 0 00 0 0

⎞⎟⎠

However C2 = C, so r = s = 0, t = rs = 0. We conclude that C = 0.

Also solved by Pop Ovidiu Florin, C.N. Dragos-Voda, Sighetu Marmatiei, Ramania; Daniel Lasaosa, Pam-plona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim, Fort Lee High School, NJ, USA; Yejin Kim,The Taft School, Watertown, CT, USA; Seo Yeong Kwag, Academy, Blairstown, NJ, USA; Fong Ho Leung,Hoi Ping Chamber of Commerce Secondary School, Hong Kong; Michael Tang, MN, USA; P.V.Swaminathan,Smart Minds Academy, Chennai, India; Naïm Mégarbané, UPMC, Paris, France; Nermin Hodžic, Dobošnica,Bosnia and Herzegovina; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Moubinool Omarjee,Lycée Henri IV, Paris, France.


U431. Evaluatelimt→0

1

t∫

t

0

√1 + exdx and lim

t→0

1

t∫

t

0ee

x

dx.


Solution by Henry Ricardo, Westchester Area Math Circle, NY, USASince limt→0 ∫ t

0

√1 + exdx = 0, L’Hospital’s Rule and the Fundamental Theorem of Calculus give us

limt→0

∫ t0

√1 + exdxt

= limt→0

√1 + et1

=√

2.

Similarly, we see that

limt→0

∫ t0 e

exdx

t= lim

t→0

eet

1= e.

Also solved by Daniel Lasaosa, Pamplona, Spain; Akash Singha Roy, Kolkata, India; Joehyun Kim,Fort Lee High School, NJ, USA; Erica Choi, Blair Academy, Blairstown, NJ, USA; Yejin Kim, The TaftSchool, Watertown, CT, USA; Bekhzod Kurbonboev, NUUz, Tashkent, Uzbekistan; Luca Ferrigno, Universitádegli studi di Tor Vergata, Roma, Italy; Albert Stadler, Herrliberg, Switzerland; Naïm Mégarbané, UPMC,Paris, France; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Arkady Alt, San Jose, CA, USA; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Ángel Plaza, University of Las Palmas de GranCanaria, Spain; Moubinool Omarjee, Lycée Henri IV, Paris, France; Narayanan P, Vivekananda College,Chennai, India; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Giurgi Vasile, C.N.Dragos-Voda, Sighetu Marmatiei, Romania; Xingze Xu, Hangzhou Foreign Languages School A-level Centre,China.


U432. For every point P (x, y, z) on the unit sphere, consider the points Q(y, z, x) and R(z, x, y). For everypoint A on the sphere, denote ∠(AOP ) = p, ∠(AOQ) = q and ∠(AOR) = r. Prove that

∣ cos q − cos r∣ ≤ 2√

3 sinp

2.


Solution by Daniel Lasaosa, Pamplona, SpainBy the Cosine Law, AQ2 = OA2 + OQ2 − 2OA ⋅ OQ cos q = 2 − 2 cos q, or cos q = 1 − AQ2

2 , and similarlycos r = 1 − AR2

2 . Moreover, since OP = OA = 1, triangle AOP is isosceles at O, and sin p2 = OA sin p

2 = AP2 .

The proposed inequality rewrites asAR2 −AQ2 ≤ 2

√3AP.

Note now that, under this form, we may perform a variable change since the problem has been stated interms of distances. Now, in order to perform a correct variable change, note that

PQ2 = QR2 = RP 2 = (x − y)2 + (y − z)2 + (z − x)2,

or we may choose a coordinate system with origin at the center of the unit sphere, such that P ≡ (1,0,0),the equation y = 0 bisects segment QR, or Q ≡ (d, `, h) and R ≡ (d,−`, h), where 2` = PQ = QR = RP . Notefurther that d2 + `2 + h2 = 1 because Q,R are on the unit sphere, whereas

4`2 = PQ2 = RP 2 = (1 − d)2 + `2 + h2 = 2 − 2d, d = 1 − 2`2, h = `√

3 − 4`2.

Now, the expression found for h suggests that ` ≤√32 , which indeed holds, and equality means that the

circumcircle of PQR is an equator of the unit sphere, which is consistent with d = −12 and h = 0. Having

this in mind, note further that A is any point on this unit sphere, with coordinates A ≡ (u, v,w) such thatu2 + v2 +w2 = 1, or

AR2 −AQ2 = (v + `)2 − (v − `)2 = 4v`, AP 2 = (1 − u)2 + v2 +w2 = 2 − 2u.

The proposed inequality then rewrites as2v` ≤

√6(1 − u),

which holds trivially if v ≤ 0. If v ≥ 0, we may square both sides, resulting in the equivalent inequality4v2`2 ≤ 6(1 − u), where since 4`2 ≤ 3, it suffices to show that

v2 ≤ 2(1 − u) = 1 − 2u + u2 + v2 +w2 = (1 − u)2 +w2 + v2,

trivially true and with equality iff u = 1 and w = 0, and consequently v = 0. Note further that, regardless ofthe value of `, if u = 1 and v = w = 0, then A = P , resulting in AP = 0, and since PQ = RP , also in AQ = AR,or equality holds regardless of `. The conclusion follows, equality holds iff A = P , in which case p = 0 andq = r.

Also solved by Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; Akash Singha Roy, Kolkata, India.


Olympiad problems

O427. Let ABC be a triangle and ma,mb,mc be the lengths of its medians. Prove that√

3 (ama + bmb + cmc) ≤ 2s2.

Proposed by Dragoljub Miloševic̀, Gornji Milanovac, Serbia

First solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, ItalyThe inequality is

√3(a + b + c)∑

cyc

a

(a + b + c)

√2b2 + 2c2 − a2

4≤ (a + b + c)2

2

The concavity of√x yields

√3(a + b + c)∑

cyc

a

(a + b + c)

√2b2 + 2c2 − a2

4≤√

3(a + b + c)¿ÁÁÀ∑

cyc

a

(a + b + c)2b2 + 2c2 − a2

4

so it suffices to prove

√3(a + b + c)

¿ÁÁÀ∑

cyc

a

(a + b + c)2b2 + 2c2 − a2

4≤ (a + b + c)2

2

or squaring

3∑cyc

a(2b2 + 2c2 − a2) ≤ (a + b + c)3

Now let’s set a = y + z, b = x + z, c = x + y, x, y, z ≥ 0and the inequality becomes

∑sym

(x3 + 3x2y) ≥ 24xyz

and this is clearly AGM.


Second solution by Daniel Lasaosa, Pamplona, SpainNote first that ma

2 = b2+c2

2 − a2

4 = a2+b2+c2

2 − 3a2

4 . Define therefore

f(x) =√K − x2,

where x ≤√K. Note that for K = a2+b2+c2

2 , we have ma = f (√3a2 ), and similarly for b, c. Now, the first and

second derivatives of f(x) are

f ′(x) = − x√K − x2

, f ′′(x) = − K

(√K − x2)

3< 0,

or f is strictly concave, and by Jensen’s inequality,

√3 (ama + bmb + cmc) ≤

√3(a + b + c)f

⎛⎝

√3 (a2 + b2 + c2)2(a + b + c)

⎞⎠=√

12Ks2 − 9K2,

with equality iff a = b = c, or it suffices to show that√

12Ks2 − 9K2 ≤ 2s2, 0 ≤ 4s4 − 12Ks2 + 9K2 = (2s2 − 3K)2 ,

clearly true. Note that condition a = b = c, necessary for equality in Jensen’s inequality, also results in3K = 2s2 = 9a2

2 . The conclusion follows, equality holds iff ABC is equilateral.

Also solved by Albert Stadler, Herrliberg, Switzerland; Akash Singha Roy, Kolkata, India; Nermin Hodžic,Dobošnica, Bosnia and Herzegovina; Arkady Alt, San Jose, CA, USA; Nicus,or Zlota‚ Traian Vuia TechnicalCollege, Focs,ani, Romania; Scott H. Brown, Auburn University Montgomery, Montgomery, AL, USA.


O428. Determine all positive integers n for which the equation

x2 + y2 = n(x − y)

is solvable in positive integers. Solve the equation

x2 + y2 = 2017(x − y).

Proposed by Dorin Andrica, Cluj-Napoca, Romania and Vlad Crişan, Göttingen, Germany

Solution by Ricardo Largaespada, Universidad Nacional de Ingeniería, Managua, Nicaragua

x2 + y2 = n(x − y)⇔ 4x2 − 4nx + n2 + 4y2 + 4ny + n2 = 2n2

Then if u = 2x − n and v = 2y + n, the original equation is equivalent to solve the Pythagorean equation:u2 + v2 = 2n2.

If gcd(u,2) = 2, then u = 2u1 and 4u21 + v2 = 2n2 implies that v = 2v1 ⇒ 2(u21 + v21) = n2 ⇒ n = 2n1 ⇒u21 + v21 = 2n21, and this equation is equivalent to the original. So we can assume that u and v are odds.

Now consider Z[i], let z = u + vi, then zz = N(z) = u2 + v2 = 2n2 = (1 + i)(1 − i)n2. By unique facto-rization in Z[i] implies that 1 + i∣z or 1 − i∣z. If 1 + i∣z, then w = z

1+i =z(1−i)

2 = u+v2 + v−u

2 i. Then we have(u+v

2)2 + (v−u

2)2 = N(w) = n2.

Then n2 is the sum of two squares because u+v2 and ∣v−u∣

2 are integers, then solving this equation inpositive integers we get by the Pythagorean equation that n = a2 + b2 where a > b. We can also get u+v

2 and∣v−u∣2 from {2ab, a2 − b2}.

Finally the equation x2 +y2 = 2017(x−y) has solution, because 2017 = 442 +92 (this pair is unique), thenu+v2 and ∣v−u∣2 take the values of 2 ⋅ 44 ⋅ 9 or 442 − 92. Using that u+v

2 = x + y and ∣v−u∣2 = ∣y − x + n∣. We havefour cases:

{ x + y = 1855y − x + 2017 = 792

⇒ (x, y) = (1540,315)

{ x + y = 1855−y + x − 2017 = 792

⇒ (x, y) = (2332,−477)

{ x + y = 792y − x + 2017 = 1855

⇒ (x, y) = (477,315)

{ x + y = 792−y + x − 2017 = 1855

⇒ (x, y) = (2332,−1540)

And the positive integers solutions are (x, y) ∈ {(1540,315), (477,315)}.

Also solved by Daniel Lasaosa, Pamplona, Spain; Luca Ferrigno, Universitá degli studi di Tor Vergata,Roma, Italy; Akash Singha Roy, Kolkata, India; Nermin Hodžic, Dobošnica, Bosnia and Herzegovina; AlbertStadler, Herrliberg, Switzerland; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia.


O429. Let ABC be non-obtuse triangle.Prove that

mamb +mbmc +mcma ≤ (a2 + b2 + c2) (5

8+ r

4R)


Solution by Nermin Hodžic, Dobošnica, Bosnia and HerzegovinaSince the triangle is nonobtuse we have a2 + b2 − c2 ≥ 0.Then we have

(a2 + b2)[(a + b)2 − c2] − 2ab(2a2 + 2b2 − c2) = (a − b)2(a2 + b2 − c2) ≥ 0⇒

√(a2 + b2) ⋅ (a + b)

2 − c22ab

≥√

2a2 + 2b2 − c2 (1)

Using formulas

mc =1

2

√2a2 + 2b2 − c2

r

R= cosA + cosB + cosC − 1 = −1 + 1

2∑cyc

a2 + b2 − c2ab

we have

mamb +mbmc +mcma ≤ (a2 + b2 + c2) (5

8+ r

4R)⇔

1

4∑cyc

√(2a2 + 2b2 − c2)(2b2 + 2c2 − a2) ≤ (a2 + b2 + c2)

⎛⎝

5

8− 1

4+ 1

8∑cyc

a2 + b2 − c2ab

⎞⎠⇔

2∑cyc

√(2a2 + 2b2 − c2)(2b2 + 2c2 − a2) ≤ (a2 + b2 + c2)

⎛⎝

3 +∑cyc

a2 + b2 − c2ab

⎞⎠⇔

2∑cyc

√(2a2 + 2b2 − c2)(2b2 + 2c2 − a2) ≤ (a2 + b2 + c2) ⋅∑

cyc

a2 + b2 + ab − c2ab

⇔

∑cyc

(2a2 + 2b2 − c2) + 2∑cyc

√(2a2 + 2b2 − c2)(2b2 + 2c2 − a2) ≤ (a2 + b2 + c2) ⋅∑

cyc

a2 + b2 + ab − c2ab

+ 3∑cyc

a2⇔

⎛⎝∑cyc

√2a2 + 2b2 − c2

⎞⎠

2

≤ (a2 + b2 + c2) ⋅⎛⎝

3 +∑cyc

a2 + b2 + ab − c2ab

⎞⎠⇔

(∑cyc

√2a2 + 2b2 − c2)2 ≤ (a2 + b2 + c2) ⋅∑

cyc

(a + b)2 − c2ab

Since a + b > c using Cauchy-Shwarz inequality we have

(a2 + b2 + c2) ⋅∑cyc

(a + b)2 − c2ab

= 1

2∑cyc

(a2 + b2) ⋅∑cyc

(a + b)2 − c2ab

=

=∑cyc

(a2 + b2) ⋅∑cyc

(a + b)2 − c22ab

≥⎛⎝∑cyc

√(a2 + b2) ⋅ (a + b)

2 − c22ab

⎞⎠

2

≥(1)⎛⎝∑cyc

√2a2 + 2b2 − c2

⎞⎠

2

Equality holds if and only if a = b = c

Also solved by Albert Stadler, Herrliberg, Switzerland; Kevin Soto Palacios, Huarmey, Perú; Akash SinghaRoy, Kolkata, India; Nicus,or Zlota‚ Traian Vuia Technical College, Focs,ani, Romania; Scott H. Brown,Auburn University Montgomery, Montgomery, AL, USA.


O430. Find the number of positive integers n ≤ 106 such that 5 divides (2nn).

Proposed by Enrique Trevinio, Lake Forest College, USA

Solution by Nermin Hodžic, Dobošnica, Bosnia and HerzegovinaLemma:Let x, y ∈ R, then we have

⌊x⌋ + ⌊y⌋ ≤ ⌊x + y⌋

Proof:W.L.O.G let 0 ≤ x, y < 1.Then {x} = x,{y} = y

⌊x⌋ + ⌊y⌋ ≤ ⌊x + y⌋⇔ {x} + {y} ≥ {x + y}

If x + y < 1 then

{x} + {y} = x + y = {x + y}

so equality holds.Let x + y ≥ 1 then

⌊x + y⌋ = 1⇔ x + y = 1 + {x + y}⇒ {x} + {y} = 1 + {x + y} > {x + y}

so the inequality holds, with equality only if x + y < 1. ∎

As a corolary we have

2⌊t⌋ ≤ ⌊2t⌋ (1)

With equality only if 0 < {x} < 1

2.

Let ep(n) be the greatest exponent of p dividing n.Then ep(n!) =∞

∑k=1

⌊ npk

⌋, so

e5((2n

n)) = e5(

(2n)!(n!)2 ) =

∞

∑k=1

(⌊2n

5k⌋ − 2⌊ n

5k⌋)

Using (1) we have

⌊2n

5k⌋ − 2⌊ n

5k⌋ ≥ 0⇒

∞

∑k=1

(⌊2n

5k⌋ − 2⌊ n

5k⌋) ≥ 0

Equality holds if and only if

⌊2n

5k⌋ − 2⌊ n

5k⌋ = 0,∀k ≥ 1⇒ 0 < { n

5k} < 1

2,∀k ≥ 1

Let (2nn) not divisible by 10.Then e5((2nn )) = 0.Then we have

e10((2n

n)) =

∞

∑k=1

(⌊2n

5k⌋ − 2⌊ n

5k⌋) = 0

So we have

{ n5k

} < 1

2,∀k ≥ 1


Letn =

m

∑k=0

ck ⋅ 5k, ck ∈ {0,1,2,3,4}

Now we have

1

2> {n

5} = c0

5⇒ c0 <

5

2⇒ c0 ∈ {0,1,2}

1

2> { n

52} = 5c1 + c0

25⇒ c1 <

25 − 2c010

⇒ c1 ∈ {0,1,2}

1

2> { n

53} = 25c2 + 5c1 + c0

125⇒ c2 <

125 − 10c1 − 2c050

⇒ c2 ∈ {0,1,2}

⋮

1

2> { n

5m} = 5m−1cm−1 + ... + c0

5m⇒ cm−1 <

5m − 2∑m−2k=0 ck ⋅ 5k

2 ⋅ 5m−1⇒ cm−1 ∈ {0,1,2}

1

2> { n

5m+1} = 5mcm + ... + c0

5m+1⇒ cm < 5m+1 − 2∑m−1

k=0 ck ⋅ 5k2 ⋅ 5m ⇒ cm ∈ {0,1,2}

So, positive integer n , such that (2nn) is not divisible by 5 is of the form

n =m

∑k=0

ck ⋅ 5k, ck ∈ {0,1,2}

Since

106 = 2 ⋅ 58 + 2 ⋅ 57 + 4 ⋅ 56 + 0 ⋅ 55 + 0 ⋅ 54 + 0 ⋅ 53 + 0 ⋅ 52 + 0 ⋅ 51 + 0 ⋅ 50

All the numbers of the form

n =8

∑k=0

ck ⋅ 5k, ck ∈ {0,1,2}

are included, and excluding zero the total number of such is 38 − 1.Hence the total number of n such that (2n

n) is divisible by 5 is

106 − 38 + 1 = 993440

Also solved by Albert Stadler, Herrliberg, Switzerland; Daniel Lasaosa, Pamplona, Spain; Akash SinghaRoy, Kolkata, India.


O431. Let a, b, c, d be positive real numbers such that a + b + c + d = 3. Prove that

a2 + b2 + c2 + d2 + 64

27abcd ≥ 3.

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Nermin Hodžic, Dobošnica, Bosnia and HerzegovinaUsing Schurs inequality we have

a3 + b3 + c3 + 3abc ≥ a2(b + c) + b2(c + a) + c2(a + b)b3 + c3 + d3 + 3bcd ≥ b2(c + d) + c2(d + b) + d2(b + c)a3 + c3 + d3 + 3acd ≥ a2(c + d) + c2(d + a) + d2(a + c)a3 + b3 + d3 + 3abd ≥ a2(b + d) + b2(d + a) + d2(a + b)

⇒

a3d + b3d + c3d + 3abcd ≥ a2d(b + c) + b2d(c + a) + c2d(a + b)b3a + c3a + d3a + 3abcd ≥ b2a(c + d) + c2a(d + b) + d2a(b + c)a3b + c3b + d3b + 3abcd ≥ a2b(c + d) + c2b(d + a) + d2b(a + c)a3c + b3c + d3c + 3abcd ≥ a2c(b + d) + b2c(d + a) + d2c(a + b)

⇒

∑cyc

a3(b + c + d) + 12abcd ≥ 2∑cyc

a2(bc + cd + db)⇔

5∑cyc

a3(b + c + d) + 60abcd ≥ 10∑cyc

a2(bc + cd + db) (1)

where the cyclic sum runs over the set {a, b, c, d}.Also using Schurs inequality we have

a4 + b4 + c4 + abc(a + b + c) ≥ a3(b + c) + b3(c + a) + c3(a + b)b4 + c4 + d4 + bcd(b + c + d) ≥ b3(c + d) + c3(d + b) + d3(b + c)a4 + c4 + d4 + acd(a + c + d) ≥ a3(c + d) + c3(d + a) + d3(a + c)a4 + b4 + d4 + abd(a + b + d) ≥ a3(b + d) + b3(d + a) + d3(a + b)

⇒

3∑cyc

a4 +∑cyc

a2(bc + cd + db) ≥ 2∑cyc

a3(b + c + d) (2)

Adding (1) and (2) we have

3∑cyc

a4 + 3∑cyc

a3(b + c + d) + 60abcd ≥ 9∑cyc

a2(bc + cd + db)⇔

2∑cyc

a4 + 2∑cyc

a3(b + c + d) + 40abcd ≥ 6∑cyc

a2(bc + cd + db)⇔

3[∑cyc

a4 + 2∑cyc

a3(b + c + d) + 2∑cyc

a2(bc + cd + db) +∑cyc

a2(b2 + c2 + d2)] + 64abcd ≥

≥∑cyc

a4 + 4∑cyc

a3(b + c + d) + 12∑cyc

a2(bc + cd + db) + 3∑cyc

a2(b2 + c2 + d2) + 24abcd⇔

3∑cyc

a2 ⋅ (∑cyc

a)2 + 64abcd ≥ (∑cyc

a)4⇔

27∑cyc

a2 + 64abcd ≥ 81⇔∑cyc

a2 + 64

27abcd ≥ 3

Equality holds if and only if a = b = c = d = 3

4.

Also solved by Daniel Lasaosa, Pamplona, Spain; AN-anduud Problem Solving Group, Ulaanbaatar, Mon-golia; Albert Stadler, Herrliberg, Switzerland; Akash Singha Roy, Kolkata, India; Arkady Alt, San Jose, CA,USA; Ioan Viorel Codreanu, Satulung, Maramures,, Romania; Nicus,or Zlota‚ Traian Vuia Technical College,Focs,ani, Romania; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy.


O432. Let ABCDEF be a cyclic hexagon which contains an inscribed circle. Denote by ωA, ωB, ωC , ωD, ωE

and ωF the inscribed circle in the triangle FAB,ABC,BCD,CDE,DEF and EFA, respectively. LetÀB be the external common tangent of ωA and ωB, other than the line AB; lines `BC ,`CD,`DE ,ÈF and`FA are defined analogously. Let A1 be the intersection of te lines `FA and ÀB, B1 the intersection ofthe lines ÀB and `BC ; points C1,D1,E1 and F1 are defined analogously. Suppose that A1B1C1D1E1F1

is a convex hexagon. Prove that its diagonals A1D1,B1E1 and C1F1 are concurrent.

Proposed by Nairi Sedrakian, Yerevan, Armenia

Solution by the authorWe will prove that the hexagon A1B1C1D1E1F1 is centrally symmetric and therefore the main diago-nals A1D1,B1E1 and C1F1 pass though the symmetry center of the hexagon. We denote the centers ofωA, ωB, ωC , ωD, ωE and ωF by IA, IB, IC , ID, IE and IF respectively.

Claim 1:ÀB ∣∣CF ∣∣`DE , `BC ∣∣AD∣∣ÈF , and `CD ∣∣BE∣∣`FA. Proof:By the symmetry it suffices to prove that ÀB ∣∣CF. Let M be the midpoint of the arc AB of the circumcirclenot containing C and F , and let m be the tangent to the circumcircle at M , which is parallel to AB.

Then we have

∠(FM,CF ) =∠(m,CM) =∠(AB,CM). (1)

It is well known that the incenter IA of the triangle FAB satisfies MIA =MA =MB; similarly MIB =MA = MB, so MIA = MIB and therefore ∠(IAIB,CM) = ∠(FM, IAIB). The lines AB and ÀB aresymmetric about the line IAIB, so

∠(AB,CM) =∠(AB, IAIB) +∠(IAIB,CM) =∠(IAIB, ÀB) +∠(FM, IAIB) =∠(FM, ÀB).

This equation combined with (1) yields ∠(FM,CF ) =∠(FM, ÀB), so indeed CF ∣∣ÀB. ∎


Claim 2:A1B1 +C1D1 +E1F1 = B1C1 +D1E1 + F1A1.

Proof:Let TA, UA, VA, and WA be the points where ωA touches the lines AB,FA, ÀB, and `FA, respectively, anddefine the points TB, . . . ,WF analogously.

Since the hexagon ABCDEF is tangential, we have

AB +CD +EF = BC +DE + FA (2)

Furthermore we have

ATA = AUA, . . . , FTF = FUF and A1VA = A1WA, . . . , F1VF = F1WF , (3)

because these pairs of segments are tangents drawn to the circles ωA, . . . , ωF .Finally, from the symmetry about the lines IAIB, . . . , IF IA, we can wee that

TAUB = VAWB, . . . , TFUA = VFWA. (4)


By combining (3) and (4),

A1B1 = VAWB −A1VA −B1WB = TAUB −A1VA −B1WB

= (AB −ATA −BUB) −A1VA −B1WB

= AB −ATA −BTB −A1VA −B1VB.

Analogously,B1C1 = BC −BTB −CTC −B1VB −C1VC ,

. . .

F1A1 = FA − FTF −ATA − F1VF −A1VA.

Now the clain can be achieved by plugging these formulae into (2) and cancelling identical terms. ∎

Claim 3:ÐÐÐ→A1B1 =

ÐÐÐ→E1D1,

ÐÐÐ→B1C1 =

ÐÐÐ→F1E1, and

ÐÐÐ→C1D1 =

ÐÐÐ→A1F1.

Proof:Let X,Y, and Z be those points for which the quadrilaterals F1A1B1X, B1C1D1Y , and D1E1F1Z areparallelograms. By Claim 1 we fave F1X ∣∣A1B1∣∣E1D1∣∣F1Z, so the points F1,X,Z are collinear; it can beseen similarly that B1,X,Y are collinear and so are D1, Y,Z. We will show that the points X,Y,Z coincide.

The points X,Y,Z either coincide or form a triangle. Suppose that XY Z is a triagnle with the sameorientation as the haxagon A1B1C1D1E1F1. Then

F1A1 +B1C1 +D1E1 =XB1 + Y D1 +ZF1 >Y B1 +ZD1 +XF1 = C1D1 +E1F1 +A1B1,

contradicting Claim 2.If XY Z is a triangle with the opposite orientation from the hexagon, we get another contradiction

F1A1 +B1C1 +D1E1 <C1D1 +E1F1 +A1B1

∎Claim 3 shows that the hexagon A1B1C1D1E1F1 is indeed centrally symmetric, as requred. Done!


Documents

Junior problems - AwesomeMath · J430. IntriangleABC,∠C>90 and3a+ 15ab+5b=7c. Provethat∠C≤120 . ProposedbyTituAndreescu,UniversityofTexasatDallas,USA SolutionbyPolyahedra,PolkStateCollege,USA