Junior problems
J325. For positive real numbers a and b, define their perfect mean to be half of the sum of their arithmetic
and geometric means. Find how many unordered pairs of integers (a, b) from the set {1, 2, . . . , 2015}
have their perfect mean a perfect square.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
Solution by Alessandro Ventullo, Milan, Italy
We have
i.e.
a+b
2
√
+
2
√
ab
= n2
√
a+
b = 2n,
n ∈ N,
n ∈ N.
As a, b ∈ {1, 2, . . . , 2015}, then a and b must be perfect squares and n ≤ 44. Assume that a ≤ b. If a is an odd
perfect square, then b is an odd perfect square, and making a case by case analysis for a ∈ {1, 32 , . . . , 432 }
we get 22 + 21 + . . . + 2 + 1 = 253 unordered pairs. Similarly, if a is an even perfect square, then b is an
even perfect square and we get 22 + 21 + . . . + 2 + 1 = 253 unordered pairs. So, there are 253 + 253 = 506
unordered pairs which satisfy the given conditions.
Also solved by Yujin Kim, The Stony Brook School, NY, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Arber Igrishta, Eqrem Qabej, Vushtrri, Kosovo; Behzod Ibodullaev, Lyceum TCTU, Tashkent,
Uzbekistan; David E. Manes, Oneonta, NY, USA; Dilshoda Ibragimova, Academic Lyceum Nr.2 under the
SamIES, Samarkand, Uzbekistan; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Francesc Gispert Sanchez, CFIS, Universitat Politecnica de Catalunya, Barcelona,
Spain; Haroun Meghaichi, University of Science and Technology Houari Boumediene, Algiers, Algeria; Joseph Lee, Loomis Chaffee School in Windsor, CT, USA; Luiz Ernesto, M.A.Leitão, Brasil; Sardor Bozorboyev,
Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Titu Zvonaru and Neculai Stanciu, Romania; Ji Eun Kim,
Tabor Academy, MA, USA; Misiakos Panagiotis, Athens College(HAEF), Greece; Michael Tang, Edina High
School, MN, USA; Kyoung A Lee, The Hotchkiss School, Lakeville, CT, USA; Myong Claire Yun, Seoul
International School, South Korea.
Mathematical Reflections 1 (2015)
1
J326. Let a, b, c be nonnegative integers. Prove that
p
p
p
√
√
√
2a2 + 3b2 + 4c2 + 3a2 + 4b2 + 2c2 + 4a2 + 2b2 + 3c2 ≥ ( a + b + c)2 .
Proposed by by Titu Andreescu, University of Texas at Dallas, USA
Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USA
Note that
r
p
2a2 3b2 4c2
1
2a2 + 3b2 + 4c2 = 3
+
+
≥ (2a + 3b + 4c)
9
9
9
3
by Jensen’s inequality. Similarly,
p
1
3a2 + 4b2 + 2c2 ≥ (3a + 4b + 2c)
3
and
p
1
4a2 + 2b2 + 3c2 ≥ (4a + 2b + 3c).
3
Therefore,
p
p
p
2a2 + 3b2 + 4c2 + 3a2 + 4b2 + 2c2 + 4a2 + 2b2 + 3c2 ≥ 3(a + b + c)
√
√
√
( a)2 + ( b)2 + ( c)2
= 9
3
√
√
√ !2
a+ b+ c
≥ 9
3
√
√
√
= ( a + b + c)2 ,
where the final inequality follows from the quadratic mean-arithmetic mean inequality.
Also solved by George Gavrilopoulos, High School Of Nea Makri, Athens, Greece; Yong Xi Wang,East
China Institute Of Technology, China; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Utsab
Sarkar, Chennai Mathematical Institute, India; Titu Zvonaru and Neculai Stanciu, Romania; Stefan Petrevski, NOVA International School in Skopje, Macedonia; Albert Stadler, Herrliberg, Switzerland; Sardor
Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Rade Krenkov, SOU "Goce Delcev", Valandovo, Macedonia; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Paolo Perfetti, Universitá
degli studi di Tor Vergata Roma, Roma, Italy; Odilbek Utamuratov, Uzbekistan; Nicuşor Zlota, “Traian Vuia”
Technical College, Focşani, Romania; Haroun Meghaichi, University of Science and Technology Houari Boumediene, Algiers, Algeria; Francesc Gispert Sanchez, CFIS, Universitat Politecnica de Catalunya, Barcelona,
Spain; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Dilshoda
Ibragimova, Academic Lyceum Nr.2 under the SamIES, Samarkand, Uzbekistan; Behzod Ibodullaev, Lyceum
TCTU, Tashkent, Uzbekistan; David E. Manes, Oneonta, NY, USA; Arkady Alt, San Jose, CA, USA; Arber
Igrishta, Eqrem Qabej, Vushtrri, Kosovo; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Myong Claire
Yun, Seoul International School, South Korea; Kyoung A Lee, The Hotchkiss School, Lakeville, CT, USA;
Michael Tang, Edina High School, MN, USA; Misiakos Panagiotis, Athens College(HAEF), Greece; Ji Eun
Kim, Tabor Academy, MA, USA; Yujin Kim, The Stony Brook School, NY, USA.
Mathematical Reflections 1 (2015)
2
J327. A jeweler makes a circular necklace out of nine distinguishable gems: three sapphires, three rubies and
three emeralds. No two gems of the same type can be adjacent to each other and necklaces obtained
by rotation and reflection (flip) are considered to be identical. How many different necklaces can she
make?
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
Solution by Myong Claire Yun, Seoul International School, South Korea
Start by placing the three sapphires S1 , S2 , S3 as shown in the figure below. There will be three gaps in
between to fill with emeralds and rubies.
The distribution of the mix jewels in the 3 gaps can be categorized as:
Type 1 : (4 + 1 + 1) or (1 + 4 + 1) or (1 + 1 + 4)
Type 2 : (3 + 2 + 1) and the permutations
Type 3 : (2 + 2 + 2)
Let’s count separately the number of necklaces of each type:
Type 1 : (4 + 1 + 1): We need 2 rubies and 2 emeralds to fill in gap 1.
3 3
=9
2 2
ways to select these 4 jewels.
They can be arranged in 2 · 2 · 2 = 8 ways. The remaining two gaps can be filled out in 2 ways. Hence
these are 9 · 8 · 2 = 144 necklaces of type (4 + 1 + 1).
Type 2 : (3 + 2 + 1): We need 2 rubies and 1 emerald or 1 ruby and 2 emeralds to fill in gap 1.
3 3
2
= 18
2 1
ways.
Arranging can be done in 2 ways. Gap 3 can be gilled out in 2 ways. We are left with one ruby abd one
emerald to place in gap 2 → 2 ways.
Therefore there is 18 · 2 · 2 · 2 = 144 necklaces of type (3 + 2 + 1).
Mathematical Reflections 1 (2015)
3
Type 3 : (2 + 2 + 2):
3 3
2
= 18
1 1
ways to fill gap 1.
Once the first gap is taken care of, there is
2 3
2=8
1 1
ways to fill out gap 2. Therefore, there are 18 · 8 · 2 = 288 necklaces of type (2 + 2 + 2).
In conclusion, there is 3 · 144 + 6 · 144 + 1 · 288 = 1584 different necklaces.
Also solved by Yujin Kim, The Stony Brook School, NY, USA; Ji Eun Kim, Tabor Academy, MA, USA;
Misiakos Panagiotis, Athens College(HAEF), Greece; Kyoung A Lee, The Hotchkiss School, Lakeville, CT,
USA; Arber Igrishta, Eqrem Qabej, Vushtrri, Kosovo; Joseph Lee, Loomis Chaffee School in Windsor, CT,
USA.
Mathematical Reflections 1 (2015)
4
J328. Let a, b, cbe postive real numbers such that a + b + c = 2,Prove that
p
p
p
a2 + bc + b2 + ca + c2 + ab ≤ 3
Proposed by An Zhen-ping, Xianyang Normal University, China
Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia
Without loss of generality, assume that a ≥ b ≥ c . By the AM-GM inequality, we have
r
p
a2 + bc
1 a2 + bc
2
a + bc =
· (a + c) ≤
+ (a + c)
a+c
2
a+c
1
1 a2 + ac
+ (a + c) = (2a + c),
≤
2
a+c
2
and
r
b2 + ca
+ (b + c) ,
b+c
r
p
c2 + ab
1 c2 + ab
2
· (b + c) ≤
+ (b + c) .
c + ab =
b+c
2
b+c
p
b2 + ca =
1
b2 + ca
· (b + c) ≤
b+c
2
Summing up the above inequalities and using c ≤ b, we have
p
p
p
1
(b2 + c2 ) + a(b + c)
2
2
2
a + bc + b + ca + c + ab ≤
2a + 2b + 3c +
2
b+c
2
b + bc
1
)
≤ (3a + 2b + 3c +
2
b+c
3
= (a + b + c) = 3.
2
Equality if and only if {a, b, c} = {1, 1, 0}.
Also solved by Yujin Kim, The Stony Brook School, NY, USA; Ji Eun Kim, Tabor Academy, MA, USA;
Misiakos Panagiotis, Athens College(HAEF), Greece; Kyoung A Lee, The Hotchkiss School, Lakeville, CT,
USA; Myong Claire Yun, Seoul International School, South Korea; Yong Xi Wang,East China Institute Of
Technology, China; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Ángel Plaza, University of Las Palmas
de Gran Canaria, Spain; Arkady Alt, San Jose, CA, USA; Farrukh Mukhammadiev, Academic Lyceum Nr.1
under the SamIES, Samarkand, Uzbekistan; Francesc Gispert Sanchez, CFIS, Universitat Politecnica de
Catalunya, Barcelona, Spain; Jean Heibig, Lycée Stanislas, Paris, France; Joseph Lee, Loomis Chaffee School
in Windsor, CT, USA; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania; Paolo Perfetti,
Universitá degli studi di Tor Vergata Roma, Roma, Italy; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov,
Tashkent, Uzbekistan; Titu Zvonaru and Neculai Stanciu, Romania; Utsab Sarkar, Chennai Mathematical
Institute, India.
Mathematical Reflections 1 (2015)
5
J329. Let a1 , a2 , . . . , a2015 be positive integers such that
a1 + a2 + · · · + a2015 = a1 a2 · · · a2015 .
Prove that among numbers a1 , a2 , . . . , a2015 at most nine are greater than 1.
Proposed by Titu Zvonaru, Comaneşti and Neculai Stanciu, Buzău, Romania
Solution by G.R.A.20 Problem Solving Group, Roma, Italy
We show that at most six are greater than 1.
Let us assume that a1 ≥ a2 ≥ · · · ≥ ad ≥ 2 and ad+1 = · · · = a2015 = 1 where d ≥ 2 is the maximal number
of terms greater than 1. Then
2015a1 ≥ a1 2d−1
which implies that d ≤ 1 + blog2 (2015)c = 11. Moreover
da1 + 2015 − d ≥ a1 a2 2d−2
which implies that given a1 ≥ 2 and 2 ≤ d ≤ 11, then a2 should satisfies
1
2015 − d
2 ≤ a2 ≤ min(F (a1 , d), a1 ) where F (a1 , d) := d−2
+d .
a1
2
Note that F is decreasing (not strictly) with respect to a1 and we can easily solve the problem by considering
several cases.
1) d < 11 because F (2, 11) = 1.
2) d < 10 because F (3, 10) = 2 and F (4, 10) = 1 implies that then a1 ≤ 3, a2 = · · · = a10 = 2 and
a1 + 2022 = a1 + 9 · 2 + 2015 − 10 6= a1 · 29 ≤ 1536.
3) d < 9 because F (8, 9) = 2 and F (9, 9) = 1 implies that a1 ≤ 8 and by examining all the finite cases we
have no solutions.
4) d < 8 because F (16, 8) = 2 and F (17, 8) = 1 implies that a1 ≤ 16 and by examining all the finite cases
we have no solutions.
5) d < 7 because F (35, 7) = 2 and F (36, 7) = 1 implies that a1 ≤ 35 and by examining all the finite cases
we have no solutions.
6) d = 6 because F (77, 6) = 2 and F (78, 6) = 1 implies that a1 ≤ 77 and by examining all the finite cases
we have precisely one solution
a1 = 17, a2 = 5, a3 = 3, a4 = a5 = a6 = 2, a7 = · · · = a2015 = 1
and
17 + 5 + 3 + 2 + 2 + 2 + 2015 − 6 = 17 · 5 · 3 · 23 = 2040.
Also solved by Yujin Kim, The Stony Brook School, NY, USA; Ji Eun Kim, Tabor Academy, MA, USA;
Michael Tang, Edina High School, MN, USA; Kyoung A Lee, The Hotchkiss School, Lakeville, CT, USA;
Myong Claire Yun, Seoul International School, South Korea; Adnan Ali, Student in A.E.C.S-4, Mumbai,
India; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; George
Gavrilopoulos, High School Of Nea Makri, Athens, Greece; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov,
Tashkent, Uzbekistan.
Mathematical Reflections 1 (2015)
6
J330. Let ABCD be a quadrilateral with centroid G, inscribed in a circle with center O, and diagonals
intersecting at P . Prove that if O, G, P are collinear, then either ABCD is an isosceles trapezoid or
an orthogonal quadrilateral.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
Solution by Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan
Let M and N be the midpoints of AC and BD. Observe that G is midpoint of M N and quadrilateral
OM N P is cyclic. If AC ⊥ BD, then OM N P is a rectangle. The diagonals of OM N P bisect each other, so
G lies on OP .
Assume that ABCD is not orthogonal, then consider Ω, the center of the circumcircle of OM N P .
Observe that Ω 6= G and Ω lies on OP . Then ΩM = ΩN , ΩG ⊥ M N and M O = ON . It follows that
triangles AM O is congruent to triangle BN O, so AC = BD. Therefore ∠CDA = ∠BAD and ABCD is an
isosceles trapezoid.
Also solved by Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Titu Zvonaru and Neculai Stanciu, Romania.
Mathematical Reflections 1 (2015)
7
Senior problems
S325. Let S be the incenter of the triangle formed by the midpoints of the triangle ABC. Cevians AS, BS,
CS intersect the sides BC, CA, AB in points M , N , P , respectively. Prove that if the perpendiculars
to the sides of triangle ABC at points M , N , P are concurrent, then triangle ABC is isosceles.
Proposed by Roxana Mihaela Stanciu and Nela Ciceu, Bacău, Romania
Solution by Andrea Fanchini, Cantú, Italy
The incenter of the triangle DEF formed by the midpoints of the triangle ABC is known as Spieker
center and it have barycentric coordinates S = X10 (b + c : c + a : a + b).
So the line AS have equation AS ≡ (a + b)y − (c + a)z = 0 and then point M = AS ∩ |{z}
BC , have coordinates
x=0
M (0 : c + a : a + b).
Line BS have equation BS ≡ (a + b)x − (b + c)z = 0 and then point N = BS ∩ |{z}
CA , so it have coory=0
dinates N (b + c : 0 : a + b).
Line CS have equation CS ≡ (c + a)x − (b + c)y = 0 and then point P = CS ∩ |{z}
AB , so it have coorz=0
dinates P (b + c : c + a : 0).
The infinite point of the perpendicular to the side BC is BC∞⊥ (−a2 : SC : SB ), so the perpendicular
to the side BC at point M have equation M BC∞⊥ ≡ [(c + a)SB − (a + b)SC ]x − a2 (a + b)y + a2 (c + a)z = 0
The infinite point of the perpendicular to the side CA is CA∞⊥ (SC : −b2 : SA ), so the perpendicular
to the side CA at point N have equation N CA∞⊥ ≡ b2 (a + b)x − [(b + c)SA − (a + b)SC ]y − b2 (b + c)z = 0
The infinite point of the perpendicular to the side AB is AB∞⊥ (SB : SA : −c2 ), so the perpendicular
to the side AB at point P have equation P AB∞⊥ ≡ −c2 (c + a)x + c2 (b + c)y + [(b + c)SA − (c + a)SB ]z = 0
Mathematical Reflections 1 (2015)
8
Now three lines pi x + qi y + ri z = 0, with i = 1, 2, 3,
p1
p2
p3
are concurrent if and only if
q1 r1 q2 r2 = 0
q3 r3 In our case the perpendiculars to the sides of triangle ABC at points M , N , P are
only if
(c + a)SB − (a + b)SC
−a2 (a + b)
a2 (c + a)
b2 (a + b)
(a + b)SC − (b + c)SA
−b2 (b + c)
2
2
−c (c + a)
c (b + c)
(b + c)SA − (c + a)SB
concurrent if and
=0
It’s easy to prove that this is true if and only if a = b or b = c or c = a and that’s when triangle ABC is
isosceles, q.e.d.
2
2
For example when a = b ⇒ SA = SB = c2 , SC = a2 − c2 and so we have
(c + 3a)SA − 2a3
−2a3
a2 (c + a)
2a3
2a3 − (c + 3a)SA −a2 (c + a)
2
−c (c + a)
c2 (c + a)
0
summing the first two rows we obtain
(c + 3a)SA − 2a3
−2a3
a2 (c + a)
(c + 3a)SA
−(c + 3a)SA
0
−c2 (c + a)
c2 (c + a)
0
=0
similarly for the other two cases.
Also solved by Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan.
Mathematical Reflections 1 (2015)
9
S326. Let a, b, c be positive reals such that, a3 + b3 + c3 + abc = 13 . Prove that,
abc + 9
X
cyc
1
a5
≥
2
2
4b + bc + 4c
4 (a + b + c) (ab + bc + ca)
Proposed by Titu Andreescu, University of Texas at Dallas, USA
Solution by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia
Applying the Cauchy-Schwarz inequality for the sequences
√
√
x1 = qabc,
y1 = 9abc
p
9a5
x2 = 4b2 +bc+4c
y2 = a(4b2 + bc + 4c2 )
2,
q
p
9b5
x3 = 4c2 +ca+4a
y3 = b(4c2 + ca + 4a2 )
2,
q
p
9c5
x4 = 4a2 +ab+4b
y4 = c(4a2 + ab + 4b2 )
2,
we obtain
a5
b5
c5
abc + 9
+
+
4b2 + bc + 4c2 4c2 + ca + 4a2 4a2 + ab + 4b2
(3abc + 3a3 + 3b3 + 3c3 )2
≥
9abc + 4(ab2 + bc2 + ca2 ) + 3abc + 4(ac2 + ba2 + cb2 )
1
=
2
2
2
4[(ab + abc + ba ) + (bc + abc + c2 b) + (ca2 + abc + ac2 )]
1
=
.
4(a + b + c)(ab + bc + ca)
q
1
Equality if and only if a = b = c = 3 12
.
Also solved by Utsab Sarkar, Chennai Mathematical Institute, India; Arkady Alt, San Jose, CA, USA;
Behzod Ibodullaev, Lyceum TCTU, Tashkent, Uzbekistan; Farrukh Mukhammadiev, Academic Lyceum Nr.1
under the SamIES, Samarkand, Uzbekistan; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania; Odilbek Utamuratov, Uzbekistan; Paolo Perfetti, Universitá degli studi di Tor Vergata Roma, Roma,
Italy; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Titu Zvonaru and Neculai Stanciu,
Romania.
Mathematical Reflections 1 (2015)
10
S327. Let H be the orthocenter of an acute triangle ABC and let D and E be the feet of the altitudes from
vertices B and C, respectively. Line DE intersects the circumcircle of triangle ABC in points F and
G (F lies on the smaller arc of AB and G lies on the smaller arc of AC). Denote by S the projection
of H onto line DE. Prove that EF + DS = DG + ES.
Proposed by İlker Can Çiçek, Instanbul, Turkey
First solution by Prithwijit De, HBCSE, Mumbai, India
Assume ∠B ≥ ∠C. Let HS = x and DE = y. Angle chasing gives us ∠DHS = ∠B and ∠EHS = ∠C.
Also x = 2R cos A cos B cos C, where R is the circumradius of triangle ABC and y = a cos A. Thus
DS − ES = x(tan B − tan C) = 2R cos A sin(B − C).
(1)
By the intersecting chord theorem it follows that DG · DF = AD · DC and EG · EF = AE · EB. Subtracting
one from the other leads to
DG − EF =
AD · DC − AE · EB
a cos A(c cos C − b cos B)
=
= 2R cos A sin(B − C).
y
a cos A
(2)
Hence DS − ES = DG − EF and the result follows.
Second solution by Ivan Borsenco, Massachusetts Institute of Technology, USA
Let M be the midpoint of F G and N be the midpoint of DE. Denote by O and Ω the circumcenter
and the center of the nine-point circle of triangle ABC. Observe that M is the projection of O, N is the
projection of Ω, and S is the projection of H onto F G. Because OΩ = HΩ, we have M N = N S, DS = EM
and ES = DM . From the fact that F M = M G, we have EF + EM = DG + DM , which is equivalent to
EF + DS = DG + ES, as desired.
Also solved by Misiakos Panagiotis, Athens College(HAEF), Greece; Kyoung A Lee, The Hotchkiss School,
Lakeville, CT, USA; George Gavrilopoulos, High School Of Nea Makri, Athens, Greece; Adnan Ali, Student
in A.E.C.S-4, Mumbai, India; Andrea Fanchini, Cantú, Italy; Arkady Alt, San Jose, CA, USA; Behzod
Ibodullaev, Lyceum TCTU, Tashkent, Uzbekistan; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under
the SamIES, Samarkand, Uzbekistan; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan;
Saturnino Campo Ruiz, Salamanca, Spain; Titu Zvonaru and Neculai Stanciu, Romania.
Mathematical Reflections 1 (2015)
11
S328. Let k be an odd positive integer. There are k positive integers written on a circle that add up to 2k.
Prove that for any 1 ≤ m ≤ k, among the k given numbers, we can find one or more consecutively
placed numbers that add up to 2m.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
Solution by the author
Let a1 , . . . , ak be numbers written on a circle. Consider k sums of 2m consecutive numbers on a circle:
si = ai + ai+1 + · · · + ai+2m−1 , where i ∈ {1, . . . , k} and the indices are taken modulo k. We have
s1 + s2 + · · · + sk = 2m(a1 + a2 + · · · + ak ) = 4km.
If all si are equal to 4m, then ai = a2m+i for all i ∈ {1, . . . , k}. Observe that because k is odd, we have
d = gcd(k, 2m) | m. We find that in this case our k numbers on a circle, consist of kd congruent blocks of d
numbers. The sum of numbers in each block is 2d, so if we consider m
d consecutive blocks, their sum is 2m,
as desired.
If not all si are equal, then there exists sj such that sj ≤ 4m − 1. Consider 2m sums:
aj , aj + aj+1 , . . . , aj + aj+1 + · · · + aj+2m−1 .
If any of these sums is equal to 2m, we are done. Otherwise, each of these sums is less than 4m and by the
pigeonhole principle, two of these sums give the same residue module 2m. Taking their difference, we find a
sum of several consecutive numbers on circle that is equal to 2m.
Mathematical Reflections 1 (2015)
12
S329. Given a quadrilateral ABCD with AB + AD = BC + DC let (I1 ) be the incircle of triangle ABC
which is tangent to AC at E, and let (I2 ) be the incircle of triangle ADC which is tangent to AC at
F . The diagonals AC and BD intersect at point P . Suppose BI1 and DE intersect at point S and
DI2 and BF intersect at point T . Prove that S, P, T are collinear.
Proposed by Khakimboy Egamberganov, Tashkent, Uzbekistan
Solution by Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan
At first we use these following lemmas:
Lemma 1 : Let ABCD be a quadrilateral. Prove that AB + AD = BC + BD if and only if ABCD is
B− extangential or D− extangential.
Proof: Assume that ABCD is B or D− extangential. It is enough to prove in case of ABCD is B−
extangential.
Let B excircle of ABCD tangents to the sides BA, CD, AD, BC at points X, Y, Z, T , respectively. We
have BX = BT or BA + AD + DZ = BC + CD + DY and DZ = DY , so BA + AD = BC + CD
Assume that BA + AD = BC + CD (1) and let Ω be the circle tangent to lines BA, CD, AD. It is
enough to prove that CB is tangent to Ω.Let tangent line from C to Ω intersects BA at B0
Then quadrilateral AB0 CD - B0 - extangencial. =⇒ AB0 + AD = CD + CD0 (2). We have from (1)
and (2) that B = B0 , hence Lemma1 is proved.
Lemma 2 : The lines I1 I2 , AC, BD are concurrent.
Proof : From Lemma 1 ABCD is B− extangential. Let I1 I2 ∩ AC = P0 is insimilicenter of (I1 ) and (I2 ).
However B is exsimilicenter of (I1 ) and Ω, and D is insimilicenter of (I2 ) and Ω.
From Monge-D’Alembert cirle theorem we have B, D, P0 are collinear, so P0 = P .
Lemma 3 : Let Γb is B− excircle of 4ABC and Γd be is D− excircle of 4ADC. Then AC is tangent to
Γb at F and is also tangent to Γd at E.
Proof:It is obviously, we have CA+CB−AB
= AD+AC−DC
.
2
2
Hence F is exsimilicenter of (I2 ) and Γb , E is exsimilicenter of (I1 ) and Γd .
We prove the problem by applying abovementioned lemmas: Let T0 be the exsimilicenter of (I2 ) and Ω.
We have by Monge-D’Alembert cirle theorem:
(1) For {(I1 ), (I2 ), Γb } we obtain T0 , F, B are collinear , so T0 ∈ BF .
(2) For {(I2 ), Ω, Γb } we get T0 , I, I2 are collinear , so T0 ∈ DI2 .
From (1) and (2) we have T = T0 or T is exsimilicenter of (I2 ) and Ω.
Similarly we have S is insimilicenter of (I1 ) and Ω.
Finally applying by Monge-D’Alembert cirle theorem for (I1 ), (I2 ), Ω we obtain that T, S, P are collinear.
Mathematical Reflections 1 (2015)
13
S330. Let x1 , x2 , . . . , xn be real numbers, n ≥ 2, such that x21 + x22 + · · · + x2n = 1. Prove that,
n q
X
X
1 − x2i + kn ·
xj xk ≥ n − 1
i=1
Where kn = 2 − 2
q
1+
1≤j<k≤n
1
n−1 .
Proposed by Marius Stănean, Zalău, Romania
Solution by the author
We havePtwo cases:
xj xk ≤ 0 then
1. if
1≤j<k≤n
n q
X
1 − x2i + kn ·
i=1
P
2. if
X
xj xk ≥
n
X
i=1
1≤j<k≤n
xj xk ≥ 0 then noting q =
1≤j<k≤n
P
1 − x2i = n − 1;
xj xk ≥ 0 the inequality we can write as
1≤j<k≤n
n q
X
1 − x2i ≥ n − 1 − kn · q.
i=1
By squaring both sides we get
n
X
X
(1 − x2i ) + 2
i=1
q
q
1 − x2j · 1 − x2k ≥ (n − 1)2 − 2(n − 1)kn q + kn2 q 2 ⇐⇒
1≤j<k≤n
2
X
q
q
1 − x2j · 1 − x2k ≥ (n − 1)(n − 2) − 2(n − 1)kn q + kn2 q 2 .
1≤j<k≤n
Using Cauchy-Schwarz inequality we have for 1 ≤ j < k ≤ n
q
q
q
q
1 − x2j · 1 − x2k = x21 + . . . + x2j−1 + x2j+1 + . . . + x2n · x21 + . . . + x2k−1 + x2k+1 + . . . + x2n ≥
| x21 + . . . + x2j−1 + x2j+1 + . . . + x2k−1 + x2k+1 + . . . + x2n + xj xk |
and summing this for all j, k such that 1 ≤ j < k ≤ n we get
q
X q
X
2
1 − xj · 1 − x2k ≥
| 1 − x2i − x2k + xj xk |≥
1≤j<k≤n
1≤j<k≤n
(n − 1)(n − 2)
n
2
2
2
+q
2 − (n − 1)(x1 + x2 + . . . + xn ) + q =
2
Therefore to prove the inequality (1) it is enough to show that
(n − 1)(n − 2) + 2q ≥ (n − 1)(n − 2) − 2(n − 1)kn q + kn2 q 2 ⇐⇒
q(2 + 2(n − 1)kn − kn2 q) ≥ 0.
But kn is a root of the quadratic equation x2 − 4x −
2+2(n−1)kn
2
kn
=
n−1
2
4
n−1
= 0 ⇐⇒ (n − 1)x2 − 4(n − 1)x − 4 = 0, so
therefore (2) is equivalent with the following inequality
n−1
q
−q ≥0
2
Mathematical Reflections 1 (2015)
14
which is true because q ≥ 0 and q ≤
X
n−1
2 .
The last inequality follows from
(xj − xk )2 ≥ 0 ⇐⇒ (n − 1)(x21 + x22 + . . . + x2n ) ≥ 2
1≤j<k≤n
X
xj xk .
1≤j<k≤n
The equality hold when q = 0 which means that one
q of the numbers x1 , x2 , . . . , xn is ±1 and the other 0
n−1
or q = 2 which means that x1 = x2 = . . . = xn = ± n1 .
Also solved by Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov,
Tashkent, Uzbekistan, Utsab Sarkar, Chennai Mathematical Institute, India
Mathematical Reflections 1 (2015)
15
Undergraduate problems
U325. Let A1 B1 C1 be a triangle with circumradius R1 . For each n ≥ 1, the incircle of triangle An Bn Cn is
tangent to its sides at points An+1 , Bn+1 , Cn+1 . The circumradius of triangle An+1 Bn+1 Cn+1 , which
Rn+1
is also the inradius of triangle An Bn Cn is Rn+1 . Find limn→∞
.
Rn
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
Solution by Yong Xi Wang,East China Institute Of Technology, China
Suppose ∆An Bn Cn has an incircle with radius rn ,then we have Rn+1 = rn and use well known
rn
An
Bn
Cn
= 4 sin
sin
sin
Rn
2
2
2
on the other hand,we have
∠An+1 =
π ∠An
π ∠Bn
π ∠Cn
−
, ∠Bn+1 = −
∠Cn+1 = −
2
2
2
2
2
2
Because
∠An + ∠Bn+1 ICn+1 = π, ∠Cn+1 An+1 Bn+1 = 2∠Bn+1 ICn+1
so we have
n−1
π
1
π
π π
1
=⇒ ∠An = + ∠A1 −
−
∠An+1 − = − ∠An −
3
2
3
3
3
2
simaler we have
n−1
n−1
π π
1
π π
1
∠Bn = + ∠B1 −
−
, ∠Cn = + ∠C1 −
−
3
3
2
3
3
2
so we have
lim An = lim Bn = lim Cn =
n→∞
so
lim
n→∞
n→∞
n→∞
π
3
π
π
π
1
Rn+1
rn
= lim
= lim 4 sin sin sin =
n→∞ Rn
n→∞
Rn
6
6
6
2
Also solved by Michael Tang, Edina High School, MN, USA; Ji Eun Kim, Tabor Academy, MA, USA;
Kyoung A Lee, The Hotchkiss School, Lakeville, CT, USA; AN-anduud Problem Solving Group, Ulaanbaatar,
Mongolia; Arkady Alt, San Jose, CA, USA; Saturnino Campo Ruiz, Salamanca, Spain; Titu Zvonaru and
Neculai Stanciu, Romania.
Mathematical Reflections 1 (2015)
16
U326. Find
∞
X
a2
n=0 n
an + 2
+ an + 1
where a0 > 1 and 3an+1 = a3n + 2 for all integers n ≥ 1.
Proposed by Arkady Alt, San Jose, CA, USA
Solution by G. C. Greubel, Newport News, VA , USA
Consider the series
∞
X
n=0
an + 2
a2n + an + 1
where 3an+1 = a3n + 2. The reduction of the series is as follows.
S=
=
∞
X
n=0
∞
X
n=0
an + 2
a2n + an + 1
(an − 1)(an + 2)
a3n − 1
where 3an+1 = a3n + 2 was used. Now,
∞
X
a2n + an − 2
S=
3(an+1 − 1)
=
n=0
∞
X
n=0
∞
X
(an − 1)(a2n + an − 2)
3(an − 1)(an+1 − 1)
an+1 − an
(an − 1)(an+1 − 1)
n=0
∞ X
1
1
=
−
.
an − 1 an+1 − 1
=
n=0
The collapse of this telescopic series leads to the desired result, namely,
∞
X
a2
n=0 n
an + 2
1
=
.
+ an + 1
a0 − 1
Also solved by Ji Eun Kim, Tabor Academy, MA, USA; Michael Tang, Edina High School, MN, USA;
Kyoung A Lee, The Hotchkiss School, Lakeville, CT, USA; AN-anduud Problem Solving Group, Ulaanbaatar,
Mongolia; Yong Xi Wang,East China Institute Of Technology, China; Brian Bradie, Christopher Newport
University, Newport News, VA, USA; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Cailan Li; Haroun
Meghaichi, University of Science and Technology Houari Boumediene, Algiers, Algeria; Moubinool Omarjee
Lycée Henri IV, Paris, France; Paolo Perfetti, Universitá degli studi di Tor Vergata Roma, Roma, Italy;
Prithwijit De, HBCSE, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Alessandro Ventullo, Milan,
Italy.
Mathematical Reflections 1 (2015)
17
U327. Let (an )n≥0 be a sequence of real numbers with a0 = 1 and
an+1 =
n2 a
an
.
2
n + an + 1
Find the limit lim n3 an .
n→∞
Proposed by Khakimboy Egamberganov, Tashkent, Uzbekistan
Solution by Alessandro Ventullo, Milan, Italy
Observe that an+1
∞
X
1
an
< 2 = 2 , so
an converges by the Comparison Test. Let
n an
n
n=0
∞
X
an = c ∈ R.
n=0
We have
1
an+1
= n2 + an +
1
.
an
So,
1
1
= 02 + a0 +
a1
a0
1
1
= 12 + a1 +
a2
a1
.. .. ..
. . .
1
1
= (n − 1)2 + an−1 +
,
an
an−1
and summing the two columns, we get
n−1
(n − 1)n(2n − 1) X
1
=
+
ak + 1.
an
6
k=0
Hence,
1
(n − 1)n(2n − 1)
=
+
3
n an
6n3
Since
Pn−1
ak + 1
.
n3
k=0
Pn−1
lim
n→∞
ak + 1
c+1
= lim
= 0,
n→∞ n3
n3
k=0
we get
lim
n→∞
1
(n − 1)n(2n − 1)
1
= lim
= ,
n3 an n→∞
6n3
3
so
lim n3 an = 3.
n→∞
Also solved by Arkady Alt, San Jose, CA, USA; Ji Eun Kim, Tabor Academy, MA, USA; AN-anduud
Problem Solving Group, Ulaanbaatar, Mongolia; Yong Xi Wang,East China Institute Of Technology, China;
Brian Bradie, Christopher Newport University, Newport News, VA, USA; Haroun Meghaichi, University of
Science and Technology Houari Boumediene, Algiers, Algeria; Moubinool Omarjee Lycée Henri IV, Paris,
France; Paolo Perfetti, Universitá degli studi di Tor Vergata Roma, Roma, Italy; Albert Stadler, Herrliberg,
Switzerland.
Mathematical Reflections 1 (2015)
18
U328. Let (an )n≥1 be an increasing sequence of positive real numbers such that lim an = ∞ and the sequence
n→∞
(an+1 − an )n≥1 is monotonic. Evaluate
lim
n→∞
a1 + . . . + an
√
n an
Proposed by Mihai Piticari and Sorin Rădulescu, Romania
Solution by Paolo Perfetti, Universitá degli studi di Tor Vergata Roma, Roma, Italy
The sequence (an+1 − an )n≥1 converges to a limit L because of its monotonicity. Suppose L > 0 is finite.
By Cesaro–Stolz we have
an
L = lim (an+1 − an ) =⇒ L = lim
n→∞
n→∞ n
Moreover
√
L
an+1 − an
√
=0
lim ( an+1 − an ) = lim √
√ =
n→∞
n→∞
an+1 + an
∞
Also by Cesaro–Stolz we consider
lim
n→∞
(a1 + . . . + an+1 ) − (a1 + . . . + an )
an+1
= lim
√
√
√
√
n→∞ (n + 1) an+1 − n an
(n + 1) an+1 − n an
The reciprocal gives
√
√
√ (n + 1) an+1 − n an
an
√
n+1 √
lim
= lim
( an+1 − an ) +
n→∞
n→∞
an+1
an+1
an+1
and
√
√
n+1 √
1
lim
( an+1 − an ) = · 0 = 0,
n→∞ an+1
L
lim
n→∞
an
=0
an+1
The conclusion is
lim
n→∞
an+1
√ =∞
√
(n + 1) an+1 − n an
an+1
√ is positive via the increasing monotonicity of (an ) and an > 0.
√
(n + 1) an+1 − n an
an
Now let L = lim (an+1 − an ) = 0 and then 0 = lim
n→∞
n→∞ n
by Cesaro–Stolz. As above we consider
because
(a1 + . . . + an+1 ) − (a1 + . . . + an )
an+1
= lim
√
√
√
√
n→∞
n→∞ (n + 1) an+1 − n an
(n + 1) an+1 − n an
lim
that is
lim √
n→∞
an+1
an+1 n
√
√
an+1 − an
√
an+1
+1
Now
√ that
√ we prove
an+1 − an
n
is bounded above. We write
√
an+1
n
an+1 − an
n(an+1 − an )
√ ≤
√
√
an+1 an+1 + an
2an
Mathematical Reflections 1 (2015)
19
Lemma: Under the assumptions of the statement of the problem, the positive sequence
n(an+1 − an )
is
2an
bounded above.
Proof: Let’s argue by contradiction supposing that for any positive p there exists np such that
panp
nP (anp +1 − anp )
or anp +1 − anp ≥
. The monotonicity of
2anp
np
an+1 − an yields
an+1 − an ≥ anp +1 − anp ≥
panp
∀ 1 ≤ n ≤ np − 1
np
This implies
a1 = anp
np
X
panp
+
(ak − ak+1 ) < anp − np
= np (1 − p) < 0
np
k=1
and this is a contradiction with the positivity of the sequence {an }.
The consequence of the Lemma is the existence of a constant C such that 0 <
implies
lim
n→∞ √
an+1
√
an+1 n
√
an+1 − an
√
an+1
+1
≥√
n(an+1 − an )
≤ C. This
2an
an+1
=∞
an+1 (C + 1)
The last step is the case L = ∞.
This means that for all p > 0 there exists np such that n > np implies an+1 − an > p. It follows
np −1
an = a1 +
X
(ak+1 − ak ) +
k=1
≥ a1 + (np − 1)
n
X
(ak+1 − ak ) ≥
k=np
min
(ak+1 − ak ) + p(n − np + 1)
1≤k≤np −1
= a1 + (np − 1)(a2 − a1 ) + p(n − np + 1) ≥ p(n − np + 1) ≥ p
n
2
if n ≥ 2(np − 1) The monotonicity increasing of {an } has been used.
Based on this we can write
n
a1 + . . . + a2np −3 + a2np −2 + . . . + an ≥ (2np − 3)a1 + (n − 2np + 3)p ≥
2
2
n
n
≥ (n − 2np + 3)p ≥ p
2
4
a1 + . . . + an
for any n ≥ 4np + 6. Finally we put the above lover bounds in
and write
√
n an
√
√
√
a1 + . . . + an
a1 + . . . + an a1 + . . . + an
a1 + . . . + an √ n 1 √
=
≥
≥ p
= p
√
√
n an
n
an
n
2n
Since p can be as large as we want, this shows that
a1 + . . . + an
=∞
√
n→∞
n an
lim
also when L = ∞.
Also solved by Moubinool Omarjee Lycée Henri IV, Paris, France.
Mathematical Reflections 1 (2015)
20
U329. Let a1 ≤ a2 ≤ . . . ≤ a n(n−1) be the distances between n distinct points lying on the plane. Prove that
2
there is a constant c such that for any n we can find indices i and j such that
ai
− 1 < c ln n .
aj
n2
Proposed by Nairi Sedrakyan, Yerevan, Armenia
Solution by Ivan Borsenco, Massachusetts Institute of Technology, USA
ai
Assume that for any pair of indices (i, j) we have aj − 1 ≥ d and d is the greatest such number. Then
a n(n−1) ≥ (1 + d)a n(n−1) −1 ≥ . . . ≥ (1 + d)
2
n(n−1)
−1
2
a1 .
2
Let AB = a n(n−1) and CD = a1 . From the fact that AC + BC > AB, we can assume without loss of
2
generality that AC ≥ AB
2 . Then
min(AC, AD)
CD
2CD
2
CD
d≤
− 1 =
≤
≤
=
n(n−1)
max(AC, AD)
max(AC, AD)
AC
AB
(1 + d) 2 −1
and d(1 + d)
n(n−1)
−1
2
≤ 2.
If in our configuration AB = a n(n−1) and AC = a1 , then
2
BC
AC
1
− 1 ≤
=
d ≤ .
n(n−1)
AB
AB
(1 + d) 2 −1
In any case, we obtain d(1 + d)
n(n−1)
−1
2
≤ 2.
Assume that n ≥ 4 and consider f (x) = x(1 + x)
f
100 ln n
n2
n(n−1)
−1
2
100 ln n
=
n2
1+
Because
1+
1
!
. Observe that
! n(n−1) −1
2
1
.
n2
100 ln n
n2
+1
100 ln n
≥ e,
n2
100 ln n
we get
1+
Hence
f
100 ln n
n2
1
100 ln n 10nln2 n
≥
·e
n2
Therefore for c = 100, we always have d <
Mathematical Reflections 1 (2015)
100 ln n
≥ e n2 +100 ln n > e
n2
100 ln n
n(n−1)
−1
2
≥
10 ln n
n2
.
100 ln n 3 ln n
·e
= 100 ln n · n > 2.
n2
c ln n
.
n2
21
U330. Find all functions f : R → R with the properties
(i)
f is an antiderivative
(ii)
f is integrable on any compact interval
Z x
2
(iii) f (x) =
f (t)dt for all x ∈ R
0
Proposed by Mihai Piticari, Câmpulung Moldovenesc, Romania
Solution by Paolo Perfetti, Universitá degli studi di Tor Vergata Roma, Roma, Italy
We claim that f1 (x) ≡ 0 and f2 (x) = x/2 are solutions. We are going to prove that there are no other
solutions. Clearly f (0) = 0. If f is an antiderivative, it is differentiable so by (iii)
2f (x)f 0 (x) = f (x)
It follows that if f (x0 ) 6= 0 then f 0 (x0 ) = 1/2. Suppose now that there exists a function g(x) satisfying
(i)–(iii) but g(x) 6= f1 and g 6= f2 . This means that there exists a point ξ > 0 (or ξ < 0) such that g(ξ) 6= 0
and g(ξ) 6= ξ/2.
By the mean value theorem
g(ξ) − g(0)
= g 0 (η),
ξ−0
If g(η) 6= 0 we have g 0 (η) = 1/2 and then g(ξ) = ξ/2 which is impossible so we are forced to sat that
g(η) = 0.
Now, consider the following two Lemmas.
Lemma 1: If f satisfying (i)–(iii) is such that f (x1 ) = 0, x1 6= 0, (say x1 > 0) then f (x) = 0 for any
x ∈ [0, x1 ].
Proof: Suppose f (x2 ) 6= 0, 0 < x2 < x1 . By the mean value theorem it would follow that a value x3 would
exist such that f (x3 ) 6= 0 and f 0 (x3 ) < 0. but this is impossible.
Lemma 2 : If f satisfies (i)–(iii) and f (x) ≡ 0, for x ∈ [a, b] with a ≤ 0 ≤ b, a 6= b, the f (x) = 0 for any
x ∈ R.
Proof: Suppose indeed f (x2 ) 6= 0 and x2 > b. By f 0 (x2 ) = 1/2 and the fact that any derivative is a
"Darboux–function", it follows that f 0 (x) assumes all the value between the value 0 and 1/2 but this is
impossible.
The conclusion is that the function g(x) described above, does not exist.
Mathematical Reflections 1 (2015)
22
Olympiad problems
O325. The taxicab distance between points P1 = (x1 , y1 ) and P2 (x2 , y2 ) in a coordinate plane is given by
d(P1 P2 ) = |x1 − x2 | + |y1 − y2 |.
The taxicab disk with center O and radius R is the set of points P such that d(P, O) ≤ R. Given n
points that are pairwise at most R taxicab distance apart, find the smallest constant c such that any
such set of points can be covered by a taxicab disk of radius cR.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
Solution by the author
Assume that points are at most a unit taxicab distance apart. We are looking for the smallest c such that
any such set of points can be covered by the taxicab disk of radius c. Consider four points with coordinates
(± 41 , ± 14 ). Then they are at most one taxicab unit apart and only can be covered by a taxicab disk with
radius at least 21 . We prove that c = 21 is the smallest constant.
Consider a rotation of the plane with center at the origin and angle 45◦ . Any
taxicab disk becomes a
1
1
0
0
0
square and points Pi = (xi , yi ) become Pi = (xi , yi ) = √2 (xi + yi ), √2 (xi − yi ) . We are looking for the
least square that can cover our rotated set. Note that x-coordinates of any two points in the rotated set are
at most √12 apart:
1
1
1
√ |xi + yi − xj − yj | ≤ √ (|xi − xj | + |yi − yj |) ≤ √ .
2
2
2
Similarly, any two y-coordinates are at most
min(yi0 ) ≤
√1 .
2
√1
2
apart. We have max(x0i ) − min(x0i ) ≤
Thus rotated set can be covered by a
√1
2
√1
2
and max(yi0 ) −
× √12 square, which initially corresponds to a taxicab
disk of radius 12 .
Mathematical Reflections 1 (2015)
23
−−→
O326. Let ABC be a non-isosceles triangle. Let D be a point on the ray AB, but not on the segment AB,
−→
and let E be a point on the ray AC, but not on the segment AC, such that AB · BD = AC · CE.
The circumcircles of triangles ABE and ACD intersect in points A and F . Let O1 and O2 be the
circumcenters of triangles ABC and ADE. Prove that lines AF , O1 O2 , and BC are concurrent.
Proposed by İlker Can Çiçek, Instanbul, Turkey
Solution by Adnan Ali, Student in A.E.C.S-4, Mumbai, India
We first prove two Lemmas:
Lemma 1: O1 O2 is the perpendicular bisector of BC.
Proof: It is sufficient to show that O2 lies on the perpendicular bisector of BC. Let r be the radius of
(ADE). Then, power of B, C w.r.t to (ADE) is expressed as
AB · BD = r2 − O2 B 2 = AC · CE = r2 − O2 C 2
Thus O2 B = O2 C, proving that O2 lies on the perpendicular bisector of BC.
Lemma 2: AF bisects BC.
−−→
−−→
Proof: Let M = AF ∩ BC, BC ∩ (ABE) = Q, CB ∩ (ADC) = P . Then, power of C w.r.t to (ABE)
and power of B w.r.t (ADC) can be expressed as respectively:
AC · CE = QC · CB
= AB · BD = P B · BC
thus showing that P B = QC (1).
Now expressing the power of M w.r.t to (ABE)
AM · M F = QM · M B
and power of M w.r.t to (ADC) as
AM · M F = P M · M C.
Hence
P M · M C = QM · M B ⇒ P B · M C + BM · M C = QC · M B + M B · M C
⇒ P B · M C = QC · M B ⇒ M B = M C
(from (1)).
Hence AF bisects BC.
Now from Lemma 1 and Lemma 2 we conclude that AF, O1 O2 , BC concur at M , the midpoint of BC.
Also solved by Misiakos Panagiotis, Athens College(HAEF), Greece; Andrea Fanchini, Cantú, Italy; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Titu Zvonaru and Neculai Stanciu, Romania.
Mathematical Reflections 1 (2015)
24
O327. Let a, b, c be positive reals such that a2 + b2 + c2 + abc = 4. Prove that,
√
√
√
a+b+c≤ 2−a+ 2−b+ 2−c
Proposed by An Zhen-ping, Xianyang Normal University, China
Solution by Utsab Sarkar, Chennai Mathematical Institute, India
Notice that a, b, c ∈ (0, 2) and hence we’ll make the substitution for a = 2 cos A, b = 2 cos B, c = 2 cos C.
Therefore our constraint becomes to,
cos2 A + cos2 B + cos2 C + 2 cos A cos B cos C = 1 =⇒ A + B + C = π
Thus for a triangle ∆ABC we need only to prove,
X √
cyc
r
⇐⇒
X
cyc
X
A
sin − cos A ≥ 0
2 − a − a ≥ 0 ⇐⇒
2
cyc
X p
X
(a + b − c)(a + c − b) X b2 + c2 − a2
≥
⇐⇒
a bc(a + b − c)(a + c − b) ≥
(a2 b+a2 c−a3 )
bc
bc
cyc
cyc
cyc
Now using Ravi Transformation we see, a = y +z, b = z +x, c = x+y which provides an equivalent inequality
to prove,
X
X
p
(y + z) (x + y)(x + z)yz ≥
(y + z)2 x
cyc
cyc
Where x, y, z are positive reals.
By Cauchy–Schwarz Inequality we get,
(x + y)(x + z) ≥ (x +
√
yz)2
Now,it only remains to show,
X
X
X
√
√
√
(y + z) yz (x + yz) ≥
(y + z)2 x ⇐⇒
x(y + z) yz ≥ 6xyz
cyc
cyc
cyc
And the last inequality is true by AM-GM, since
X
X
X
√
√
√
x(y + z) yz ≥
x · 2 yz · yz =
2xyz = 6xyz
cyc
cyc
cyc
.
Also solved by Ji Eun Kim, Tabor Academy, MA, USA; Misiakos Panagiotis, Athens College(HAEF),
Greece; AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Yong Xi Wang,East China Institute Of Technology, China; Adnan Ali, Student in A.E.C.S-4, Mumbai, India; Arkady Alt, San Jose, CA,
USA; Behzod Ibodullaev, Lyceum TCTU, Tashkent, Uzbekistan; Farrukh Mukhammadiev, Academic Lyceum
Nr.1 under the SamIES, Samarkand, Uzbekistan; Nicuşor Zlota, “Traian Vuia” Technical College, Focşani, Romania; Paolo Perfetti, Universitá degli studi di Tor Vergata Roma, Roma, Italy; Piriyakorn, Triam
Udom Suksa School, Thailand; Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Shatlyk
Mamedov, School Nr. 21, Dashoguz, Turkmenistan; Titu Zvonaru and Neculai Stanciu, Romania.
Mathematical Reflections 1 (2015)
25
O328. The diagonals AD, BE, CF of the convex hexagon ABCDEF intersect at point M . Triangles ABM ,
BCM , CDM , DEM , EF M , F AM are acute. Prove that circumcenters of these triangles are concyclic
if and only if the areas of quadrilaterals ABDE, BCEF , CDF A are equal.
Proposed by Nairi Sedrakyan, Yerevan, Armenia
Solution by the author
Proof of necessity: Let O1 , O2 , O3 , O4 , O5 and O6 be the centers of the circumcircles of triangles
ABM, BCM, CDM, DEM, EF M and F AM , respectively. Notice that O1 , O2 and O4 , O5 are laying on the
perpendicular bisectors of BM and EM , respectively.
Therefore, O1 O2 ⊥ BE and O4 O5 ⊥ BE ⇒ O1 O2 k O4 O5 . Moreover, the projection of O1 O4 onto
BE has length 21 BE. Since O1 , O2 , O4 , O5 are concyclic by condition of the problem and O1 O2 k O4 O5 ⇒
O1 O4 = O2 O5 , and BE forms equal angles with O1 O4 and O2 O5 . By analogy we get O1 O4 = O3 O6 .
Consider T1 T4 , T2 T5 , T3 T6 sharing the same midpoint and T1 T4 = O1 O4 , T2 T5 = O2 O5 , T3 T6 = O3 O6 .
Note that T1 T2 T4 T5 is a rectangle, therefore, T1 T5 k BE and T1 T5 is the projection of T1 T4 onto T1 T5 ⇒
T1 T5 = 21 BE.
Similarly we get T1 T3 k AD, T1 T3 = 21 AD. Finally,
1
1
SABDE = AD · BE sin ∠AM B = · 2T1 T3 · 2T1 T5 sin ∠T3 T1 T5 = 4ST1 T3 T5 .
2
2
Proof of sufficiency: Again, let O1 , O2 , O3 , O4 , O5 and O6 be the centers of the circumcircles of triangles
ABM, BCM, CDM, DEM, EF M and F AM , respectively. Notice that O1 , O2 and O4 , O5 are laying on the
perpendicular bisectors of BM and EM , respectively.
Therefore, O1 O2 ⊥ BE and O4 O5 ⊥ BE ⇒ O1 O2 k O4 O5 . Let AD = 2a, BE = 2b, CF = 2c, ∠AM B =
γ, ∠BM C = α, ∠CM D = β. Since
α + β + γ = 180o
SABDE = 2ab sin γ, SBCEF = 2bs sin α, SCDF A = 2ac sin β
and
SABDE = SBCEF = SCDF A ⇒
sin β
sin γ
sin α
=
=
.
a
b
c
We deduce that α, β, γ are angles of a triangle with sides a, b and c. The lengths of projections of O1 O4
onto BE and AD are equal to 12 BE and 21 AD, respectively.
Mathematical Reflections 1 (2015)
26
Hence, O1 O4 = 2R, where R - radius of the circumcircle of the triagle with sidelengths a, b, c.
By analogy we get that O2 O5 = 2R, O3 O6 = 2R ⇒ O1 O4 = O2 O5 = O3 O6 .
Finally,
O1 O2 k O4 O5 , O2 O3 k O5 O6 , O3 O4 k O6 O1
and
O1 O4 = O2 O5 = O3 O6
yields that by condition of the problem, O1 , O2 , O3 , O4 , O5 and O6 are concyclic.
Also solved by Sardor Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan.
Mathematical Reflections 1 (2015)
27
O329. For any positive integer r, denote by Pr the path on r vertices. For any positive integer g, prove that
there exists a graph G with no cycles of length less than g with the following property: any two coloring
of the vertices of G contains a monochromatic copy of Pr .
Proposed by Cosmin Pohoata, Columbia University, USA
Solution by the author
Recall the celebrated result by Erdös that there exists graphs with arbitrarily large girth and arbitrarily
large chromatic number. More precisely, for any positive integers k and g, there is some graph G which
has no cycle of length less than g and for which χ(G) ≥ k. This was one of the first applications of the
probabilistic method in combinatorics. For the proof, see for example Lecture 11 from Daniel Spielman’s
graph theory class at Yale: http://www.cs.yale.edu/homes/spielman/462/2007/.
For our problem, we choose k = 4r + 1. Next, notice that if some vertex x ∈ V (G) has degree < 4r, then
we can use induction to argue that we can in fact color G using 4r colors. Indeed, remove x from G and
notice that any 4n-coloring of the new graph can be augmented to a coloring of G (since we can assign x
one of the 4n colors, provided that deg x < 4r). This contradicts the fact that χ(G) ≥ 4r + 1. Consequently,
every graph G with chromatic number at least 4n + 1 has a subgraph with a minimum degree vertex at least
4r. It follows that any two coloring of G contains a monochromatic subgraph of minimum degree at least r.
Within this monochromatic subgraph, it is easy to see that we have a path of length r; hence we found a
monochromatic copy of Pr .
Mathematical Reflections 1 (2015)
28
O330. Four segments divide a convex quadrilateral into nine quadrilaterals. The points of intersections of
these segments lie on the diagonals of the quadrilateral (see figure below).
It is known that quadrilaterals Q1 , Q2 , Q3 , Q4 admit inscribed circles. Prove that the quadrilateral Q5
also has an inscribed circle.
Proposed by Nairi Sedrakyan, Yerevan, Armenia
Solution by Cosmin Pohoata, Columbia University, USA
The following is an exposition of the proof from Cosmin Pohoata and Titu Andreescu, 110 Geometry
Problems for the International Mathematical Olympiad, XYZ Press, 2014.
The key idea is to make use of the following result due to Marius Iosifescu repeatedly.
Iosifescu’s Theorem. A convex quadrilateral ABCD admits an inscribed circle if and only if
tan
x
z
y
w
· tan = tan · tan ,
2
2
2
2
where x = ∠ABD, y = ∠ADB, z = ∠BDC and w = ∠DBC.
Proof. Using the trigonometric formula
u
1 − cos u
=
,
2
1 + cos u
we get that the equality in the theory is equivalent to
tan2
(1 − cos x)(1 − cos z)(1 + cos y)(1 + cos w)
= (1 − cos y)(1 − cos w)(1 + cos x)(1 + cos z).
Let a = AB, b = BC, c = CD, d = DA and q = BD. From the Law of Cosines, we have
cos x =
so that
a2 + q 2 − d2
,
2aq
1 − cos x =
d2 − (a − q)2
(d + a − q)(d − a + q)
=
2aq
2aq
1 + cos x =
(a + q)2 − d2
(a + q + d)(a + q − d)
=
.
2aq
2aq
and
In the same way
1 − cos y =
(a + d − q)(a − d + q)
,
2dq
Mathematical Reflections 1 (2015)
1 + cos y =
(d + q + a)(d + q − a)
,
2dq
29
1 − cos z =
(b + c − q)(b − c + q)
,
2cq
1 + cos z =
(c + q + b)(c + q − b)
,
2cq
1 − cos w =
(c + b − q)(c − b + q)
,
2bq
1 + cos w =
(b + q + c)(b + q − c)
.
2bq
Thus, after some rearranging, the identity we started with becomes
P (d − a + q)2 (b − c + q)2 − (a − d + q)2 (c − b + q)2 = 0,
where
d + a − q)(b + c − q)(d + q + a)(b + q + c)
16abcdq 4
is a positive expression according to the triangle inequality applied in triangles ABD and BCD. Factoring
the above and cancelling out terms in the parantheses yields
4qP (b + d − a − c) (d − a)(b − c) + q 2 = 0,
P =
where the expression in the second paranthesis can never be equal to zero - since, by the triangle inequality,
q > a − d and q > b − c, so q 2 is always either (strictly) less or greater than (a − d)(b − c). It thus follows
that
x
z
y
w
tan · tan = tan · tan
2
2
2
2
holds if and only if b + d − a − c = 0, i.e. b + d = a + c. By Pithot’s theorem, we therefore conclude that this
happens if and only if ABCD admits an inscribed circle.
In order to use Iosifescu’s theorem for quadrilateral 5, we aim to prove that
tan
∠F HG
∠BDC
∠ADB
∠EHF
· tan
= tan
· tan
.
2
2
2
2
However, quadrilateral 3 has an incircle, so
tan
∠EF H
∠F HG
∠EHF
∠HF G
· tan
= tan
· tan
.
2
2
2
2
It thus suffices to show that
tan
∠EF H
∠ADB
∠HF G
∠BDC
· tan
= tan
· tan
.
2
2
2
2
However, by applying Iosifescu’s lemma once again to quadrilateral 1, the above happens if and only if
tan
∠ABD
∠BDC
∠ADB
∠DBC
· tan
= tan
· tan
,
2
2
2
2
which we know is true if and only if ABCD has inscribed circle.
This is where the magic comes in. We show that ABCD has an inscribed circle by using Iosifescu’s
theorem for the other diagonal AC of ABCD. Namely, it suffices to show that
tan
∠BAC
∠ACD
∠BCA
∠CAD
· tan
= tan
· tan
.
2
2
2
2
This is however easy to prove in the same style as above. Indeed, by writing out Iosifescu’s lemma for
quadrilaterals 2, 3 and 4, and keeping in mind the vertical angles from the vertices of quadrilateral 3, we get
precisely the above statement. We leave the details for the reader.
Also solved by Adnan Ali, Mumbai, India, AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Farrukh Mukhammadiev, Academic Lyceum Nr.1 under the SamIES, Samarkand, Uzbekistan; Sardor
Bozorboyev, Lyceum S.H.Sirojjidinov, Tashkent, Uzbekistan; Shatlyk Mamedov, School Nr. 21, Dashoguz,
Turkmenistan; Titu Zvonaru and Neculai Stanciu, Romania.
Mathematical Reflections 1 (2015)
30