Juniorproblems - awesomemath.org · Juniorproblems J475.Let ABC be a triangle with ∠Band ∠C acute and diﬀerent from 45 . Let Dbe the foot of the altitudefromA. Provethat∠Aisrightifandonlyif

Junior problems

J475. Let ABC be a triangle with ∠B and ∠C acute and different from 45○. Let D be the foot of thealtitude from A. Prove that ∠A is right if and only if

1

AD −BD+

1

AD −CD=

1

AD.

Proposed by Adrian Andreescu, University of Texas at Austin, USA

First solution by Albert Stadler, Herrliberg, SwitzerlandLet AD = h,BD = x,CD = y. The stated equation then reads as h2 = xy.If ∠A is right then by the right triangle altitude theorem or geometric mean theorem we have h2 = xy.

Suppose that h2 = xy. Clearly, cot∠B =x

h, cot∠C =

y

h. Then

cot(∠B +∠C) =cot(∠B) cot(∠C) − 1

cot(∠B) + cot(∠C)= 0,

implying that ∠B +∠C =π

2and thus ∠A =

π

2.

Second solution by Joel Schlosberg, Bayside, NY, USAAD ≠ BD since otherwise △ABD would be an isosceles right triangle with ∠B = 45○; AD ≠ CD sinceotherwise △ACD would be a right triangle with ∠C = 45○. Thus 1

AD−BD + 1AD−CD = 1

AD is equivalent to

0 = AD(AD −CD) +AD(AD −BD) − (AD −BD)(AD −CD)

= AD2 −BD ⋅CD

and thus to AD ∶ BD = CD ∶ AD. This in turn is equivalent to △BAD ∼ △ACD since both triangleshave a right angle at vertex D; for the same reason it is further equivalent to ∠BAD and ∠ACD beingcomplementary (since ∠ACD is complementary to ∠BAD iff it is congruent to ∠ABD) and thus to ∠BACbeing right.

Also solved by Polyahedra, Polk State College, FL, USA; Daniel Lasaosa, Pamplona, Spain; Yi WonKim, Taft School in Watertown, CT, USA; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain;Corneliu Mănescu-Avram, Ploies,ti, Romania; Dumitru Barac, Sibiu, Romania; Ioannis D. Sfikas, Athens,Greece; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania; Miguel Amengual Covas, CalaFiguera, Mallorca, Spain; Santosh Kumar, Hyderabad, India; Titu Zvonaru, Comănes,ti, Romania; TakujiGrigorovich Imaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, Chennai Mathematical Institute, India;Nikos Kalapodis, Patras, Greece; Joonsoo Lee, Dwight Englewood School, NJ, USA; Min Jung Kim, Boston,MA, USA; Sebastian Foulger, Charters Sixth Form, Sunningdale, England, UK.

Mathematical Reflections 2 (2019) 1

J476. Let x, y, z be positive numbers such that x + y + z ≥ S. Prove that

6 (x3 + y3 + z3) + 9xyz ≥ S3.

Proposed by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain

Solution by Daniel Lasaosa, Pamplona, SpainLet σ = x+ y + z ≥ S, or it suffices to show that the LHS is not smaller than σ3. In turn, this is equivalent to

5 (x3 + y3 + z3) + 3xyz ≥ 3 (x2y + y2z + z2x + xy2 + yz2 + zx2) .

Schur’s inequality may be written in the form

x3 + y3 + z3 + 3xyz ≥ x2y + y2z + z2x + xy2 + yz2 + zx2.

Moreover, the AM-GM inequality produces x3 + y3 + z3 − 3xyz ≥ 0, with equality iff x = y = z, which is also asufficient condition for equality in Schur’s inequality. Adding thrice the first inequality and twice the second,results in the proposed result. The conclusion follows, equality holds iff x = y = z.

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Brian Bradie, Christopher New-port University, Newport News, VA, USA; Akash Singha Roy, Chennai Mathematical Institute, India; NikosKalapodis, Patras, Greece; Joonsoo Lee, Dwight Englewood School, NJ, USA; Min Jung Kim, Boston, MA,USA; Polyahedra, Polk State College, FL, USA; Duy Quan Tran, University of Medicine and Pharmacy, HoChi Minh, Vietnam; Arkady Alt, San Jose, CA, USA; Albert Stadler, Herrliberg, Switzerland; Aron Ban-Szabo, Budapest, Hungary; Dumitru Barac, Sibiu, Romania; Henry Ricardo, Westchester Area Math Circle,NY, USA; Ioannis D. Sfikas, Athens, Greece; Nguyen Viet Hung, Hanoi University of Science, Vietnam;Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Pantelis.N, Junior High School Athens,Greece; Titu Zvonaru, Comănes,ti, Romania.


J477. Find the area of a kite ABCD with AB −CD = (√

2 + 1) (√

3 + 1) and 11∠A =∠C.


Solution by the authorLet E be the intersection of diagonals AC and BD. Then ∠EAB = (15/2)○, so AE = BE cot(15/2)○. But

cot(x

2) =

cos(x/2)

sin(x/2)=

2 cos2(x/2)

2 sin(x/2)cos(x/2)=

1 + cosx

sinx,

so

cot(15

2)○

=1 + cos (15)

sin(15)=

1 +√

6+√

24

√

6−√

24

= 2 +√

2 +√

3 +√

6.

It follows that

1

2(√

2 + 1)(√

3 + 1) = AE −BE = (2 +√

2 +√

3 +√

6)BE −BE,

implying BE =1

2and AE =

1

2(2 +

√2 +

√3 +

√6).

Hence the area of the kite is1

2(2 +

√2 +

√3 +

√6).

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Daniel Lasaosa, Pamplona,Spain; Albert Stadler, Herrliberg, Switzerland.


J478. Prove that in any triangle ABC the following inequality holds:

4 (`a2 + `b

2 + `c2) ≤ (a + b + c)2.

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

Solution by Arkady Alt, San Jose, CA, USAWe have that

la =2bc

b + ccos

A

2=

2bc

b + c

√s(s − a)

bc=

2√bc

b + c

√s(s − a) ≤

√s(s − a).

Therefore,la ≤

√s(s − a)

and similarly lb ≤√s(s − b) and lc ≤

√s(s − c).

Hence,4(l2a + l

2b + l

2c) ≤ 4s(3s − a − b − c) = 4s2 = (2s)2 = (a + b + c)2.

Also solved by Telemachus Baltsavias, Keramies Junior High School, Kefalonia, Greece; Polyahedra, PolkState College, USA; Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Akash Singha Roy, Chennai Mathematical Institute, India; Duy Quan Tran, University of Medicineand Pharmacy, Ho Chi Minh, Vietnam; Arkady Alt, San Jose, CA, USA; Albert Stadler, Herrliberg, Swi-tzerland; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania; Aron Ban-Szabo, Budapest,Hungary; Corneliu Mănescu-Avram, Ploies,ti, Romania; Dumitru Barac, Sibiu, Romania; Ioan Viorel Co-dreanu, Satulung, Maramureş, Romania; Ioannis D. Sfikas, Athens, Greece; Miguel Amengual Covas, CalaFiguera, Mallorca, Spain; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Titu Zvonaru,Comănes,ti, Romania; Kevin Soto Palacios, Huarmey, Perú.


J479. Let a, b, c be nonzero real numbers, not all equal, such that

(a2

bc− 1)

3

+ (b2

ca− 1)

3

+ (c2

ab− 1)

3

= 3(a2

bc+b2

ca+c2

ab−bc

a2−ca

b2−ab

c2) .

Prove that a + b + c = 0.

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

Solution by Polyahedra, Polk State College, USALet x = a2

bc , y =b2

ca , and z =c2

ab . Then x, y, z are not all equal, xy = abc2, yz = bc

a2, zx = ca

b2, and xyz = 1. Thus,

the given equation becomes

0 = x3 + y3 + z3 − 3 (x2 + y2 + z2) − 3xyz + 3(xy + yz + zx)

=1

2(x + y + z − 3) [(x − y)2 + (y − z)2 + (z − x)2] .

Hence,

0 = x + y + z − 3 =a3 + b3 + c3 − 3abc

abc=

(a + b + c) [(a − b)2 + (b − c)2 + (c − a)2]

2abc,

so a + b + c = 0.

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Daniel Lasaosa, Pamplona,Spain; Akash Singha Roy, Chennai Mathematical Institute, India; Nikos Kalapodis, Patras, Greece; JoonsooLee, Dwight Englewood School, NJ, USA; Min Jung Kim, Boston, MA, USA; Arkady Alt, San Jose, CA,USA; Albert Stadler, Herrliberg, Switzerland; Aenakshee Roy, Mumbai, India; Dumitru Barac, Sibiu, Roma-nia; Ioannis D. Sfikas, Athens, Greece; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti, Romania;Santosh Kumar, Hyderabad, India; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; TituZvonaru, Comănes,ti, Romania.


J480. Let m,n be integers greater than 1. Find the number of ordered systems (a1, a2, . . . , am) where ai arenonnegative integers less than n and such that

a1 + a3 +⋯ ≡ a2 + a4 +⋯ (mod n + 1).

Proposed by Mircea Becheanu, Montreal, Canada

Solution by Daniel Lasaosa, Pamplona, SpainFor each ordered system (a1, a2, . . . , am), consider the number whose representation in base n is a1a2 . . . am.Clearly, there is a one-to-one correspondence between the ordered systems of length m and the nonnegativenumbers less than nm. Now, the condition given in the problem statement is clearly equivalent to a1a2 . . . ambeing a multiple of n+1, since n2u ≡ 1 (mod n+1) and n2u+1 ≡ −1 (mod n+1) for all nonnegative integer u.The problem is therefore equivalent to finding how many numbers in {0,1, . . . , nm −1} are multiples of n+1.

• If m is even, then nm − 1 is a multiple of n + 1, or since 0 is also a multiple of n + 1, the total numberof such ordered systems is nm

−1n+1 + 1.

• If m is odd, then nm + 1 is a multiple of n + 1, or the total number of such ordered systems is nm+1

n+1 .

Both forms may be combined into one, namely that the total number of such ordered system for any m is

⌊nm + n

n + 1⌋ .

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, ChennaiMathematical Institute, India; Sebastian Foulger, Charters Sixth Form, Sunningdale, England, UK; AlbertStadler, Herrliberg, Switzerland.


Senior problems

S475. Let a and b be positive real numbers such that

a3

b2+b3

a2= 5

√5ab.

Prove that√a

b+

√b

a=√

5.


Solution by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, JapanLet x =

√ab and y = x + 1

x , then we obtain

a3

b2+b3

a2= 5

√5ab

⇔ x5 +1

x5= 5

√5

⇔ (x +1

x)(x4 − x2 + 1 −

1

x2+

1

x4) = 5

√5

⇔ y(y4 − 5y2 + 5) = 5√

5

⇔ (y −√

5)(y4 +√

5y + 5) = 0.

Since y4 +√

5y + 5 > 0, y =√

5.

Also solved by Daniel Lasaosa, Pamplona, Spain; Brian Bradie, Christopher Newport University, NewportNews, VA, USA; Akash Singha Roy, Chennai Mathematical Institute, India; Joonsoo Lee, Dwight EnglewoodSchool, NJ, USA; Min Jung Kim, Boston, MA, USA; Arkady Alt, San Jose, CA, USA; Albert Stadler,Herrliberg, Switzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Dumitru Barac, Sibiu, Romania; JoelSchlosberg, Bayside, NY, USA; Ioannis D. Sfikas, Athens, Greece; Marin Chirciu, Colegiul Nat,ional ZincaGolescu, Pites,ti, Romania; Miguel Amengual Covas, Cala Figuera, Mallorca, Spain; Nicuşor Zlota, TraianVuia Technical College, Focşani, Romania; Oana Prajitura, College at Brockport, SUNY, NY, USA; SantoshKumar, Hyderabad, India; Spyridon Kalloniatis, Shape American High School; Titu Zvonaru, Comănes,ti,Romania; Kevin Soto Palacios, Huarmey, Perú.


S476. Prove that in any triangle ABC,

4 cosA + π

4cos

B + π

4cos

C + π

4≥

√r

2R.

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

Solution by Arkady Alt, San Jose, CA, USANoting that

4∏cyc

cosA + π

4= 4∏

cyc

sin(π −A

4) = 4∏

cyc

sinB +C

4= sin

A

2+ sin

B

2+ sin

C

2− 1

and√

r

2R=

√

2 ⋅r

4R=

√

2 sinA

2sin

B

2sin

C

2

we can rewrite original inequality as follows:

sinA

2+ sin

B

2+ sin

C

2− 1 ≥

√

2 sinA

2sin

B

2sin

C

2. (1)

Let α ∶=π −A

2, β ∶=

π −B

2, γ ∶=

π −C

2.

Then α,β, γ ∈ (0,π

2) , α + β + γ = π and (1)⇐⇒

cosα + cosβ + cosγ − 1 ≥√

2 cosα cosβ cosγ. (2)

Let A1B1C1 be some triangle with angles α,β, γ and let s,R and r from nowbe, respectively, semiperimeter, circumradius and inradius of △A1B1C1.

Then, since cosα + cosβ + cosγ = 1 +r

Rand cosα cosβ cosγ =

s2 − (2R + r)2

4R2

we obtain that

(2) ⇐⇒r

R≥

√s2 − (2R + r)2

2R2⇐⇒

r2

R2≥s2 − (2R + r)2

2R2⇐⇒

s2 − (2R + r)2 ≤ 2r2 ⇐⇒ s2 ≤ 4R2 + 4Rr + 3r2 (Gerretsen’s Inequality).

Also solved by Daniel Lasaosa, Pamplona, Spain; Nikos Kalapodis, Patras, Greece; Takuji GrigorovichImaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, Chennai Mathematical Institute, India; IoannisD. Sfikas, Athens, Greece; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania; Kevin SotoPalacios, Huarmey, Perú.


S477. Let ABCD be a kite inscribed in a circle such that

2AB2 +AC2 + 2AD2 = 4BD2.

Prove that ∠A = 4∠C.


Solution by Daniel Lasaosa, Pamplona, SpainClearly, AC = 2r is a diameter of the circle circumscribed to the kite, ∠ABC = ∠CDA = 90○, and denotingγ = ∠DCA = ∠ACB, we have ∠BAC = ∠CAD = 90○ − γ, or denoting by P the intersection of bothdiagonals,

AB = AD = 2r sinγ, BD = 2PB = 2AB cosγ = 4r sinγ cosγ,

and the proposed condition rewrites as

0 = 1 − 12 sin2 γ + 16 sin4 γ.

Now, by the de Moivre formulae, any angle γ verifies that 5γ is an odd multiple of π2 iff

0 = cos(5γ) = cosγ (cos4 γ − 10 cos2 γ sin2 γ + 5 sin4 γ) =

= cosγ (1 − 12 sin2 γ + 16 sin4 γ) ,

or since 0 < γ < π2 , we have γ ∈ { π

10 ,3π10

}. If γ = π10 , then C = 2γ = π

5 and A = π −C = 4π5 = 4C, or it suffices to

rule out the case γ = 3π10 . Note that, in this case, ∠BAD = π − 2γ = γ = 2π

5 < π2 , we have AB2 = AD2 > 2r2,

or 2AB2 +AC2 +2AD2 > 12r2, whereas since BD = 2r sin(2γ) < 2r sin 60○ =√

3r, then 4BD2 < 12r2, and theproposed condition is not met when γ = 3π

10 . The conclusion follows.

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, ChennaiMathematical Institute, India; Joonsoo Lee, Dwight Englewood School, NJ, USA; Min Jung Kim, Boston,MA, USA; Albert Stadler, Herrliberg, Switzerland; Marin Chirciu, Colegiul Nat,ional Zinca Golescu, Pites,ti,Romania; Santosh Kumar, Hyderabad, India; Titu Zvonaru, Comănes,ti, Romania.


S478. Let a, b, c be positive numbers. Prove that

a

b + c+

b

c + a+

c

a + b+ 4(

a

2a + b + c+

b

a + 2b + c+

c

a + b + 2c) ≥

9

2.

Proposed by Titu Zvonaru, Comăneşti, Romania

Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USAUsing the Cauchy-Schwarz inequality,

∑cyclic

a

b + c+ 4 ∑

cyclic

a

2a + b + c= ∑

cyclic

a2

a(b + c)+ ∑cyclic

4a2

2a2 + a(b + c)

≥9(a + b + c)2

2(a2 + b2 + c2) + 4(ab + bc + ca)

=9(a + b + c)2

2(a + b + c)2=

9

2.

Also solved by Pantelis.N, Junior High School Athens, Greece; Daniel Lasaosa, Pamplona, Spain; TakujiGrigorovich Imaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, Chennai Mathematical Institute, India;Nikos Kalapodis, Patras, Greece; Arkady Alt, San Jose, CA, USA; Albert Stadler, Herrliberg, Switzerland;Adamopoulos Dionysios, 3rd High School, Pyrgos, Greece; Ioan Viorel Codreanu, Satulung, Maramureş,Romania; Ioannis D. Sfikas, Athens, Greece; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain;Maalav Mehta, Ahmedabad, India; Nguyen Viet Hung, Hanoi University of Science, Vietnam; Nicuşor Zlota,Traian Vuia Technical College, Focşani, Romania; Kevin Soto Palacios, Huarmey, Perú.


S479. Let a1, a2, . . . , an be nonnegative real numbers and let (i1, i2, . . . , in) be a permutation of the numbers(1,2, . . . , n) such that ik /= k, for all k = 1,2, . . . , n. Prove that

an1ai1 + an2ai2 + ⋅ ⋅ ⋅ + a

nnain ≥ a1a2 . . . an(a1 + a2 + ⋅ ⋅ ⋅ + an).


Solution by the authorFor convenience, we introduce the following notations: P = a1a2 . . . an and Pi = P /ai. The inequality isequivalent to the following:

an1Pi1

+a2Pi2

+ ⋅ ⋅ ⋅ +anPin

≥ a1 + a2 + ⋅ ⋅ ⋅ + an. (1)

For every i, Pi is a product of n − 1 factors and we can write it as

Pi = ( n−1√a1...ai−1ai+1...an)

n−1.

Now, we use the following lemma:Lemma. For every positive integer n > 1 and every positive numbers a1, . . . , ak,b1, . . . , bk, the following inequality holds:

an1bn−11

+ ⋅ ⋅ ⋅ +ankbn−1k

≥(a1 + ⋅ ⋅ ⋅ + ak)

n

(b1 + ⋅ ⋅ ⋅ + bk)n−1.

This lemma is more or less known and it can be proven elementary by induction on k, or by using Holderinequality.Using Lemma in (1) we obtain the inequality:

an1Pi1

+a2Pi2

+ ⋅ ⋅ ⋅ +anPin

≥(a1 + ⋅ ⋅ ⋅ + an)

n

( n−1√Pi1 + ⋅ ⋅ ⋅ +

n−1√Pin)

n−1. (2)

For evrey i we have by AM-GM

n−1√Pi ≤

a1 + ⋅ ⋅ ⋅ + ai−1 + ai+1 + ⋅ ⋅ ⋅ + ann − 1

.

Using these in (2)we obtain

(a1 + ⋅ ⋅ ⋅ + an)n

( n−1√Pi1 + ⋅ ⋅ ⋅ +

n−1√Pin)

n−1≥

(a1 + ⋅ ⋅ ⋅ + an)n

(a1 + ⋅ ⋅ ⋅ + an)n−1= a1 + ⋅ ⋅ ⋅ + an. (3)

Also solved by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan.


S480. Let a, b, c be positive real numbers. Prove that

a (a3 + b3)

a2 + ab + b2+b (b3 + c3)

b2 + bc + c2+c (c3 + a3)

c2 + ca + a2≥

2

3(a2 + b2 + c2) .

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Nicuşor Zlota, Traian Vuia Technical College, Focşani, RomaniaFirst, we guess the equality holds when a = b = c. We want to find a result in the following form

a(a3 + b3)

a2 + ab + b2≥ma2 + nb2

This inequality is equivalent to

a4 + ab3 − (ma4 +ma3b +ma2b2 + na2b2 + nab3 + nb4) ≥ 0⇔

(1 −m)a4 −ma3b − (m + n)a2b2 + (1 − n)ab3 − nb4 ≥ 0

Consider the polynomial f(a) = (1 −m)a4 −ma3b − (m + n)a2b2 + (1 − n)ab3 − nb4

We will find two real numbers m,n for which a = b is a double root of f(a).This happens when f ′(a) = 4(1 −m)a3 − 3ma2b − 2(m + n)ab2 + (1 − n)b3

f(b) = 0 results in 3m + 3n = 2f ′(b) = 0 results in 9m + 3n = 5

Therefore, m =1

2, n =

1

6. We will show that

a(a3 + b3)

a2 + ab + b2≥

1

2a2 +

1

6b2

Indeed, this inequality is equivalent to

3a4 − 3a3b − 4a2b2 + 5ab3 − b4 ≥ 0⇔ (a − b)2(3a2 + 3ab − b2) ≥ 0

b(b3 + c3)

b2 + bc + c2≥

1

2b2 +

1

6c2

Similarly, we havec(c3 + a3)

c2 + ca + a2≥

1

2c2 +

1

6a2

Summing up these relations concludes the proof.

Also solved by Daniel Lasaosa, Pamplona, Spain; Albert Stadler, Herrliberg, Switzerland; Ioannis D.Sfikas, Athens, Greece; Pantelis.N, Junior High School Athens, Greece.


Undergraduate problems

U475. Evaluatelimx→0

sin(x sinx) + sin(x sin(x sinx))

x sin(sinx) + sin(sin(x sinx))


First solution by Thiago Landim de Souza Leão, Federal University of Pernambuco, Recife, BrazilSince sinx = x + o(x), then x sin(sinx) = sin(x sinx) = sin(sin(x sinx)) = x2 + o(x2) and sin(x sin(x sinx)) =x3 + o(x3) = o(x2). Therefore

limx→0

sin(x sinx) + sin(x sin(x sinx))

x sin(sinx) + sin(sin(x sinx))=x2 + o(x2)

2x2 + o(x2)=

1

2.

Second solution by Mircea Becheanu, Montreal, CanadaWe divide the numerator and denominator by x2 and we obtain the following limit:

l = limx→0

sin(x sinx)x2

+sin(x sin(x sinx))

x2

x sin(sinx)x2

+sin(sin(x sinx))

x2

One evaluates separately:

limx→0

sin(x sinx)

x2= limx→0

sin(x sinx)

x sinx.x sinx

x2= 1,

limx→0

sin(x sin(x sinx))

x2= limx→0

sin(x sin(x sinx))

x sin(x sinx).x sin(x sinx)

x sinx.x sinx

x2= 0,

limx→0

x sin(sinx)

x2= limx→0

sin(sinx)

x= limx→0

sin(sinx)

sinx.sinx

x= 1,

limx→0

sin(sin(x sinx))

x2= limx→0

sin(sin(x sinx))

sin(x sinx).sin(x sinx)

x sinx.x sinx

x2= 1.

When we gather together these limits we have:

l =1 + 0

1 + 1=

1

2.

Also solved by Daniel Lasaosa, Pamplona, Spain; Brian Bradie, Christopher Newport University, NewportNews, VA, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Akash Singha Roy, Chennai Ma-thematical Institute, India; Joonsoo Lee, Dwight Englewood School, NJ, USA; Min Jung Kim, Boston, MA,USA; Sebastian Foulger, Charters Sixth Form, Sunningdale, England, UK; Kevin Soto Palacios, Huarmey,Perú; Arkady Alt, San Jose, CA, USA; Albert Stadler, Herrliberg, Switzerland; Yi Won Kim, Taft Schoolin Watertown, CT, USA; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Dumitru Barac,Sibiu, Romania; G. C. Greubel, Newport News, VA, USA; Henry Ricardo, Westchester Area Math Circle,NY, USA; Joel Schlosberg, Bayside, NY, USA; Ioannis D. Sfikas, Athens, Greece; Lukas Seier, ChartersSixth Form, Sunningdale, England, UK; Moubinool Omarjee, Lycée Henri IV, Paris, France; Oana Prajitu-ra, College at Brockport, SUNY, NY, USA; Santosh Kumar, Hyderabad, India; Sarah B. Seales, Prescott,AZ, USA; Khakimboy Egamberganov, ICTP.


U476. Evaluate

∫x(x + 1)(4x − 5)

x5 + x − 1dx


Solution by Daniel Lasaosa, Pamplona, SpainNote that x5 + x − 1 = (x3 + x2 − 1) (x2 − x + 1), or

x(x + 1)(4x − 5)

x5 + x − 1=

3

2⋅

2x − 1

x2 − x + 1+√

3 ⋅2

√3⋅

1

1 + (2x−1√

3)2−

3x2 + 2x

x3 + x2 − 1.

Therefore,

∫x(x + 1)(4x − 5)

x5 + x − 1dx =

3

2log ∣x2 − x + 1∣ +

√3 arctan(

2x − 1√

3) − log ∣x3 + x2 − 1∣ +C,

where C is an integration constant.

Also solved by Pedro Acosta De Leon, Massachusetts Institute of Technology, Cambridge, MA, USA;Brian Bradie, Christopher Newport University, Newport News, VA, USA; Takuji Grigorovich Imaiida, Fu-jisawa, Kanagawa, Japan; Akash Singha Roy, Chennai Mathematical Institute, India; Joonsoo Lee, DwightEnglewood School, NJ, USA; Min Jung Kim, Boston, MA, USA; Sebastian Foulger, Charters Sixth Form,Sunningdale, England, UK; Kevin Soto Palacios, Huarmey, Perú; Khakimboy Egamberganov, ICTP; ArkadyAlt, San Jose, CA, USA; Albert Stadler, Herrliberg, Switzerland; Yi Won Kim, Taft School in Watertown,CT, USA; Corneliu Mănescu-Avram, Ploies,ti, Romania; G. C. Greubel, Newport News, VA, USA; HenryRicardo, Westchester Area Math Circle, NY, USA; Ioannis D. Sfikas, Athens, Greece; Lukas Seier, ChartersSixth Form, Sunningdale, England, UK; Moubinool Omarjee, Lycée Henri IV, Paris, France; Nicuşor Zlota,Traian Vuia Technical College, Focşani, Romania; Oana Prajitura, College at Brockport, SUNY, NY, USA;Sarah B. Seales, Prescott, AZ, USA.


U477. Evaluate

limn→∞

π2

6 −∑nk=11n2

log (1 + 1n)

Proposed by Tiago Landim de Sousa Leão, University of Pernambuco, Brasil

Solution by Ángel Plaza, University of Las Palmas de Gran Canaria, SpainLet L be the proposed limit. We shall show that L = 1.

L = limn→∞

n (π2

6 −∑nk=11n2 )

n log(1 + 1n)

= limn→∞

n(π2

6−

n

∑k=1

1

n2)

= limn→∞

π2

6 −∑nk=11n2

1n

= limn→∞

1n2

1n −

1n+1

(by Stolz-Cezaro Lemma)

= 1

Also solved by Daniel Lasaosa, Pamplona, Spain; Brian Bradie, Christopher Newport University, NewportNews, VA, USA; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan; Sebastian Foulger, Charters SixthForm, Sunningdale, England, UK; Khakimboy Egamberganov, ICTP; Albert Stadler, Herrliberg, Switzerland;Dumitru Barac, Sibiu, Romania; G. C. Greubel, Newport News, VA, USA; Joel Schlosberg, Bayside, NY,USA; Ioannis D. Sfikas, Athens, Greece; Nicuşor Zlota, Traian Vuia Technical College, Focşani, Romania.


U478. Let n be a positive integer. Prove that

n

∏k=1

(1 + tan4 kπ

2n + 1)

is a sum of two perfect squares.

Proposed by Dorin Andrica, Babeş-Bolyai University, Cluj-Napoca, Romania

Solution by Daniel Lasaosa, Pamplona, SpainBy the De Moivre formulae, we have

sin ((2n + 1)α) = sinα cos2n αn

∑u=0

(2n + 1

2u + 1)(−1)u tan2u α = sinα cos2n αp (tan2 α) ,

where p(x) is a polinomial with integer coefficients. Note that when α = kπ2n+1 with k = 1,2, . . . ,2n, the LHS

vanishes, or tan2 kπ2n+1 is a root of p(x) for k = 1,2, . . . ,2n. Moreover, k and 2n + 1 − k produce the same

absolute value of tanα with opposite signs, or p(x) has exactly n roots, this n roots being equal to the nvalues of tan2 kπ

2n+1 for k = 1,2, . . . , n. In other words, and since the coefficient of tan2n α is (−1)n, we maywrite

p(x) = (−1)nn

∏k=1

(x − tan2 kπ

2n + 1) .

Taking the modulus squared of p(i), where i is the imaginary unit, we obtain

∣p(i)∣2 =n

∏k=1

∣i − tan2 kπ

2n + 1∣ =

n

∏k=1

(1 + tan4 kπ

2n + 1) .

But p(x) has integer coefficients, or p(i) = r + is where r, s are integers, and indeed

n

∏k=1

(1 + tan4 kπ

2n + 1) = ∣p(i)∣2 = r2 + s2.

The conclusion follows.

Also solved by Ioannis D. Sfikas, Athens, Greece; Albert Stadler, Herrliberg, Switzerland; Akash SinghaRoy, Chennai Mathematical Institute, India; Khakimboy Egamberganov, ICTP.


U479. Evaluate∞

∑n=1

∞

∑m=1

(−1)n+mx2(n+m)

(2n + 2m)!.

Proposed by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain

Solution by Brian Bradie, Christopher Newport University, Newport News, VA, USABecause the series is absolutely convergent, we may sum terms in any order. Summing the terms withn +m = k for increasing k yields

∞

∑n=1

∞

∑m=1

(−1)m+nx2(n+m)

(2n + 2m)!=

∞

∑k=2

(−1)k(k − 1)x2k

(2k)!

=x

2

∞

∑k=2

(−1)kx2k−1

(2k − 1)!−

∞

∑k=2

(−1)kx2k

(2k)!

= −x

2

∞

∑k=1

(−1)kx2k+1

(2k + 1)!−

∞

∑k=2

(−1)kx2k

(2k)!

= −x

2(sinx − x) − (cosx − 1 +

x2

2)

= 1 − cosx −x

2sinx.

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Akash Singha Roy, Chennai Mathematical Institute, India; Joonsoo Lee, Dwight Englewood School,NJ, USA; Min Jung Kim, Boston, MA, USA; Khakimboy Egamberganov, ICTP; Albert Stadler, Herrliberg,Switzerland; G. C. Greubel, Newport News, VA, USA; Joe Simons, Utah Valley University, UT, USA; IoannisD. Sfikas, Athens, Greece; Thiago Landim de Souza Leão, Federal University of Pernambuco, Recife, Brazil.


U480. Let⎛⎜⎝

4 −3 215 −10 610 −6 3

⎞⎟⎠

Find the least possible n for which one entry of An is 2019.


First solution by Takuji Grigorovich Imaiida, Fujisawa, Kanagawa, Japan

⎛⎜⎝

4 −3 215 −10 610 −6 3

⎞⎟⎠

n

=1

4

⎛⎜⎝

0 4 02 12 03 8 2

⎞⎟⎠

⎛⎜⎝

−1 0 00 −1 10 0 −1

⎞⎟⎠

n⎛⎜⎝

−6 2 01 0 05 −3 2

⎞⎟⎠

=1

4

⎛⎜⎝

0 4 02 12 03 8 2

⎞⎟⎠

⎛⎜⎝

(−1)n 0 00 (−1)n (−1)n−1n0 0 (−1)n

⎞⎟⎠

⎛⎜⎝

−6 2 01 0 05 −3 2

⎞⎟⎠

=⎛⎜⎝

5(−1)n−1n + (−1)n 3(−1)nn 2(−1)n−1n15(−1)n−1n (−1)n − 9(−1)n−1n 6(−1)n−1n10(−1)n−1n −6(−1)n−1n 4(−1)n−1n + (−1)n

⎞⎟⎠.

Since 2 ∤ 2019 and 5 ∤ 2019, it is sufficient to investigate 4 entries, 5(−1)n−1n + (−1)n,3(−1)nn, (−1)n −9(−1)n−1n and 4(−1)n−1n + (−1)n. After annoying calculation, we obtain that only when n = 505, 2019appears as an entry of the matrix. Therefore the desired n = 505.

Second solution by the authorLet A = B − I3. We notice that B2 = 03, hence

An = (B − I3)n = (−1)(n−1)nB + (−1)nI3.

The conclusion follows as in the first solution.

Also solved by Daniel Lasaosa, Pamplona, Spain; Joonsoo Lee, Dwight Englewood School, NJ, USA;Min Jung Kim, Boston, MA, USA; Sebastian Foulger, Charters Sixth Form, Sunningdale, England, UK;Khakimboy Egamberganov, ICTP; Albert Stadler, Herrliberg, Switzerland; Dumitru Barac, Sibiu, Romania;G. C. Greubel, Newport News, VA, USA; Joe Simons, Utah Valley University, UT, USA; Ioannis D. Sfikas,Athens, Greece; Lukas Seier, Charters Sixth Form, Sunningdale, England, UK; Oscar Brown, ChartersSchool, Sunningdale, UK.


Olympiad problems

O475. Let a, b, c be nonnegative real numbers such that ab+c ≥ 2. Prove that

(ab + bc + ca) [1

(a + b)2+

1

(b + c)2+

1

(c + a)2] ≥

49

18.

Proposed by Marius Stănean, Zalău, Romania

Solution by Daniel Lasaosa, Pamplona, SpainTaking the first derivative of the LHS with respect to a we find it to be

1

b + c−ab + bc + ca − b2

(a + b)3−ab + bc + ca − c2

(c + a)3.

Now, when ab+c ≥ 2, the first term is at least 2

a ≥1a+b +

1c+a , or the first derivative is at least

1

a + b−ab + bc + ca − b2

(a + b)3+

1

c + a−ab + bc + ca − c2

(c + a)3=

=(a + b)(a − c) + 2b2

(a + b)3+

(c + a)(a − b) + 2c2

(c + a)3,

clearly non negative, being zero only when a = b = c = 0, which results in the proposed expression not havinga real value. It thus follows that we just need to prove the inequality when a = 2(b + c), and a necessarycondition for equality is a = 2(b + c).

When a = 2(b + c), and denoting s = b + c and p = bc, the inequality rewrites as

98s6 + 69ps4 + 12p2s2 + p3

36s6 + 12ps4 + p2s2≥

49

18,

which rewrites asp (654s4 + 167ps2 + 18p2) ≥ 0,

clearly true and with equality iff p = 0, ie iff bc = 0. Note that since a = 2(b + c), if b = c = 0 the proposedexpression does not have a real value, or equality holds iff either (a, b, c) = (2k, k,0) or (a, b, c) = (2k,0, k),for any positive real k. The conclusion follows.

Also solved by Albert Stadler, Herrliberg, Switzerland; Akash Singha Roy, Chennai Mathematical Institute,India; Ioannis D. Sfikas, Athens, Greece.


O476. Let a, b, c be nonnegative real numbers such that a + b + c = 3. Prove that

(a2 − ab + b2)(b2 − bc + c2)(c2 − ca + a2) + 11abc ≤ 12.

Proposed by An Zhenping, Xianyang Normal University, China

Solution by Corneliu Mănescu-Avram, Ploies,ti, Romania and Nicuşor Zlota, Traian Vuia TechnicalCollege, Focşani, Romania

Denote a + b + c = p = 3, ab + bc + ca = q, abc = r. After expanding, we can rewrite the given inequality as

P = (a2 − ab + b2)(b2 − bc + c2)(c2 − ca + a2) = a2b2c2 − abc∑a3 +∑a2b2(a2 + b2) −∑a3b3

P = r2 − r(p3 − 3pq + 3r) + (p2 − 2q)(q2 − 2pr) − 3r2 − (q3 − 3pqr + 3r2) =

−8r2 − 810r + 30qr − 3q3 + 9q2.

The inequality equivalent to

−8r2 − 810r + 30qr − 3q3 + 9q2 + 11r − 12 ≤ 0 or

8r2 + 10r(7 − 3q) + 3(q3 − 3q2 + 4) ≥ 0

Puttingf(r) = 8r2 + 10r(7 − 3q) + 3(q3 − 3q2 + 4)

Consider two cases:

0 ≤ q ≤35

8⇒ rct ≤ 0. (1)

Then f(0) = 3(q3 − 3q2 + 4) = 3(q − 2)2(q + 1) ≥ 0

∆ = 100(7 − 3q)2 − 96(q3 − 3q2 + 4) = 4(−24q3 + 297q2 − 1050q + 1127) < 0. (2)

So, f(r) ≥ 0 and the conclusion follows.

Also solved by Albert Stadler, Herrliberg, Switzerland; Akash Singha Roy, Chennai Mathematical Institute,India; Ioannis D. Sfikas, Athens, Greece.


O477. Solve in integer numbers the equation

(x2 − 3)(y3 − 2) + x3 = 2(x3y2 + 2) + y2.

Proposed by Konstantinos Metaxas, Athens, Greece

Solution by the authorThe only solutions of the Diophantine equation

(x2 − 3)(y3 − 2) + x3 = 2(x3y2 + 2) + y2. (1)

are (x, y) = (−2,−1) and (x, y) = (1,−1).Assume y is even. Then x is even by (3), which according to (3) implies 22∣2. This contradiction means

y is odd. Equation (1) is equivalent to

(2y2 − 1)x3 + (2 − y3)x2 + 3y3 + y2 − 2 = 0 (2)

and(x2 − 3)y3 − (2x3 + 1)y2 + x3 − 2x2 + 2 = 0. (3)

If x = 0, then 3y3 + y2 = 2, yielding y2∣2, implying y = ±1 and 2 = 3(±1)3 + (±1)2 = ±3 + 1, which isimpossible. Hence x ≠ 0, meaning ∣x∣, ∣y∣ ≥ 1 (since y is odd). According to (2) and (3) we have

x2∣3y3 + y2 − 2 (4)

andy2∣x3 − 2x2 + 2. (5)

Apllying (2) - (5), wend that the only solutions of equation (1) when ∣x∣, ∣y∣ ≤ 4 are (x, y) = (−2,−1) and (x, y) = (1,−1).

Assume ∣x∣, ∣y∣ ≥ 5. By (4) and (5) we obtain

x2 ≤ ∣3y3 + y2 − 2∣ < 3∣y∣3 + ∣y∣2 < 4∣y∣3

andy2 ≤ ∣x3 − 2x2 + 2∣ < ∣x∣3 + 2∣x∣2 < 2∣x∣3

respectively, implying∣x∣ < 2∣y∣

32 (6)

and∣y∣ <

√2∣x∣

32 . (7)

Now, equation (1) can be expressed as

(y − 2x)(xy)2 = −x3 + 2x2 + 3y3 + y2 − 2. (8)

Dividing both sides of equation (8) with (xy)2, the result is

y − 2x = −x

y2+

3y

x2+

1

x2+

2

y2−

2

(xy)2

yielding

∣y − 2x∣ <∣x∣

∣y∣2+

3∣y∣

∣x∣2+

1

∣x∣2+

2

∣y∣2(9)

yielding∣y − 2x∣ < 3. (10)

Now, y − 2x is odd since y is odd. Hence y − 2x = ±1 by (10), i.e. y = 2x ± 1, which inserted in (8) result in

±x2(2x ± 1)2 = −x3 + 2x2 + 3(2x ± 1)3 + (2x ± 1)2 − 2 (11)

Consequently by (11)x∣ ± 3 + 1 − 2 = −1 ± 3

meaning ∣x∣ ≤ 3 + 1 = 4. This contradiction (since ∣x∣ ≥ 5) means there are no solutions to equation (1) when∣x∣, ∣y∣ ≥ 5. Conclusion: Equation (1) has exactly two solutions, namely (x, y) = (−2,−1) and (x, y) = (1,−1).


O478. Let ABCDEF be a cyclic hexagon inscribed in a circle of radius 1. Suppose that the diagonalsAD,BE,CF are concurrent in the point P . Prove that

AB +CD +EF ≤ 4.

Proposed by Miguel Amengual Covas, Cala Figuera, Mallorca, Spain

Solution by the author

C

B

A

F E

DP

Suppose w.l.o.g. that EF ≥max {AB,CD}. Then EF ≥ AB and EF ≥ CD.Consider the case EF ∥ AD; see Figure 1. Since AB ≤ EF , we have arc AB ≤ arc EF implying α ≤ γ

and henceδ = α + β (by the external angle theorem applied to △AEP at P )

≤ γ + β.

C

B

A

F E

DP

β

β

β

α

γ

δ

EF ∥ AD

Figure 1

Thusδ ≤ γ + β

which implies (in △ABP )

AB ≤ AP.

Similarly,CD ≤ PD.


Therefore,AB +CD ≤ AP + PD = AD ≤ 2,

where the last inequality holds because AD ≤ diameter.

Next consider that the points A, B, C, and D are fixed and points P , E, and F are variable; see Figure2. Then EF has its maximum value when EF ∥ AD.

Indeed, we observe that the length of EF is maximum when θ is maximum i.e., when φ (= θ +∠BFC)

is maximum, that is, when AD is tangent at P to the circumcircle of △BPC. When this happens we haveβ = γ, and therefore α = β = γ = δ.

C

B

A

F E

DP

δ

θ

β

γ

φ

α

Figure 2

Finally we may choose points E′ and F ′ on the given unit circle so that E′F ′ is parallel to AD and E′Bmeets F ′C at P ′ on AD.

By the above, it follows thatEF ≤ E′F ′.

Because E′F ′ ≤ diameter, we have the desired result

AB +CD +EF ≤ (AB +CD) +E′F ′ ≤ 2 + 2 = 4.


O479. Let ABCD be a quadrilateral with AB = CD = 4, AD2 +BC2 = 32 and ∠(ABD) +∠(BDC) = 51○.If BD =

√6 +

√5 +

√2 + 1, find AC.


First solution by Daniel Lasaosa, Pamplona, SpainFirst, note that

cos2 (15○) =1 + cos(30○)

2=

2 +√

3

4=

8 + 4√

3

16=

(√

6)2+ (

√2)

2+ 2

√6√

2

16=

= (

√6 +

√2

4)

2

.

Note also that for α = 18○, we have by the de Moivre formulae that

0 = cos (90○) = cos(5α) = cosα (1 − 12 sin2 α + 16 sin4 α) ,

sin2 α =6 ± 2

√5

16=

(√

5)2+ 12 ± 2

√5

16= (

√5 ± 1

4)

2

.

Now, of both solutions of cos(5α) = 0 with positive sine, the least angle, and thus the least sine, correspondsto α = 18○, or

cos (36○) = 1 − 2 sin2 (18○) = 1 − 23 −

√5

8=

√5 + 1

4.

It follows that BD = 4 cos (36○) + 4 cos (15○).

Denote β =∠ABD and δ =∠BDC, or by the Cosine Law,

AD2 = 16 +BD2 − 8BD cosβ, BC2 = 16 +BD2 − 8BD cos δ,

or adding and using that AD2 +BC2 = 32, we have

2 cosβ + δ

2cos

β − δ

2= cosβ + cos δ =

BD

4= cos (36○) + cos (15○) = 2 cos

51○

2cos

21○

2,

and since wlog β ≥ δ (because we may interchange A,C and simultaneously B,D without altering theproblem), and as β + δ = 36○ + 15○, we have β − δ = 36○ − 15○, or β = 36○ and δ = 15○.

Denote α =∠ADB. By the Sine Law and the Cosine Law,

sinα =4 sin (36○)

AD, cosα =

BD2 +AD2 − 16

2AD ⋅BD,

while at the same time sin (15○) =√

6−√

24 so that sin2 (15○) + cos2 (15○) = 1. Therefore,

cos∠ADC = cos (15○) cosα − sin (15○) sinα =2 +

√3

AD−

(√

3 − 1)√

5 −√

5

2AD,

where we have used that sin (36○) =√

10−2√

54 , and that by the Cosine Law,

AD2 = AB2 +BD2 − 2AB ⋅BD cosβ = 18 + 4√

3 − 2√

5.

Now, using again the Cosine Law, we finally obtain after some algebra

AC2 = AD2 + 16 − 8AD cos∠ADC = 18 − 4√

3 − 2√

5 + 2 (√

6 −√

2)√

10 − 2√

5 =

= (√

6 −√

2)2+ (

√10 − 2

√5)

2

+ 2 (√

6 −√

2)√

10 − 2√

5 =

= (√

6 −√

2 +√

10 − 2√

5)2

,

and finally,AC =

√6 −

√2 +

√10 − 2

√5.


Second solution by the authorWe have AB2 +CD2 = AD2 +BC2, so diagonals AC and BD are perpendicular. If we denote the point ofintersection of these diagonals by E, then

BE +ED = AB cosu +CD cos v = 4(cosu + cos v),

where u =∠ABD and v =∠BDC.But BE +ED = (

√6 +

√2) + (

√5 + 1) = 4 cos(36) + 4 cos(15). It follows that

cosu + cos v = cos(36) + cos(15),

implying

2 cos(u + v

2) cos(

u − v

2) = 2 cos(

51

2) cos(

21

2) .

The condition u + v = 51 implies ∣u − v∣ = 21 so, assuming u > v we get u = 36 and v = 15. HenceBE =

√5 + 1 and ED =

√6 +

√2, implying

AC = AE +EC =√

10 − 2√

5 +√

6 −√

2.

Also solved by Lukas Seier, Charters Sixth Form, Sunningdale, England, UK; Albert Stadler, Herrliberg,Switzerland; Akash Singha Roy, Chennai Mathematical Institute, India; Sebastian Foulger, Charters SixthForm, Sunningdale, England, UK.


O480. At a party, some people shake hands. The followings are known:

– Each person shakes hands with exactly 20 persons.

– For each pair of persons who shake hands with each other, there is exactly one other person whoshakes hands with both of them.

– For each pair of persons who do not shake hands with each other, there are exactly six otherpersons who shake hands with both of them.

Determine the number of people in the party.

Proposed by Navid Safaei, Sharif University of Technology, Tehran, Iran

Solution by Adamopoulos Dionysios, 3rd High School, Pyrgos, GreeceSuppose we have N students.

Denote by S, the set of the triples (s1, s2, s3), such as s1 shake hands with both s2 and s3 (think of s2and s3 as an unordered pair).

Note that (a, b, c) ≡ (a, c, b), but (a, b, c) ≠ (b, a, c).We wish to count ∣S∣.Each of the N students shake hands with exactly 20 people, who form (20

2) = 190 pairs.

So ∣S∣ = 190N .On the other hand, we have a total of (N2 ) pairs of students. We can distinguish them into two categories.

The pairs that shake hands, call them F , and the pairs that do not shake hands, call them E.Clearly, F +E = (N

2).

Moreover there is a total of 20N2 = 10N , shaking of hands (each person shakes hands with 20 other people,

but we count twice!).As a result F = 10N and hence E = (N

2) − 10N .

From the last two conditions of the problem we have that

∣S∣ = F + 6E = 10N + 6 [(N

2) − 10N] = 3N(N − 1) − 50N

We have just counted ∣S∣ in two ways and now we can get that

3N(N − 1) − 50N = 190N ⇔ 3N(N − 1) = 240N ⇔ N − 1 = 80⇔ N = 81

Also solved by Daniel Lasaosa, Pamplona, Spain; Takuji Grigorovich Imaiida, Fujisawa, Kanagawa,Japan; Akash Singha Roy, Chennai Mathematical Institute, India; Joonsoo Lee, Dwight Englewood School,NJ, USA; Min Jung Kim, Boston, MA, USA; Albert Stadler, Herrliberg, Switzerland.


Documents

Juniorproblems - awesomemath.org · Juniorproblems J475.Let ABC be a triangle with ∠Band ∠C acute and diﬀerent from 45 . Let Dbe the foot of the altitudefromA. Provethat∠Aisrightifandonlyif