mathematical reflections

Junior problems

J199. Prove that there are infinitely many pairs (p, q) of primes such that p6 + q4 has twopositive divisors whose difference is 4pq.

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

Solution by Arkady Alt, San Jose, USA

Let p = 2 and let q be odd prime number. Then

p6 + q4 = 64 + q4 =(8 + 4q + q2

) (8− 4q + q2

)and we have (

8 + 4q + q2)−(8− 4q + q2

)= 8q = 4pq.

Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Prithwijit De,HBCSE, Mumbai, India; Roberto Bosch Cabrera, Havana, Cuba; Tigran Hakobyan, Ar-menia.

Mathematical Reflections 4 (2011) 1

J200. Let x, y, z be positive real numbers with x ≤ 2, y ≤ 3 and x + y + z = 11. Prove thatxyz ≤ 36.

Proposed by Mircea Lascu, Zalau, and Marius Stanean, Zalau, Romania

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

By AM-GM inequality, we have

xyz =1

6· (3x) · (2y) · z ≤ 1

6

(3x+ 2y + z

3

)3

=1

6

(2x+ y + 11

3

)3

≤ 1

6

(2 · 2 + 3 + 11

3

)3

= 36.

We have equality if and only if x = 2, y = 3, z = 6.

Also solved by Arkady Alt, San Jose, USA; Daniel Lasaosa, Universidad Publica deNavarra, Spain; Omran Kouba, Damascus, Syria; Perfetti Paolo, Dipartimento di Matem-atica, Universita degli studi di Tor Vergata Roma, Italy; Prithwijit De, HBCSE, Mum-bai, India; Roberto Bosch Cabrera, Havana, Cuba; Salem Malikic, Sarajevo, Bosnia andHerzegovina; Tigran Hakobyan, Armenia; Christopher Wiriawan, Jakarta, Indonesia.


J201. Let ABC be an isosceles triangle with AB = AC. Point D lies on side AC such that∠CBD = 3∠ABD. If

1

AB+

1

BD=

1

BC,

find ∠A.


Solution by Roberto Bosch Cabrera, Florida, USA

Let ∠ABD = x, and note that we have ∠CBD = 3x and ∠ACB = 4x. By the Lawof Sines in triangles ABC and BDC, we obtain that AB

sin 4x = BCsin 8x and BD

sin 4x = BCsin 7x ,

respectively. Thus, the relation 1AB + 1

BD = 1BC is equivalent to sin 8x+ sin 7x = sin 4x,

where we can consider x to lie in (0, π8 ) since ∠A = π− 8x > 0. Now, using that sin 4x−sin 8x = −2 sin(2x) cos 6x the equation can be rewritten as sin 7x = (−2 cos 6x) sin(2x),and so we spot the solution x = π

9 . We will finish by proving that it is unique onthis interval. We split this discussion into two cases. First, suppose x > π

9 is someother solution, and observe that 6x > 6π

9 , so −2 cos 6x > 1; hence sin 7x > sin 2x. But2x ∈ (0, π4 ) and 7x ∈ (7π9 ,

7π8 ), so sin 7x = sin(π − 7x) yields π − 7x > 2x, i.e. x < π

9 ,which is a contradiction. Now, suppose x < π

9 is a solution, and see that sin 7x < sin 2x;if 7x ≤ π

2 , then 7x < 2x, and we got a contradiction, whereas if 7x > π2 , then π−7x < 2x,

i.e .x > π9 , again contradiction. We conclude that x = π

9 is the only solution, and so∠A = π − 8π

9 = π9 .

Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; Omran Kouba,Damascus, Syria; Perfetti Paolo, Dipartimento di Matematica, Universita degli studi diTor Vergata Roma, Italy.


J202. Let ABC be a triangle with incenter I and let A1, B1, C1 be the symmetric points ofI with respect to the midpoints of sides BC,CA,AB. If Ia, Ib, Ic denote the excenterscorresponding to sides BC,CA,AB, respectively, prove that lines IaA1, IbB1, IcC1 areconcurent.

Proposed by Dorin Andrica, Babes-Bolyai University, Romania

First solution by Cosmin Pohoata, Princeton University, USA

Consider the homothety H(I, 1/2) with pole I and ratio 1/2. It is well-known that themidpoints of the segments IIa, IIb, IIc are in fact the midpoints of the arcs BC, CA,AB not containing the vertices of the triangle; therefore, the lines IaA1, IbB1, IcC1 areconcurrent since the mediators of the triangle ABC are concurrent.

Second solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain

In exact trilinear coordinates, I ≡ (r, r, r), r being the inradius, and the midpoint of

BC is Ma ≡(

0, hb2 ,hc2

), where ha, hb, hc are the respective lengths of the altitudes from

A,B,C. Thus A1 ≡ (−r, hb − r, hc − r). Denoting by ra, rb, rc the exradii correspondingto sides BC,CA,AB, clearly Ia ≡ (−ra, ra, ra), or in trilinear coordinates (α, β, γ), thepoints on line IaA1 satisfy

0 =

∣∣∣∣∣∣α β γ−1 1 1−r hb − r hc − r

∣∣∣∣∣∣ =

∣∣∣∣ α+ β α+ γhb − 2r hc − 2r

∣∣∣∣ ,or by cyclic permutations, the point that satisfies (ha − 2r)(β + γ) = (hb − 2r)(γ + α) =(hc − 2r)(α+ β) is simultaneously on lines IaA1, IbB1, IcC1. The conclusion follows.

Also solved by Felipe Ignacio Arbulu Lopez, Instituto Nacional General Jose Migue Car-rera, Santiago, Chile; Roberto Bosch Cabrera, Florida, USA.


J203. Let ABCD be a trapezoid (AB||CD) with acute angles at vertices A and B. Line BC andthe tangent lines from A and E to the circle of center D tangent to AB are concurrentat F. Prove that AC bisects the segment EF if and only if AF + EF = AB.



We begin with the following preliminary result:

Lemma. Let ABC be an arbitrary triangle with incenter I centroid G. Then, the linesIG and BC are parallel if and only if b+ c = 2a.

Proof. Let la,ma be the bisector and median from A, respectively. We have that IG isparallel to BC if and only if AI

la= AG

ma. However, AI2 = (s − a)2 + r2 by Pythagoras

theorem, where s denotes the semiperimeter and r the inradius of triangle ABC. Besides,we have that

r2 =(s− a)(s− b)(s− c)

sand l2a =

bc(a+ b+ c)(b+ c− a)

(b+ c)2,

so using AGma

= 23 , we obtain

9

[(s− a)2 +

(s− a)(s− b)(s− c)s

]=

4bc(a+ b+ c)(b+ c− a)

(b+ c)2,

which after several simple algebraic manipulations becomes (b+c−2a)(2a+5b+5c) = 0,i.e. b+ c = 2a.

Returning to the original problem we have that AF + EF = 2AE and note that Gis the midpoint of EF if and only if C is the centroid of triangle AEF , so by applyingthe Lemma for this triangle, we arrive to desired conclusion.

Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain


J204. Give a straightedge and compass construction of a triangle ABC starting with its incenterI, the foot of the altitude from A, and the midpoint of the side BC.

Proposed by Cosmin Pohoata, Princeton University, USA

Solution by Cosmin Pohoata, Princeton University, USA

We are given the foot of the altitude from A and the midpoint M of BC, so the linedetermined by the two is precisely the line BC. Now, knowing the incenter we can drawthe perpendicular from I to BC and get the incircle of ABC. Recall the well-known factthat if D is the tangency point of the incircle with the side BC, and D′ the antipode ofD with respect to the incircle, then the points A, D′, X are collinear, where X is thetangency point of the A-excircle with BC. However, in order to use this, we need to findX. This is not a problem whatsoever, since MD = MX; so we have the construction ofX as the reflection of D in M . Now, just draw the lines XD′ and the altitude from A(which we can draw since we have the foot of the altitude on BC and the line BC); theyintersect at the vertex A. Afterwards, just take the tangents from A to the incircle andintersect them with BC; this will give us the vertices B and C. Hence our constructionis complete.

Also solved by Arkady Alt, San Jose, USA; Miguel Amengual Covas, Cala Figuera, Mal-lorca, Spain; Daniel Lasaosa, Universidad Publica de Navarra, Spain; Raul A. Simon,Chile.


Senior problems

S199. In triangle ABC let BB′ and CC ′ be the angle bisectors of ∠B and ∠C. Prove that

B′C ′ ≥ 2bc

(a+ b)(a+ c)

[(a+ b+ c) sin

A

2− a

2

].


Solution by Daniel Campos Salas, Costa RicaFirst, note that by Ptolemy’s inequality for quadrilateral BCB′C ′,

B′C ′ ≥ BB′ · CC ′ −BC ′ · CB′

BC.

Now, BB′ = 2aca+c cos B2 and that cos B2 cos C2 = s

a sin A2 , and these imply that

BB′ · CC ′ = 4a2bc

(a+ b)(a+ c)cos

B

2cos

C

2=

2abc(a+ b+ c)

(a+ b)(a+ c)sin

A

2.

In addition, BC ′ = aca+b and this yields BC ′ ·CB′ = a2bc

(a+b)(a+c) ; therefore by just pluggingin these identities in the inequality above, the desired conclusion follows.

Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Arkady Alt,San Jose, USA; Daniel Lasaosa, Universidad Publica de Navarra, Spain; EvangelosMouroukos, Agrinio, Greece; Roberto Bosch Cabrera, Florida, USA.


S200. On each vertex of the regular hexagon A1A2A3A4A5A6 we place a rod. On each rodwe have ai rings, where ai corresponds to the vertex Ai. Taking a ring from any threeadjacent rods we can create chains of three rings. What is the maximum number of suchchains that we can create?

Proposed by Arkady Alt, San Jose, USA


If by adjacent rods, we understand consecutive rods, we have the following possibilities forthree adjacent vertices: A1A2A3, A2A3A4, A3A4A5, A4A5A6, A5A6A1, A6A1A2. Hence,the maximum number of chains is

M := a1a2a3 + a2a3a4 + a3a4a5 + a4a5a6 + a5a6a1 + a6a1a2.


S201. Prove that in any triangle,

ra ≤ 4R sin3

(A

3+π

6

).

Proposed by Dorin Andrica, Babes-Bolyai University, Romania

Solution by Prithwijit De, HBCSE, Mumbai, India

We know that ra = R(1− cosA+ cosB + cosC). Also,

4R sin3

(A

3+π

6

)= R

(3 sin

(A

3+π

6

)− cosA

).

Therefore, to establish the inequality we need to show that

1 + cosB + cosC ≤ 3 sin

(A

3+π

6

). (*)

Now, in order to get (*), observe that it is enough to prove the inequality for acute anglesB and C, since if one of them is obtuse or right, its cosine is nonpositive. On the other

hand, cosx is concave on the interval(

0,π

2

), thus by Jensen’s inequality for concave

functions,

1 + cosB + cosC

3≤ cos

(0 +B + C

3

)= sin

(A

3+π

6

). (**)

This proves our inequality, as from (**) it follows that ra ≤ 4R sin3

(A

3+π

6

).

Also solved by Arkady Alt, San Jose, USA; Daniel Lasaosa, Universidad Publica deNavarra, Spain; Evangelos Mouroukos, Agrinio, Greece; Omran Kouba, Damascus, Syria;Roberto Bosch Cabrera, Florida, USA.


S202. Let a and b be integers such that a2m − b2n = a − b, where m and n are consecutiveintegers Prove that gcd(a, b) =

√|a− b|.


Solution by Omran Kouba, Damascus, Syria

Let δ = gcd(a, b) and write a = δa′ and b = δb′ with gcd(a′, b′) = 1. From a2m− b2n =a − b we get (ma − 1)a′ = (nb − 1)b′, so a′|(nb − 1)b′ and gcd(a′, b′) = 1. This impliesthat a′|(nb − 1), and consequently there exists an integer λ such that nb − 1 = λa′ andma− 1 = λb′ or,

nδb′ − λa′ = 1, and mδa′ − λb′ = 1 (1)

In particular, gcd(δ, λ) = 1. Now subtracting the two equalities in (1) we find that

(nδ + λ)b′ = (mδ + λ)a′,

and again, since a′|(nδ + λ)b′ and gcd(a′, b′) = 1, we conclude that a′|(nδ + λ) and thatthere exists an integer µ such that

nδ + λ = µa′, and mδ + λ = µb′. (2)

On the other hand, recalling that m and n are consecutive integers we get from (2) that

δ = εµ(a′ − b′), with ε = n−m ∈ −1,+1, (3)

Whereas from (2) we have λ = µ(a′ − nε(a′ − b′)). So, µ is a common divisor of λ andδ which are coprime, hence µ ∈ −1,+1. Finally, multiplying both sides of (3) by δand using the fact that |εµ| = 1, we conclude that δ2 = |a − b|, which is the desiredconclusion.

Also solved by Arkady Alt, San Jose, USA; Daniel Lasaosa, Universidad Publica deNavarra, Spain; Roberto Bosch Cabrera, Florida, USA; Salem Malikic, Sarajevo, Bosniaand Herzegovina; Tigran Hakobyan, Armenia.


S203. Let ABC be a triangle, and P a point not lying on its sides. Call XY Z the ceviantriangle of P with respect to ABC and consider the points Ya, Za of intersection BCwith the parallel lines to AX through Y and Z, respectively. Prove that AX,Y Za, YaZconcur in a point Q that satisfies the cross-ratio

(AXPQ) =AP

AX.

Proposed by Francisco Javier Garcia Capitan, Spain

Solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain

Triangles ACX and Y CYa are clearly similar, hence XYaAY = CX

AC and Y YaCY = AX

AC . Triangles

ABX and ZBZa are also similar, hence XZaAZ = BX

AB and ZZaBZ = AX

AB . Define Q =AX ∩ Y Za and Q′ = AX ∩ ZYa, or triangles QXZa and Y YaZa are similar yieldingQXZZa

= XYaYaZa

, and triangles Q′XYa and ZZaYa are also similar, for Q′XY Ya

= XZaYaZa

. It followsthat Q = Q′ iff XZa ·Y Ya = XYa ·ZZa, or equivalently iff AY ·BZ ·CX = AZ ·BX ·CY ,ie by Ceva’s theorem iff AX,BY,CZ concur. Since they do (at P ), it follows that Q isindeed the point where AX,Y Za, ZYa concur.

Note that (AXPQ) = APAX is by definition equivalent to AX ·QX = AQ · PX. Applying

Menelaus’ theorem to triangle APY and lines BXC, and triangle ACX and line ZaQY ,we find

AX

PX=AC ·BYPB · CY

,AQ

QX=AY · CZaCY ·XZa

.

Writing CZa = CX + XZa, and inserting XZa = ZX·BXAB , we find after some algebra

that the proposed result is equivalent to

AY ·BZ · CX = AY · CX(AB −AZ) = BX ·AZ(AC −AY ) = AZ ·BX · CY,

clearly true again by Ceva’s theorem since AX,BY,CZ concur.

Also solved by Roberto Bosch Cabrera, Florida, USA;


S204. Find all positive integers k and n such that kn − 1 and n are divisible by precisely thesame primes.

Proposed by Tigran Hakobyan, Yerevan, Armenia

Solution by Alessandro Ventullo, Milan, Italy

We immediately see that if k = 1, then kn − 1 would be divisible by every prime p, butsince n is a positive integer, n is only divisible by a finite number of primes. So, we havek > 1. If n = 1, then k − 1 ≥ 1, so we obtain k − 1 = 1, i.e, k = 2. If n = 2, thenk2 − 1 = (k − 1)(k + 1) and we want

(k − 1)(k + 1) = 2m,

with m a positive integer. Therefore k = 3 and m = 1. Now, suppose that n > 2. Wehave

n =h∏i=1

pαii , k = 1 +

h∏i=1

pβii ,

with αi, βi positive integers and pi prime for every 1 ≤ i ≤ h. Suppose αi 6= βi for every

1 ≤ i ≤ h and put a =h∏i=1

pβii . So

kn − 1 = (1 + a)n − 1

= a

(an−1 + nan−2 + . . .+

n(n− 1)

2a+ n

)= a(ab+ n)

=h∏i=1

pβi+minαi,βii

(h∏i=1

pβi−minαi,βii b+

h∏i=1

pαi−minαi,βii

)

where b =

(an−2 + nan−3 + . . .+

n(n− 1)

2

). It’s easy to see that only one between

h∏i=1

pβi−minαi,βii and

h∏i=1

pαi−minαi,βii is divisible by pi for every 1 ≤ i ≤ h, so

(h∏i=1


h∏i=1

pαi−minαi,βii

)

is not divisible by any of the pi and then is divisible by another prime p∗ which doesn’tdivide n, contradiction. So αj = βj for some j. Suppose that n is odd. By the sameargument, (

h∏i=1


h∏i=1

pαi−minαi,βii

)


is not divisible by any of the pi with i 6= j and for pj we have a = pαj

j a1 and n = pαj

j n1,where a1 and n1 are not divisible by pj . Then

a (ab+ n) = apαj

j (a1b+ n1) ,

but since b is divisible by pj , (a1b+ n1) is not divisible by pj and therefore a(ab+ n) isnot divisible by any of the prime pi for every 1 ≤ i ≤ h. So n is even and a must beeven, i.e., a = 2a1 with a1 a positive integer and

(1 + a)n − 1 = [(1 + a)n/2 − 1][(1 + a)n/2 + 1]

= [(1 + a)n/2 − 1][an/2 +

n

2an/2−1 + . . .+

n

2a+ 2

]= 2[(1 + a)n/2 − 1][2n/2−1a

n/21 + 2n/2−2a

n/2−11 + . . .+ a1 + 1]

= 2c[(1 + a)n/2 − 1]

where c = [2n/2−1an/21 + 2n/2−2a

n/2−11 + . . .+ a1 + 1]. Since c is not divisible by a1, any

prime which divides a1 (and a) doesn’t divide c and then c is not divisibile by any primewhich divides n and thus is divisible by another prime which doesn’t divide n. Thiscontradiction shows that the only positive integers which satisfy the required conditionsare n = 1, k = 2 and n = 2, k = 3.

Also solved by Albert Stadler, Switzerland.


Undergraduate problems

U199. Prove that in any triangle ABC,

3√

3 ≤ cotA+B

4+ cot

B + C

4+ cot

C +A

4≤ 3√

3

2+

s

2r,

where s and r denote the semiperimeter and the inradius of triangle ABC, respectively.



Since the function f(t) = cot t is convex in (0; π2 ), by Jensen’s inequality,

cotA+B

4+ cot

B + C

4+ cot

C +A

4≥ 3 · cot

(A+B) + (B + C) + (C +A)

12

= 3 cotπ

6= 3√

3.

So the inequality from the left hand side is proven.

For the other one, choose f(t) = cot t, x = A2 , y = B

2 , z = C2 in Popoviciu’s inequality,

and get

2

3(cot

A+B

4+ cot

B + C

4+ cot

C +A

4)

≤1

3(cot

A

2+ cot

B

2+ cot

C

2) + cot

A+B + C

6

=1

3(cot

A

2+ cot

B

2+ cot

C

2) + cot

π

6

=1

3(cot

A

2+ cot

B

2+ cot

C

2) +√

3.

It follows that

cotA+B

4+ cot

B + C

4+ cot

C +A

4≤ 1

2(cot

A

2+ cot

B

2+ cot

C

2) +

3√

3

2,

which via the well-known

cotA

2+ cot

B

2+ cot

C

2=s

r

gives us our desired inequality.

Also solved by Daniel Campos Salas, Costa Rica; Roberto Bosch Cabrera, Florida, USA.


U200. Let p be an odd prime and let n be an integer greater than 1. Find all integers k forwhich there exists an n× n matrix A of rank k such that A+A2 + · · ·+Ap = 0.

Proposed by Gabriel Dospinescu, Ecole Polythehnique, France

Solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain;

It should be specified in the problem statement that A must be a matrix with real entries,otherwise the problem becomes trivial, and a matrix of any rank k ∈ 0, 1, . . . , n maybe found; it suffices to take a diagonal matrix whose main diagonal are k p-th roots ofunity other than 1, and n− k zeros.

Let λ1, λ2, . . . , λn be the eigenvalues of A, and let P be the invertible matrix such thatP−1AP = T is triangular. Clearly, T and A have the same rank since multiplication byan invertible matrix preserves the rank. Now,

0 = P−10P = T + T 2 + · · ·+ T p,

where the RHS is clearly a triangular matrix whose main diagonal contains the terms ofthe form

λi + λi2 + · · ·+ λi

p = 0.

Note that λi 6= 1 since otherwise this result is not true, or we may multiply both sidesby λi − 1, yielding λi (λi

p − 1) = 0, ie, either λi = 0, or it is a p-th root of unity otherthan 1. If A has real entries, its characteristic polynomial has real coefficients, and itsnonzero roots can be grouped in complex conjugate pairs, ie, A has an even number ofnonzero eigenvalues, thus an even rank. Let now λ be a p-th root of unity other than1, clearly a second degree polynomial with real coefficients exists such that λ and itscomplex conjugate λ∗ are its roots; a matrix with real entries can then be found suchthat its eigenvalues are λ and λ∗. Repeat ` times this 2 × 2 matrix along the diagonalof A, and fill the rest of A with zeros, and a matrix A satisfying the conditions of theproblem is found (since A thus formed is clearly diagonalizable, and the diagonal ofT +T 2 + · · ·+T p is clearly filled with 0’s). Since this can be done for any 0 ≤ ` ≤ n

2 , weconclude that the integers k that we are looking for, are all non-negative even integersthat do not exceed n.


U201. Evaluate∞∑n=2

3n2 − 1

(n3 − n)2.


Solution by Moubinool Omarjee, Paris, France

Since3k2 − 1

(k3 − k)2= − 1

2k2+

1

2 (k − 1)2− 1

2k2+

1

2 (k + 1)2,

we have thatn∑k=2

3k2 − 1

(k3 − k)2=

1

2− 1

2n2+

1

2 (n+ 1)2− 1

8,

thus the limit is 38 .

Also solved by Alessandro Ventullo, Milan, Italy; AN-anduud Problem Solving Group,Ulaanbaatar, Mongolia; Anastasios Kotronis, Athens, Greece; Arkady Alt, San Jose,USA; Daniel Campos Salas, Costa Rica; Daniel Lasaosa, Universidad Publica de Navarra,Spain; Emanuele Tron, Pinerolo, Italy; Evangelos Mouroukos, Agrinio, Greece; G. C.Greubel, Newport News, USA; Gerhardt Hinkle, Springfield, USA; N.J. Buitrago A.,Universidad Nacional, Colombia; Jedrzej Garnek, Adam Mickiewicz University, Poz-nan, Poland; Lorenzo Pascali, Universita di Roma La Sapienza, Roma, Italy; OmranKouba, Damascus, Syria; Raul A. Simon, Chile; Roberto Bosch Cabrera, Florida, USA;Anastasios Kotronis, Athens, Greece; Tigran Hakobyan, Armenia.


U202. The interval (0, 1] is divided into N equal intervals ( i−1N , iN ], where i ∈ [1, n]. An intervalis called special if it contains at least one number from the set 1, 12 ,

13 , . . . ,

1N . Find a

good approximation for the number of special intervals.

Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA

Solution by Jedrzej Garnek, Adam Mickiewicz University, Poznan, Poland

Denote the number of special intervals by aN . We will not only approximate aN , butalso give the exact formula. The inetrval

(i−1N , iN

]is special if and only if there exists at

least one k ∈ 1, 2, . . . , N such that

i− 1

N<

1

k≤ i

N,

i.e. if and only if dNk e = i. Thus, the problem is equivalent to finding how many differentvalues does dNk e take for k = 1, 2, . . . , N .

To answer this, let us first note that the sequencedNt e

t≥1 is non-increasing. Next,

consider the function f(x) = x2 − x − N , which is strictly increasing for x ≥ 1 andtends to +∞ as x → +∞. Let t be the largest positive integer for which f(t) < 0 (i.e.f(t+ k) ≥ 0 for k ≥ 1). Then t(t− 1) < N ≤ t(t+ 1), so for 2 ≤ i ≤ t we have

N

i− 1− N

i=

N

i(i− 1)> 1,

so d Ni−1e > dNi e, and thus we get dN1 e > d

N2 e > . . . > dNt e. If i > t, then

N

i− 1− N

i=

N

i(i− 1)≤ 1,

hence we have that d Ni−1e ∈dNi e, d

Ni e+ 1

. Denote dNt e by s. Obviously, in the non-

increasing sequence of positive integers dNt e, dNt+1e, . . . , d

NN e = 1 every two consecutive

terms differ by at most 1, sodNte, d N

t+ 1e, . . . , dN

Ne

= 1, 2, . . . , s .

It follows that aN = (t− 1) + s = t+ dNt e − 1.

Now, let’s try to simply this expression by proving the formula t+ dNt e−1 = d2√ne−1.

We have two cases.

If N ≤ t2, then t2 − t < N ≤ t2, so dNt e = t. Moreover, 2√N ≤ 2t and 2t − 1 =√

4t2 − 4t+ 1 <√

4(t2 − t+ 1) ≤ 2√N, so we have that d2

√Ne = 2t and t + dNt e =

2t− 1 = d2√Ne − 1.

Otherwise, if N > t2, then t2 < N ≤ t2 + t, so dNt e = t+1 and 2t < 2√N <

√4t2 + 4t <√

4t2 + 4t+ 1 = 2t+1, so d2√Ne = 2t+1. It follows that t+dNt e−1 = 2t = d2

√Ne−1,

hence we get the same answer. This settles our discussion.

Also solved by Emanuele Tron, Pinerolo, Italy; Raul A. Simon, Chile.


U203. Let P be a polynomial of degree 5, with real coefficients, all whose zeros are real. Provethat for each real number a that is not a zero of P or P ′ there is a real number b suchthat

b2P (a) + 4bP ′(a) + 5P ′′(a) = 0.


First solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain

We will prove the following more general result: let P be an n-th degree polynomial(n ≥ 2) with real coefficients and only real zeroes, and let C,D be real numbers suchthat nC2 ≥ 4(n− 1)D ≥ (n− 1)C2. Then, there is at least one real number b such that

b2P (a) + CbP ′(a) +DP ′′(a) = 0.

The real b is unique if and only if nC2 = 4(n− 1)D and P has all zeros equal, otherwiseexactly two real b’s exist. Clearly, the originally proposed problem is a particular casewith n = 5, C = 4 and D = 5, and since 5C2 = 80 = 16D, equality holds in the proposedproblem if and only if P has five equal real zeros.

Let zi, i = 1, 2, . . . , n be the zeros of P , or P ′(a) = SP (a), P ′′(a) = 2TP (a), where wehave defined

S =n∑i=1

ui, T =∑

1≤i<j≤nuiuj , ui =

1

a− zi,

and each ui is a real number because a is not a zero of P , hence S, T are also real.Therefore, for any real C,D, the quadratic equation in b,

b2P (a) + CbP ′(a) +DP ′′(a) = 0,

has real solutions if and only if the discriminant ∆ ≥ 0, where

∆ = C2(P ′(a)

)2 − 4DP (a)P ′′(a) =(C2S2 − 8DT

)(P (a))2 ,

or since a is not a zero of P , if and only if C2S2 − 8DT ≥ 0. Now, note that

C2S2 − 8DT = C2n∑i=1

u2i − (4D − C2)∑

1≤i<j≤n2uiuj

=(nC2 − 4(n− 1)D

) n∑i=1

u2i + (4D − C2)∑

1≤i<j≤n(ui − uj)2

≥ 0,

with equality if and only if nC2 = 4(n − 1)D, and simultaneously all ui are equal; thislast condition is clearly equivalent to all zeros of P being equal.

Remark: Note that the condition P ′(a) 6= 0 is not necessary; indeed, note that if 4(n−1)D ≥ (n− 1)C2, then D ≥ 0, while when S = 0, then 2T = S2− (u21 + · · ·+u2n) < 0, or


2DT < 0, and b = ±√−2DT is indeed always real and a solution of b2P (a) +CbP ′(a) +

DP ′′(a) = 0 when P ′(a) = 0.

Second solution by Evangelos Mouroukos, Agrinio, Greece

Let a ∈ R be such that P (a) 6= 0. It suffices to show that the discriminant D =16[P ′ (a)]2 − 20P (a)P ′′ (a) of the quadratic equation x2P (a) + 4xP ′ (a) + 5P ′′ (a) = 0is nonnegative, or, equivalently, that

4[P ′ (a)

]2 ≥ 5P (a)P ′′ (a) .

More generally, we show that if n ≥ 2 is an integer, P is a polynomial of degree n withreal coefficients and real roots and a ∈ R is such that P (a) 6= 0, then we have

(n− 1)[P ′ (a)

]2 ≥ nP (a)P ′′ (a) . (1)

Let P (x) = c (x− r1) (x− r2) · · · (x− rn), where c, r1, r2, . . . rn ∈ R and c 6= 0. If a ∈ Ris such that P (a) 6= 0, then it is easy to see that

P ′ (a)

P (a)=

n∑i=1

1

a− ri(2)

andP ′′ (a)P (a)− [P ′ (a)]2

[P (a)]2= −

n∑i=1

1

(a− ri)2.

Hence,

P ′′ (a)

P (a)=

[P ′ (a)

P (a)

]2−

n∑i=1

1

(a− ri)2

or, equivalently,

P ′′ (a)

P (a)=

(n∑i=1

1

a− ri

)2

−n∑i=1

1

(a− ri)2. (3)

We apply the Cauchy-Schwarz inequality to get(n∑i=1

1

a− ri

)2

≤

(n∑i=1

12

)(n∑i=1

1

(a− ri)2

)= n

n∑i=1

1

(a− ri)2.

The above inequality can be written equivalently as

(n− 1)

(n∑i=1

1

a− ri

)2

≥ n

(n∑i=1

1

a− ri

)2

− nn∑i=1

1

(a− ri)2,

or, using (2) and (3),

(n− 1)

(P ′ (a)

P (a)

)2

≥ nP′′ (a)

P (a),

which is equivalent to (1).


Also solved by AN-anduud Problem Solving Group, Ulaanbaatar, Mongolia; Arkady Alt,San Jose, USA; Daniel Campos Salas, Costa Rica; Jedrzej Garnek, Adam MickiewiczUniversity, Poznan, Poland; Omran Kouba, Damascus, Syria; Roberto Bosch Cabrera,Florida, USA.


U204. Let A1A2 . . . An be a convex polygon and let P be a point in its interior. Prove that

mini∈1,2,...,n

∠PAiAi+1 ≤π

2− π

n.


We argue by contradiction. Accordingly, suppose that

mini∈1,2,...,n

∠PAiAi+1 >π

2− π

n.

Then, if we denote the angle ∠PAiAi+1 by αi for all i = 1, 2, . . . , n, then

di = PAi · sin∠PAiAi+1 = PAi · sinαi> PAi · sin(

π

2− π

n) = PAi · cos

π

n,

for all i = 1, 2, . . . , n, where di is distance between P and the side AiAi+1.

Summing up alll these inequalities, it follows that

d1 + d2 + · · ·+ dn > cosπ

n(PA1 + PA2 + · · ·+ PAn),

which contradicts the general version of the Erdos-Mordell inequality from H. C. Lenhard,Arch. Math., 12 (1961), pp. 311-314. This proves our problem.


Olympiad problems

O199. Prove that an acute triangle with ∠A = 20 and side-lengths a, b, c satisfying

3√a3 + b3 + c3 − 3abc = min(b, c)

is isosceles.


Solution by El Alami Mohamed

First, note that cos 20 cos 40 cos 80 = sin 160

8 sin 20 = 18 . Let α = sin 80

sin 20 ; note that the pre-vious identity yields α = 4 cos 20 cos 40 = 1

2 cos 80 , thus, since α = 4 cos 20 cos 40 =1 + 2 cos 20, it follows that α− 3 = −2(1− cos 20) = −4 sin2 10, i.e. α2(α− 3) = −1,that is α3− 3α2 + 1 = 0. At this point, we assume without loss of generality that b ≥ c.Accordingly, the equality from the hypothesis can be rewritten as a3 − 3abc + b3 = 0,which after dividing both sides by a3 becomes x3−3xy+ 1 = 0, where x = b

a and y = ca .

Now, since b ≥ c we have x ≥ y and ∠B ≥ ∠C, so 90 ≥ ∠B ≥ 80 ≥ ∠C (as the angles∠B and ∠C both add up to 160. Hence, by the Law of Sines, x ≥ sin 80

sin 20 = α. On theother hand, α > 2 and x 7−→ x3 − 3x2 + 1 is strictly increasing in ]2,+∞[; therefore0 = x3 − 3xy + 1 ≥ x3 − 3x2 + 1 ≥ α3 − 3α2 + 1 = 0, and so it follows that x = y = α.This proves that ABC is isosceles, as desired.

Also solved by Emanuele Tron, Pinerolo, Italy; Omran Kouba, Damascus, Syria; RobertoBosch Cabrera, Florida, USA.


O200. Determine all primes that do not have a multiple in the sequence an = 2nn2 + 1, n ≥ 1.

Proposed by Andrei Ciupan, Harvard University, USA

Solution by Daniel Campos Salas, Costa Rica

We will prove that 2 and the primes congruent to −1 modulo 8 are all the primes thatdo not have a multiple in the sequence.

First, note that it is clear that 2 satisfies the statement. Next, consider the primes of theform 4k+ 1. Let p be such a prime and recall that in this case there is a positive integer

q, q < p, such that p|q2 + 1 (for example, check q =(p−12

)!). Take n = (p − 1)(p − q);

by Fermat’s little theorem,

2nn2 + 1 ≡ n2 + 1 ≡ q2 + 1 ≡ 0 (mod p),

and this proves the claim in this case.

Furthermore, let p be a prime of the form 8k + 3, for some positive integer k. In thiscase, −1 and 2 are not quadratic residues modulo p, which implies that −2 is a quadraticresidue modulo p, i.e. there exists a positive integer q, q < p such that p|q2 + 2. Now,take n = (p− 1)(p− q − 1)− 1, and see that Fermat’s little theorem gives

2nn2 + 1 ≡ 2p−2q2 + 1 ≡ 0 (mod p),

which proves the claim in this second case.

Finally, consider a prime p = 8k − 1 for some positive integer k, and note that here 2is a quadratic residue, whereas −1 is not a quadratic residue. This clearly implies thatthe congruence 2nn2 + 1 ≡ 0 does not have solution, thus yielding our conclusion.

Also solved by Daniel Lasaosa, Universidad Publica de Navarra, Spain; El Alami Mo-hamed; Emanuele Tron, Pinerolo, Italy; Felipe Ignacio Arbulu Lopez, Instituto NacionalGeneral Jose Migue Carrera, Santiago, Chile; Jedrzej Garnek, Adam Mickiewicz Uni-versity, Poznan, Poland; Roberto Bosch Cabrera, Florida, USA; Tigran Hakobyan, Ar-menia.


O201. Let ABC be a triangle with circumcenter O, and let perpendiculars at B,C to BC,CAintersect the sidelines CA,AB at E,F, respectively. Proe that the perpendiculars toOB and OC at F and E, respectively intersect at a point L lying on the altitude AD,satisfying DL = LA sin2A.

Proposed by Francisco Javier Garcia Capitan, Spain

Solution by Daniel Lasaosa, Universidad Publica de Navarra, Spain

We begin by proving the following preliminary result:

Claim. Let ABC be a triangle with circumcircle O, and draw lines through B,C inter-secting at X, such that ∠ABX = π−B, ∠ACX = π−C (ie, BX,CX are the symmetricof AB,AC with respect to the external bisectors of B,C, respectively). Then, A,O,Xare collinear, and AX = b sinC

cosA = c sinBcosA .

Proof. The following proof is valid for acute triangles. However, using signed angles anddistances it is easily extended to any triangle. Clearly, ∠CBX = π − 2B, ∠BCX =π − 2C, and ∠BXC = π − 2A. By the Sine Law, BX = c cosC

cosA , while by the CosineLaw, and using that ∠ABX = π −B, we have after some algebra

AX2 = AB2 +BX2 − 2AB ·BX cos∠ABX =

(c sinB

cosA

)2

.

Again by the Sine Law,

sin(∠BAX) =BX sin(π −B)

AX= cosC = sin

(π2− C

).

Exchanging B,C we analogously obtain the symmetric results. The claim follows, since∠BAO = π

2 − C and ∠CAO = π2 −B.

Returning to the original problem note first that since BE ⊥ AF and CF ⊥ AE, thenEB,FC are the respective altitudes from E,F in triangle AEF . It also follows that B,Care on the circle with diameter EF , orABC andAEF are similar. SinceAB = AE cosA,it follows that triangle AEF is the result of the following transformation: reflect ABCon the internal bisector of angle A, then perform a scaling with factor 1

cosA and center A.Note that, since the altitude from one vertex and the line through that vertex and thecircumcircle, are symmetric with respect to the internal bisector of the correspondingangle, then the altitude from A in ABC is also the line through A and the circumcircleof AEF . Moreover, the perpendicular to OB at F clearly forms an angle equal toπ2 + ∠ABO = π − C with respect to AF . By the Claim, the point L is clearly on theline through A and the circumcircle of AEF , hence on the altitude from A in ABC, andby the relation of similarity between ABC and AEF , at a distance from A equal to

AL =1

cosA

b sinC

cosA=

AD

cos2A.

This yields DL = AL−AD = AL sin2A , which is precisely what we wanted to prove.

Also solved by El Alami Mohamed; Roberto Bosch Cabrera, Florida, USA.


O202. Find all pairs (x, y) of positive integers for which there is a nonnegative integer z suchthat (

1 +1

x

)(1 +

1

y

)= 1 +

(2

3

)z.


No solution has been received yet.


O203. Let M be an arbitrary point on the circumcircle of triangle ABC and let the tangentsfrom this point to the incircle of the triangle meet the sideline BC at X1, and X2. Provethat the second intersection of the circumcircle of triangle MX1X2 with the circumcircleof ABC (different from M) coincides with the tangency point of the circumcircle withmixtilinear incircle in angle A (As usual, the A−mixtrilinear incircle names the circletanget to AB,AC and to the circumcircle of ABC internally.).

Proposed by Cosmin Pohoata, Princeton University, Princeton, NJ

Solution by Vladimir Zajic, New York, USA

Assume without loss of generality that M lies on the same side of line BC as the vertexA. In this case, we denote by (I) the common incircle of triangles ABC, and MX1X2

with radius r, and let D, E, F , Y1, Y2 be the tangency points of (I) with the sidelinesBC, CA, AB, MX1, MX2.

Consider the inversion Ψ with center I and power r2, which takes the vertices A, B, C,M , X1, X2 to the midpoints A′, B′, C ′, M ′, X ′1, X

′2 of the sides EF , FD, DE, Y2Y1,

Y1D, DY2 of the intouch triangles DEF , and DY2Y1. The circumcircles (O), (P ) oftriangles ABC, and MX1X2 become the circumcircles (O′), (P ′) of the triangles A′B′C ′,M ′X ′1X

′2, which, because they coincide with the nine-point circles of two triangles of the

same circumcircle, are congruent and have the common radius r2 . Since the sidelines

BC, CA, AB, MX1, MX2 are tangent to the inversion circle (I), their images underΨ are the congruent circles Γa,Γb,Γc,Ω1,Ω2 with diameters ID, IE, IF , IX1, IX2,respectively. The congruent circles (O′),Γb,Γc with radii r

2 meet at point A′. Thus, acircle (A′, r) with center A′ and radius r is tangent to all three at points diametricallyopposite to A′.

The internal angle-bisector AI of the angle ∠A passing through the inversion center I iscarried into itself. The mixtilinear incircle (Ka) and the mixtilinear excircle (La) of thetriangle4ABC in the angle A are the only two circles centered on AI and simultanouslytangent to CA, AB, and (O). Since the inversion center I is the similarity center of acircle and its inversion image, only the images (K ′a), (L

′a) of (Ka), (La) are centered

on AI and tangent to Γb, Γc, (O′). The mixtilinear excircle (La), lying outside of thecircumcircle (O) and outside of the inversion circle (I), has both intersections with AIon the ray (ILa. Since the inversion in (I) has positive power r2, its image (L′a) also hasboth intersections with AI on the ray (ILa and it is centered on the ray (ILa. It cannotbe identical with the circle (A′, r) centered on the opposite ray (IA. Therefore, the imageof the the mixtilinear incircle (Ka) in angle A under Ψ is the circle (A′, r). Furthermore,the inversive image of the tangency point Z of the circles (Ka), and (O) is the tangencypoint Z ′ of (A′, r) and (O′), the antipode of A′ with respect to the circumcircle (O′).

Let now A0, B0, C0, O1, O2 be the centers of the congruent circles Γa, Γb, Γc, Ω1, Ω2.Since Γa is the reflection of (O′) in B′C ′ and since O′ and I are isogonal conjugated withrespect to triangle A′B′C ′, the quadrilateral B′Z ′C ′I is a parallelogram, and therefore,its diagonals B′C ′, IZ ′ cut each other at half at the midpoint of segment B′C ′. It nowfollows that the quadrilateral IA0Z

′O′ is also a parallelogram, and thus, IA0 = O′Z ′ = r2 .


Now since A0I = A0X′1 = O1I = O1X

′1 = r

2 , the quadrilateral A0IO1X′1 is a rhombus,

and since the segments O1X′1, IA0, O

′Z ′ are parallel and congruent, the quadrilateralO1X

′1Z′O′ is a parallelogram. In conclusion, the triangles PX ′1Z

′, and M ′O1O′ are

congruent and P ′Z ′ = M ′O′ = r2 . Hence, Z ′ lies on the circle (P ′), which means that

the second intersection of the circumcircle (P ) of triangle MX1X2 with the circumcircle(O) of ABC (different from M) coincides with the tangency point Z of the mixtilinearincircle in angle A with the circumcircle (O).

Note that the proposed problem ensures a geometric constructions of the mixtilinearincircles.

Construction. Let M be an arbitrary point on the circumcircle of a triangle ABC andlet the tangents from this point to the incircle of the triangle intersect the sideline BC atpoint X1 and X2. Denote by Z the second intersection of the circumcircles of trianglesABC and MX1X2, and let Ka be the intersection of the lines AI and ZO, where I, Oare the incenter, and circumcenter, respectively. The circle centered at Ka with radiusKaZ is the mixtilinear incircle in angle A, being simultanously tangent to AB, AC, andinternally to (O).

Also solved by El Alami Mohamed


O204. Alice and Bob play the following game: Alice has a pawn in the upper-left unit cell of a(2m + 1) × (2n + 1) board that she wants to move to the low-right cell after a infinitenumber of steps. At each step she is allowed to move the pawn to an adjacent square,while Bob chooses a particular cell to ”block”, but still so that Alice still has a path fromher current position to the low-right corner via adjacent cells. What is the maximumnumber of moves Bob can force Alice to make?

Proposed by Radu Bumbacea, Bucharest, Romania

No solution has been received yet.


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