41st Balkan Mathematical Olympiad

27th April – 2nd May 2024

Language: English
Varna, Bulgaria Monday, April 29, 2024

Problem 1. Let ABC be an acute-angled triangle with AC > AB and let D be the foot of the
A-angle bisector on BC. The reflections of lines AB and AC in line BC meet AC and AB at points
E and F respectively. A line through D meets AC and AB at G and H respectively such that G
lies strictly between A and C while H lies strictly between B and F . Prove that the circumcircles of
4EDG and 4F DH are tangent to each other.

Problem 2. Let n ≥ k ≥ 3 be integers. Show that for every integer sequence 1 ≤ a1 < a2 < . . . <
ak ≤ n one can choose non-negative integers b1 , b2 , . . . , bk , satisfying the following conditions:

(i) 0 ≤ bi ≤ n for each 1 ≤ i ≤ k,

(ii) all the positive bi are distinct,

(iii) the sums ai + bi , 1 ≤ i ≤ k, form a permutation of the first k terms of a non-constant arithmetic
progression.

Problem 3. Let a and b be distinct positive integers such that 3a + 2 is divisible by 3b + 2. Prove
that a > b2 .

Problem 4. Let R+ = (0, ∞) be the set of all positive real numbers. Find all functions f : R+ → R+
and polynomials P (x) with non-negative real coefficients such that P (0) = 0 which satisfy the equality

f (f (x) + P (y)) = f (x − y) + 2y

for all real numbers x > y > 0.

Time is 4 hours and 30 minutes

Each problem is worth 10 points
 Language: English
Problems with Solutions
Monday, April 29, 2024

Problem 1. Let ABC be an acute-angled triangle with AC > AB and let D be the foot of the
A-angle bisector on BC. The reflections of lines AB and AC in line BC meet AC and AB at points
E and F respectively. A line through D meets AC and AB at G and H respectively such that G
lies strictly between A and C while H lies strictly between B and F . Prove that the circumcircles of
△EDG and △F DH are tangent to each other.

Solution 1. Let X and Y lie on the tangent to the circumcircle of △EDG on the opposite side to
D as shown in the figure below. Regarding diagram dependency, the acute condition with AC > AB
ensures E lies on extension of CA beyond A, and F lies on extension of AB beyond B. The condition
on ℓ means the points lie in the orders E, A, G, C and A, B, H, F .

C
b

G Xb
b

b
D

Y
A b b b b

B H F
b

Using the alternate segment theorem, the condition that ⊙EDG and ⊙F DH are tangent at D can be
rewritten as
HF D = Y DH.

But using the same theorem, we get Y DH = XDG = DEG. So we can remove G, H from the
figure, and it is sufficient to prove that DEA = DF B.

The reflection property means that AD and BD are external angle bisectors in △EAB and hence D
is the E-excentre of this triangle. Thus DE (internally) bisects BEA, giving

DEA = DEB.

Now observe that the pairs of lines (BE, CE) and (BF, CF ) are reflections in BC thus E, F are
reflections in BC. Also D is its own reflection in BC. Hence DEB = DF B and so

DEA = DEB = DF B,

as required.

Problem 2. Let n ≥ k ≥ 3 be integers. Show that for every integer sequence 1 ≤ a1 < a2 < . . . <
ak ≤ n one can choose non-negative integers b1 , b2 , . . . , bk , satisfying the following conditions:

(i) 0 ≤ bi ≤ n for each 1 ≤ i ≤ k,

(ii) all the positive bi are distinct,

(iii) the sums ai + bi , 1 ≤ i ≤ k, form a permutation of the first k terms of a non-constant arithmetic
progression.

Solution 1. Let the resulting progression be Ans := {ak − (k − 1), ak − (k − 2), . . . , ak } and at be
the largest number not belonging to Ans. Clearly the set Ans \ {a1 , a2 , . . . , ak } has cardinality t; let
its members be c1 > c2 > · · · > ct . Define bj := cj − aj for 1 ≤ j ≤ t or zero otherwise. Since {cj } is
decreasing and {aj } is increasing, all bj are distinct and clearly b1 < n. After we add bj to aj we get
a permutation of Ans as desired.

Solution 2. Let the resulting progression be Ans := {ak − (k − 1), ak − (k − 2), . . . , ak }.

We proceed with the following reduction. Let δ be the smallest b we used before (in the beginning
it is n). While a1 ∈
/ Ans we map a1 to the largest element q of Ans \ {a1 , a2 , . . . , ak } and put
δnew := b1 := q − a1 . Now we rearrange the sequence of a-s. We do not touch Ans ∩ {a1 , a2 , . . . , ak }
so every b is defined at most once (in the end undefined b-s become zeros). Also b < δ and δ decreases
at each step, because q decreases and a1 grows, and hence all nonzero b-s are distinct.

Problem 3. Let a and b be distinct positive integers such that 3a + 2 is divisible by 3b + 2. Prove
that a > b2 .

Solution 1. Obviously we have a > b. Let a = bq + r, where 0 ≤ r < b. Then

3a ≡ 3bq+r ≡ (−2)q .3r ≡ −2 (mod 3b + 2)

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So 3b + 2 divides A = (−2)q .3r + 2 and it follows that

|(−2)q .3r + 2| ≥ 3b + 2 or (−2)q .3r + 2 = 0.

We make case distinction:

1. (−2)q .3r + 2 = 0. Then q = 1 and r = 0 or a = b, a contradiction.

2. q is even. Then
A = 2q .3r + 2 = (3b + 2).k.

Consider both sides of the last equation modulo 3r . Since b > r:

2 ≡ 2q .3r + 2 = (3b + 2)k ≡ 2k (mod 3r ),

so it follows that 3r |k − 1. If k = 1 then 2q .3r = 3b , a contradiction. So k ≥ 3r + 1, and therefore:

A = 2q .3r + 2 = (3b + 2)k ≥ (3b + 2)(3r + 1) > 3b .3r + 2

It follows that

2
2q .3r > 3b .3r , i.e. 2q > 3b , which implies 3b < 2bq < 3bq ≤ 3bq+r = 3a .

Consequently a > b2 .

3. If q is odd. Then
2q .3r − 2 = (3b + 2)k.

Considering both sides of the last equation modulo 3r , and since b > r, we get: k + 1 is divisible
by 3r and therefore k ≥ 3r − 1. Thus r > 0 because k > 0, and:

2q .3r − 2 = (3b + 2)k ≥ (3b + 2)(3r − 1), and therefore

3r
2q .3r > (3b + 2)(3r − 1) > 3b (3r − 1) > 3b , which shows
2
2q+1 > 3b .

But for q > 1 we have 2q+1 < 3q , which combined with the above inequality, implies that
2
3b < 2(q+1)b < 3qb ≤ 3a , q.e.d. Finally, If q = 1 then 2q .3r − 2 = (3b + 2)k and consequently
2.3r − 2 ≥ 3b + 2 ≥ 3r+1 + 2 > 2.3r − 2, a contradiction.

Solution 2. D = a − b, and we shall show D > b2 − b. We have 3b + 2|3a + 2, so 3b + 2|3D − 1. Let

D = bq + r where r < b. First suppose that r ̸= 0. We have

q+1 r−b
1 ≡ 3D ≡ 3bq+r ≡ (−2) 3 (mod 3b + 2) =⇒ 3b−r ≡ (−2)q+1 (mod 3b + 2)

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Therefore
3b + 2 ≤ |(−2)q+1 − 3b−r | ≤ 2q+1 + 3b−r ≤ 2q+1 + 3b−1

Hence
log 3
2 × 3b−1 + 2 ≤ 2q+1 =⇒ 3b−1 < 2q =⇒ (b − 1) < q
log 2
q
Which yields D = bq + r > bq > log 3
log 2 b(b − 1) ≥ b2 − b as desired. Now for the case r = 0, (−2) ≡ 1
(mod 3b + 2) and so

log 3
3b + 2 ≤ |(−2)q − 1| ≤ 2q + 1 =⇒ 3b−1 < 3b < 2q =⇒ (b − 1) < q
log 2

and analogous to the previous case, D = bq + r = bq > log 3

log 2 b(b − 1) ≥ b2 − b.

Problem 4. Let R+ = (0, ∞) be the set of all positive real numbers. Find all functions f : R+ → R+
and polynomials P (x) with non-negative real coefficients such that P (0) = 0 which satisfy the equality

f (f (x) + P (y)) = f (x − y) + 2y

for all real numbers x > y > 0.

Solution 1. Assume that f : R+ → R+ and the polynomial P with non-negative coefficients and
P (0) = 0 satisfy the conditions of the problem. For positive reals with x > y, we shall write Q(x, y)
for the relation:
f (f (x) + P (y)) = f (x − y) + 2y.

1. Step 1. f (x) ≥ x. Assume that this is not true. Since P (0) = 0 and P is with non-negative
coefficients, P (x) + x is surjective on positive reals. If f (x) < x for some positive real x, then
setting y such that y + P (y) = x − f (x) (where obviously y < x), we shall get f (x) + P (y) = x − y
and by Q(x, y), f (f (x) + P (y)) = f (x − y) + 2y, we get 2y = 0, a contradiction.

2. Step 2. P (x) = cx for some non-negative real c. We will show deg P ≤ 1 and together with
P (0) = 0 the result will follow. Assume the contrary. Hence there exists a positive l such that
P (x) ≥ 2x for all x ≥ l. By Step 1 we get

∀x > y ≥ l : f (x − y) + 2y = f (f (x) + P (y)) ≥ f (x) + P (y) ≥ f (x) + 2y

and therefore f (x − y) ≥ f (x). We get f (y) ≥ f (2y) ≥ · · · ≥ f (ny) ≥ ny for all positive integers
n, which is a contradiction.

3. Step 3. If c ̸= 0, then f (f (x) + 2z + c2 ) = f (x + 1) + 2(z − 1) + 2c for z > 1. Indeed by

Q(f (x + z) + cz, c), we get

f (f (f (x + z) + cz) + c2 ) = f (f (x + z) + cz − c) + 2c = f (x + 1) + 2(z − 1) + 2c.

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 On the other hand by Q(x + z, z), we have:

f (x) + 2z + c2 = f (f (x + z) + P (z)) + c2 = f (f (x + z) + cz) + c2 .

Substituting in the LHS of Q(f (x + z) + cz, c), we get f (f (x) + 2z + c2 ) = f (x + 1) + 2(z − 1) + 2c.

4. Step 4. There is x0 , such that f (x) is linear on (x0 , ∞). If c ̸= 0, then by Step 3, fixing x = 1, we
get f (f (1) + 2z + c2 ) = f (2) + 2(z − 1) + 2c which implies that f is linear for z > f (1) + 2 + c2 . As
for the case c = 0, consider y, z ∈ (0, ∞). Pick x > max(y, z), then by Q(x, x − y) and Q(x, x − z)
we get:
f (y) + 2(x − y) = f (f (x)) = f (z) + 2(x − z)

which proves that f (y) − 2y = f (z) − 2z and there fore f is linear on (0, ∞).

5. Step 5. P (y) = y and f (x) = x on (x0 , ∞). By Step 4, let f (x) = ax + b on (x0 , ∞). Since f
takes only positive values, a ≥ 0. If a = 0, then by Q(x + y, y) for y > x0 we get:

2y + f (x) = f (f (x + y) + P (y)) = f (b + cy).

Since the LHS is not constant, we conclude c ̸= 0, but then for y > x0 /c, we get that the RHS
equals b which is a contradiction.

Hence a > 0. Now for x > x0 and x > (x0 − b)/a large enough by P (x + y, y) we get:

ax + b + 2y = f (x) + 2y = f (f (x + y) + P (y)) = f (ax + ay + b + cy) = a(ax + ay + b + cy) + b.

Comparing the coefficients before x, we see a2 = a and since a ̸= 0, a = 1. Now 2b = b and thus
b = 0. Finally, equalising the coefficients before y, we conclude 2 = 1 + c and therefore c = 1.

Now we know that f (x) = x on (x0 , ∞) and P (y) = y. Let y > x0 . Then by Q(x + y, x) we conclude:

f (x) + 2y = f (f (x + y) + P (y)) = f (x + y + y) = x + 2y.

Therefore f (x) = x for every x. Conversely, it is straightforward that f (x) = x and P (y) = y do
indeed satisfy the conditions of the problem.

Solution 2. Assume that the function f : R+ → R+ and the polynomial with non-negative coeffi-
cients P (y) = yP1 (y) satisfy the given equation. Fix x = x0 > 0 and note that:

f (f (x0 + y) + P (y)) = f (x0 + y − y) + 2y = f (x0 ) + 2y.

Assume that g = 0. Then f (f (x + y)) = f (x) + 2y for x, y > 0. Let x > 0 and z > 0. Pick y > 0.
Then:
2y + f (x + z) = f (f (x + y + z)) = f (f (x + z + y)) = f (x) + 2(z + y).

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Therefore f (x + z) = f (x) + 2z for any x > 0 and z > 0. Setting c = f (1), we see that f (z + 1) = c + 2z
for all positive z. Therefore if x, y > 1 we have that f (x + y) = c + 2(x + y − 1) > 1. This shows that:

f (f (x + y)) = c + 2(f (x + y) − 1) = 3c + 4(x + y) − 4.

On the other hand f (x) + 2y = c + 2x + 2y. Therefore the equality f (f (x + y)) = f (x) + 2y is not
universally satisfied.

From now on, we assume that g ̸= 0. Therefore P is strictly increasing with P (0) = 0, limy→∞ P (y) =
∞, i.e. g is bijective on [0, ∞) and P (0) = 0.

Let x > 0, y > 0 and set u = f (x + y), v = P (y). From above, we have u > 0 and v > 0. Therefore:

f (f (u + v) + P (v)) = f (u) + 2v = f (f (x + y)) + 2P (y).

On the other hand f (u + v) = f (f (x + y) + P (y)) = f (x) + 2y. Therefore we obtain that:

f (f (x) + 2y + P (P (y))) = f (f (x + y)) + 2P (y).

Since g is bijective from (0, ∞) to (0, ∞) for any z > 0 there is t such that P (t) = z. Applying this
observation to z = P (P (y)) + 2y and setting x′ = x + t, we obtain that:

f (f (x+t+y))+2P (y) = f (f (x′ +y))+2P (y) = f (f (x′ )+P (P (y))+2y) = f (f (x+t)+P (t)) = f (x)+2t.

Thus if we denote h(y) = P (P (y)) + 2y, then t = P (−1) (h(y)) and the above equality can be rewritten
as:

f (f (x + P (−1) (h(y)) + y)) = f (x) + 2P (−1) (h(y)) − 2P (y) = f (x) + 2P (−1) (h(y)) + 2y − 2y − 2P (y).

Let s(y) = P (−1) (h(y))+y and note that since h is continuous and monotone increasing, g is continuous
and monotone increasing, then so are P (−1) and consequently P (−1) ◦ h and s. It is also clear, that
limy→0 s(y) = 0 and limy→∞ s(y) = ∞. Therefore s is continuously bijective from [0, ∞) to [0, ∞)
with s(0) = 0.

Thus we have:
f (f (x + s(y))) = f (x) + 2s(y) − 2y − 2P (y)

and using that s is invertible, we obtain:

f (f (x + y)) = f (x) + 2y − 2s(−1) (y) − 2P (s(−1) (y)).

Now fix x0 , then for any x > x0 and any y > 0 we have:

f (x) + 2y − 2s(−1) (y) − 2P (s(−1) (y)) = f (f (x + y)) = f (f (x0 + x + y − x0 ))

= f (x0 ) + 2(x + y − x0 ) − 2s(−1) (x + y − x0 ) − 2P (s(−1) (x + y − x0 )).

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Setting y = x0 , we get:

f (x) + 2x0 − 2s(−1) (x0 ) − 2P (s(−1) (x0 )) = f (x0 ) + 2x − 2s(−1) (x) − 2P (s(−1) (x)).

Since this equality is valid for any x > x0 we actually have that:

f (x) − 2x + 2s(−1) (x) + 2P (s(−1) (x)) = c for some fixed constant c ∈ R and all x ∈ R+ .

Let ϕ(x) = −x + 2s(−1) (x) + 2P (s(−1) (x)). Then f (x) = x − ϕ(x) + c and since ϕ is a sum of continuous
functions that are continuous at 0. Therefore f is continuous and can be extended to a continuous
function on [0, ∞). Back in the original equation we fix x > 0 and let y tend to 0. Using the continuity
of f and g on [0, ∞) and P (0) = 0 we obtain:

f (f (x)) = lim f (f (x) + P (y)) = lim (f (x − y) + P (y)) = f (x) + P (0) = f (x).

y→0+ y→0+

Finally, fixing x = 1 and varying y > 0, we obtain:

f (f (1 + y) + P (y)) = f (1) + 2y.

It follows that f takes every value on (f (1), ∞). Therefore for any y ∈ (f (1), ∞) there is z such that
f (z) = y. Using that f (f (z)) = f (z) we conclude that f (y) = y for all y ∈ (f (1), ∞).

Now fix x and take y > f (1). Hence

f (x) + 2y = f (f (x + y) + P (y)) = f (x + y + P (y)) = x + y + P (y).

We conclude f (x) − x = P (y) − y for every x an y > f (1). In particular f (x1 ) − x1 = f (x2 ) − x2 for
all x1 , x2 ∈ (0, ∞) and since f (x) = x for x ∈ (f (1), ∞), we get f (x) = x on (0, ∞).

Finally, x + 2y = f (x) + 2y = f (f (x + y) + P (y)) = f (x + y) + P (y) = x + y + P (y), which shows that

P (y) = y for every y ∈ (0, ∞).

It is also straightforward to check that f (x) = x and P (y) = y satisfy the equality:

f (f (x + y) + P (y)) = f (x + 2y) = x + 2y = f (x) + 2y.