CRMO-2015 questions and solutions
1. Let ABC be a triangle. Let B 0 and C 0 denote respectively the reflection of B and C in
the internal angle bisector of A. Show that the triangles ABC and AB 0 C 0 have the same
incentre.
Solution: Join BB 0 and CC 0 . Let the internal angle
bisector ` of A meet BB 0 in E and CC 0 in F . Since
B 0 is the reflection of B in `, we observe that BB 0 `
and BE = EB 0 . Hence B 0 lies on AC. Similarly, C 0 lies
on the line AB.
Let D be the point of intersection of BC and B 0 C 0 .
Observe that BB 0 k C 0 C. Moreover the triangles ABC
is congruent to AB 0 C 0 : this follows from the observation
that AB = AB 0 and AC = AC 0 and the included angle
A is common. Hence BC 0 = B 0 C so that C 0 CB 0 B is
an isosceles trapezium. This means that the intesection
point D of its diagonal lies on the perpendicular bisector
of its parallel sides. Thus ` passes through D. We also
observe that CD = C 0 D.
Let I be the incentre of 4ABC. This means that CI bisects C. Hence AI/ID = AC/CD.
But AC = AC 0 and CD = C 0 D. Hence we also get that AI/ID = AC 0 /C 0 D. This implies
that C 0 I bisects AC 0 B 0 . Therefore the two angle bisectors of 4AC 0 B 0 meet at I. This
shows that I is also the incentre of 4AC 0 B 0 .
2. Let P (x) = x2 + ax + b be a quadratic polynomial with real coefficients. Suppose there are
real numbers s 6= t such that P (s) = t and P (t) = s. Prove that b st is a root of the
equation x2 + ax + b st = 0.
Solution: We have
s2 + as + b = t,
t2 + at + b = s.
This gives
(s2 t2 ) + a(s t) = (t s).
Since s 6= t, we obtain s + t + a = 1. Adding the equations, we obtain
s2 + t2 + a(s + t) + 2b = (s + t).
Therefore
(s + t)2 2st + a(s + t) + 2b = (s + t).
Using s + t = (1 + a), we obtain
(1 + a)2 2st a(1 + a) + 2b = 1 a.
Simplification gives st = 1+a+b = P (1). This shows that x = 1 is a root of x2 +ax+bst = 0.
Since the product of roots is b st, the other root is b st.
3. Find all integers a, b, c such that
a2 = bc + 1, b2 = ca + 1.
Solution: Suppose a = b. Then we get one equation: a2 = ac + 1. This reduces to
a(a c) = 1. Therefore a = 1, a c = 1; and a = 1, a c = 1. Thus we get
(a, b, c) = (1, 1, 0) and (1, 1, 0).
If a 6= b, subtracting the second relation from the first we get
a2 b2 = c(b a).
This gives a + b = c. Substituting this in the first equation, we get
a2 = b(a b) + 1.
Thus a2 + b2 + ab = 1. Multiplication by 2 gives
(a + b)2 + a2 + b2 = 2.
Thus (a, b) = (1, 1), (1, 1), (1, 0), (1, 0), (0, 1), (0, 1). We get respectively c =
0, 0, 1, 1, 1, 1. Thus we get the triples:
(a, b, c) = (1, 1, 0), (1, 1, 0), (1, 1, 0), (1, 1, 0), (1, 0, 1), (1, 0, 1), (0, 1, 1), (0, 1, 1).
4. Suppose 32 objects are placed along a circle at equal distances. In how many ways can 3
objects be chosen from among them so that no two of the three chosen objects are adjacent
nor diametrically opposite?
Solution: One can choose 3 objects out of 32 objects in 32
3 ways. Among these choices
all would be together in 32 cases; exactly two will be together in 32 28 cases. Thus three
objects can be chosen such that no two adjacent in 32 3 32 (32 28) ways. Among these,
furthrer, two objects will be diametrically opposite in 16 ways and the third would be on
either semicircle in a non adjacent portion in 32 6 = 26 ways. Thus required number is
32
32 (32 28) (16 26) = 3616.
3
5. Two circles and in the plane intersect at two distinct points A and B, and the centre
of lies on . Let points C and D be on and , respectively, such that C, B and D are
collinear. Let point E on be such that DE is parallel to AC. Show that AE = AB.
Solution: If O is the centre of , then we have
1 1
AEB = AOB = (180 ACB)
2 2
1 1 1
= EDB = 180 EAB = 90 EAB .
2 2 2
But we know that AEB + EAB + EBA = 180 .
Therefore
1 1
EBA = 180 AEB EAB = 180 90 + EAB EAB = 90 EAB.
2 2
This shows that AEB = EBA and hence AE = AB.
2
6. Find all real numbers a such that 4 < a < 5 and a(a 3{a}) is an integer. (Here {a} denotes
the fractional part of a. For example {1.5} = 0.5; {3.4} = 0.6.)
Solution: Let a = 4 + f , where 0 < f < 1. We are given that (4 + f )(4 2f ) is an integer.
This implies that 2f 2 + 4f is an integer. Since 0 < f < 1, we have 0 < 2f 2 + 4f < 6.
Therefore 2f 2 + 4f can take 1, 2, 3, 4 or 5. Equating 2f 2 + 4f to each one of them and using
f > 0, we get
2 + 6 2 + 8 2 + 10 2 + 12 2 + 14
f= , , , , .
2 2 2 2 2
Therefore a takes the values:
6 8 10 12 14
a=3+ , 3+ , 3+ , 3+ , 3+ .
2 2 2 2 2
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