RMO Solved Paper
RMO Solved Paper
RMO Solved Paper
December 1, 2013
1. Let ABC be an acute angled triangle. The circle with BC as diameter intersects AB and
AC again at P and Q, respectively. Determine BAC given that the orthocenter of triangle
AP Q lies on .
Solution. Let K denote the orthocenter of triangle AP Q. Since triangles ABC and AQP
are similar it follows that K lies in the interior of triangle AP Q.
Note that KP A = KQA = 90 A. Since BP KQ is a cyclic quadrilateral it follows that
BQK = 180 BP K = 90 A, while on the other hand BQK = BQA KQA =
A since BQ is perpendicular to AC. This shows that 90 A = A, so A = 45 .
2. Let f (x) = x3 + ax2 + bx + c and g(x) = x3 + bx2 + cx + a, where a, b, c are integers with
c 6= 0. Suppose that the following conditions hold:
(a) f (1) = 0;
(b) the roots of g(x) are squares of the roots of f (x).
Find the value of a2013 + b2013 + c2013 .
Solution. Note that g(1) = f (1) = 0, so 1 is a root of both f (x) and g(x). Let p and q be the
other two roots of f (x), so p2 and q 2 are the other two roots of g(x). We then get pq = c and
p2 q 2 = a, so a = c2 . Also, (a)2 = (p + q + 1)2 = p2 + q 2 + 1 + 2(pq + p + q) = b + 2b = b.
Therefore b = c4 . Since f (1) = 0 we therefore get 1 + c c2 + c4 = 0. Factorising, we
get (c + 1)(c3 c2 + 1) = 0. Note that c3 c2 + 1 = 0 has no integer root and hence
c = 1, b = 1, a = 1. Therefore a2013 + b2013 + c2013 = 1.
3. Find all primes p and q such that p divides q 2 4 and q divides p2 1.
Solution. Suppose that p q. Since q divides (p 1)(p + 1) and q > p 1 it follows that q
divides p + 1 and hence q = p + 1. Therefore p = 2 and q = 3.
On the other hand, if p > q then p divides (q 2)(q + 2) implies that p divides q + 2 or
q 2 = 0. This gives either p = q + 2 or q = 2. In the former case it follows that that q
divides (q +2)2 1, so q divides 3. This gives the solutions p > 2, q = 2 and (p, q) = (5, 3).
4. Find the number of 10-tuples (a1 , a2 , . . . , a10 ) of integers such that |a1 | 1 and
a21 + a22 + a23 + + a210 a1 a2 a2 a3 a3 a4 a9 a10 a10 a1 = 2 .
Solution. Let a11 = a1 . Multiplying the given equation by 2 we get
(a1 a2 )2 + (a2 a3 )2 + (a10 a1 )2 = 4 .
Note that if ai ai+1 =P
2 for some i = 1, . . . , 10, then aj aj+1 = 0 for all j 6= i which
contradicts the equality 10
i=1 (ai ai+1 ) = 0. Therefore ai ai+1 = 1 for exactly two values
of i in {1, 2, . . . , 10}, ai ai+1 = 1 for two other values of i and ai ai+1 = 0 for all other
8
values of i. There are 10
2 2 = 45 28 possible ways of choosing these values. Note
that a1 = 1, 0 or 1, so in total there are 3 45 28 possible integer solutions to the given
equation.
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December 1, 2013
5. Let ABC be a triangle with A = 90 and AB = AC. Let D and E be points on the segment
BC such that BD : DE : EC = 3 : 5 : 4. Prove that DAE = 45 .
Solution. Rotating the configuraiton about A by 90 , the point B goes to the point C. Let
P denote the image of the point D under this rotation. Then CP = BD and ACP =
ABC = 45 , so ECP is a right-angled triangle with CE : CP = 4 : 3. Hence P E = ED.
It follows that ADEP is a kite with AP = AD and P E = ED. Therefore AE is the angular
bisector of P AD. This implies that DAE = P AD/2 = 45 .
6. Suppose that m and n are integers such that both the quadratic equations x2 + mx n = 0
and x2 mx + n = 0 have integer roots. Prove that n is divisible by 6.
Solution. Let a be an integer. If a is not divisible by 3 then a2 1 (mod 3), i.e., 3 divides
a2 1, and if a is odd then a2 1 (mod 8), i.e., 8 divides a2 1.
Note that the discriminants of the two quadratic polynomials are both squares of integers.
Let a and b be integers such that m2 4n = a2 and m2 + 4n = b2 . Therefore 8n = b2 a2
and 2m2 = a2 + b2 . If 3 divides m then 3 divides both a and b, so 3 divides n. On the other
hand if 3 does not divide m then 3 does not divide a or b. Therefore 3 divides b2 a2 and
hence 3 divides n.
If m is odd, then so is a, and therefore 4n = m2 a2 is divisible by 8, so n is even. On
the other hand, if m is even then both a and b are even. Further (m/2)2 n = (a/2)2 and
(m/2)2 + n = (b/2)2 , so (b a)/2 is even. In particular, n = (b2 a2 )/4 is even.
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Paper 2
December 1, 2013
1. Prove that there do not exist natural numbers x and y, with x > 1, such that
x7 1
= y5 + 1 .
x1
Solution. Simple factorisation gives y 5 = x(x3 + 1)(x2 + x + 1). The three factors on the
right are mutually coprime and hence they all have to be fifth powers. In particular, x = r5
for some integer r. This implies x3 + 1 = r15 + 1, which is not a fifth power unless r = 1 or
r = 0. This implies there are no solutions to the given equation.
2. In a triangle ABC, AD is the altitude from A, and H is the orthocentre. Let K be the centre
of the circle passing through D and tangent to BH at H. Prove that the line DK bisects AC.
Solution. Note that KHB = 90 . Therefore KDA = KHD = 90 BHD =
HBD = HAC. On the other hand, if M is the midpoint of AC then it is the circumcenter
of triangle ADC and therefore M DA = M AD. This proves that D, K, M are collinear
and hence DK bisects AC.
3. Consider the expression
20132 + 20142 + 20152 + + n2 .
Prove that there exists a natural number n > 2013 for which one can change a suitable
number of plus signs to minus signs in the above expression to make the resulting expression
equal 9999.
Solution. For any integer k we have
k 2 + (k + 1)2 + (k + 2)2 (k + 3)2 = 4 .
Note that 9999 (20132 + 20142 + 20152 + 20162 + 20172 ) = 4m for some positive integer
m. Therefore, it follows that
9999 =(20132 + 20142 + 20152 + 20162 + 20172 )
m
X
+
(4r + 2014)2 + (4r + 2015)2 + (4r + 2016)2 (4r + 2017)2 .
r=1
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5. Let n 3 be a natural number and let P be a polygon with n sides. Let a1 , a2 , . . . , an be the
lengths of the sides of P and let p be its perimeter. Prove that
a1
a2
an
+
+ +
< 2.
p a1 p a2
p an
Solution. If r and s are positive real numbers such that r < s then r/s < (r + x)/(s + x)
for any positive real x. Note that, by triangle inequality, ai < p ai , so
ai
2ai
<
,
p ai
p
for all i = 1, , 2 . . . , n. Summing this inequality over i we get the desired inequality.
6. For a natural number n, let T (n) denote the number of ways we can place n objects of weights
1, 2, . . . , n on a balance such that the sum of the weights in each pan is the same. Prove that
T (100) > T (99).
Solution. Let S(n) denote the collection of subsets A of X(n) = {1, 2, . . . , n} such that
the sum of the elements of A equals n(n + 1)/4. Then the given inequality is equivalent to
|S(100)| > |S(99)|. We shall give a map f : S(99) S(100) which is one-to-one but not
onto. Note that this will prove the required inequality.
Suppose that A is an element of S(99). If 50 A then define f (A) = (A \ {50}) {100}.
Otherwise, define f (A) = A {50}. If A and B are elements of S(99) such that f (A) = f (B)
then either 50 belongs to both these sets or neither of these sets. In either of the cases we
have A = B. Therefore f is a one-to-one function.
Note that every element in the range of f contains exactly one of 50 and 100. Let Bi =
{i, 101 i}. Then B1 B2 B24 B50 is an element of S(100). Clearly, this is not in the
range of f . Thus f is not an onto function.
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Paper 3
December 1, 2013
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Paper 3
December 1, 2013
coefficents of x in g(x)2 and in h(x)2 are both divisible by 4. In particular, the coefficient of
x in a Fermat polynomial p(x), with p(0) divisible by 4, is divisible by 4. Thus if f (x) is a
Fermat polynomial with f (0) = 1000 then f (x) + 2x cannot be a Fermat polynomial.
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Paper 4
December 1, 2013
1. Let be a circle with centre O. Let be another circle passing through O and intersecting
at points A and B. A diameter CD of intersects at a point P different from O. Prove
that
AP C = BP D .
Solution. Suppose that A0 is a point on such that A0 P C = BP D. Then the segments
OA0 and OB subtends same angle in the respective minor arcs, so OA0 = OB. This shows
that A lies on and hence A0 = A. This proves that AP C = BP D.
2. Determine the smallest prime that does not divide any five-digit number whose digits are in
a strictly increasing order.
Solution. Note that 12346 is even, 3 and 5 divide 12345, and 7 divides 12348. Consider a 5
digit number n = abcde with 0 < a < b < c < d < e < 10. Let S = (a + c + e) (b + d). Then
S = a + (c b) + (e d) > a > 0 and S = e (d c) (b a) < e 10, so S is not divisible
by 11 and hence n is not divisible by 11. Thus 11 is the smallest prime that does not divide
any five-digit number whose digits are in a strictly increasing order.
3. Given real numbers a, b, c, d, e > 1 prove that
a2
b2
c2
d2
e2
+
+
+
+
20 .
c1 d1 e1 a1 b1
Solution. Note that (a 2)2 0 and hence a2 4(a 1). Since a > 1 we have
By applying AM-GM inequality we get
a2
4.
a1
s
a2
b2
c2
d2
e2
a2 b2 c2 d2 e2
+
+
+
+
55
20 .
c1 d1 e1 a1 b1
(a 1)(b 1)(c 1)(d 1)(e 1)
1
1
4. Let x be a non-zero real number such that x4 + 4 and x5 + 5 are both rational numbers.
x
x
1
Prove that x + is a rational number.
x
Solution. For a natural number k let Tk = xk + 1/xk . Note that T4 T2 = T2 + T6 and
T8 T2 = T10 + T6 . Therefore T2 (T8 T4 + 1) = T10 . Since T2k = Tk2 + 2 it follows that T8 , T10
are rational numbers and hence T2 , T6 are also rational numbers. Since T5 T1 = T4 + T6 it
follows that T1 is a rational number.
5. In a triangle ABC, let H denote its orthocentre. Let P be the reflection of A with respect to
BC. The circumcircle of triangle ABP intersects the line BH again at Q, and the circumcircle
of triangle ACP intersects the line CH again at R. Prove that H is the incentre of triangle
P QR.
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Paper 4
December 1, 2013
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