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Team Contest
1. If x and y are positive real numbers, find the smallest value of
2 2 2 2
225 15 2− x+x + 200−20y+ y + x − 2xy+ y . Solution
Consider a pentagon OABCD such that 15
OA= , OB=x, OC = y, 10 2
OD= and
45
AOB BOC COD
∠ = ∠ = ∠ = °.
Then by the cosine theorem,
2 2 cos 45 AB= OA +OB −OA OB× × ° 2 225 x 15 2x = + − , 2 2 cos 45 BC = OB +OC −OB OC× × ° 2 2 2 x y xy = + − and CD= OC2 +OD2 −OC OD× ×cos 45° = y2 +200−20y.
By the triangle inequality, the smallest value that AB+BC+CD can have is AD, and by the cosine theorem it equals
2 2
cos135 225 200 300 725 5 29
AD= OA +OD −OA OD× × ° = + + = = . Answer: 5 29
2. The sum of 2016 real numbers is 2017 and each of them is less than 2017 2015. Prove that the sum of any two of the numbers is greater than or equal to 2017
2015. Proof
Suppose the sum of two of the numbers is less than 2017
2015. Remove them to leave behind 2014 numbers with sum at least 2017 2017 2015 2017 2017
2015 2015 2015
− = × − =
2017 2014
2015
× . By the Pigeonhole Principle, at least one of them must be greater than or equal to 2017
2015. We have a contradiction.
International Young Mathematicians’ Convention
Senior Level
O C B A D y x 10 2 15 45° 45° 45°【Marking Scheme】
Using Counter–evidence method, 10 marks.
Remove them to leave behind 2014 numbers with sum at least 2014 2017 2015 × , 10 marks.
Conclude that at least one of them must be greater than or equal to 2017 2015, 20 marks.
3. There are 2016 unit cubes, each of which can be painted black or white. How many values of n is it possible to construct an n n n× × cube with n unit 3
cubes such that each cube shares a common face with exactly three cubes of the opposite colour?
Solution
The task is possible if and only if n is even. For odd n, construct a graph with n 3
vertices representing the unit cubes. Two vertices are joined by an edge if and only if the unit cubes they represent have different colours and share a common face. Since the number of vertices is odd, it is not possible for every vertex to have degree 3, as the total degree is even. For n=2, a checkerboard colouring works. For larger even
n, the cube can be assembled from 2 2 2× × cubes in such a way that only cubes of the same colour come into contact in the merger. The possible values of n is 2, 4, 6, 8, 10 and 12 since 123 =1728<2016 14< 3 =2744. Hence there are 6 such values.
Answer: 6
4. P is the midpoint of the arc AC of the circumcircle
of an equilateral triangle ABC. M is another point on this arc and N is the midpoint of BM. K is the projection of P on the line MC, as shown in the diagram below. If the length of NA is 19 cm, find the length of NK in cm.
Solution
Let O be the circumcentre of ABC. Connect OA, ON,
OB, PA, PC and KA. Since ∠ONB= ° = ∠90 PKC,
OBN PCK
∠ = ∠ and OB = PC, triangles BON and CPK are congruent. It follows that ∠BON = ∠CPK
and ON = PK. Now
360 360
AON AOB BON APC CPK APK
∠ = ° − ∠ − ∠ = ° − ∠ − ∠ = ∠ , so that triangles
AON and APK are congruent. It follows that we have AN = AK. Now
60
NAK OAK OAN OAK PAK OAP
∠ = ∠ − ∠ = ∠ − ∠ = ∠ = °. Hence ANK is an equilateral triangle and NK =NA=19cm
Answer: 19 cm
【Marking Scheme】
Observe that triangles BON and CPK are congruent, 10 marks. Or conclude that ∠BON = ∠CPK and ON = PK, each 5 marks.
B A P O C N M K
Observe that triangles AON and APK are congruent, 10 marks. Or conclude that AN = AK, 10 marks.
Observe that ANK is an equilateral triangle, 10 marks Conclude that NK =NA=19cm, 10 marks.
Correct answer without reasons, 0 mark.
5. Let S n denote the n-th sequence so that every word in a sequence consists only of the letters A and B. The first word has only one letter A. For k ≥2, the k-th word is obtained from the (k −1)-th by simultaneously replacing every A by AAB and every B by A. Then every word is an initial part of the next word. For
example, S1= A, S2 = AAB, S3 = AABAABA and
4
S = AABAABAAABAABAAAB. Find the number of As in S10.
Solution
Now S n is obtained from Sn-1 by the replacement. We symbolize this as
1
( n ) n
t S − = S . Note that S consists of two copies of 3 S and one copy of 2 S strung 1
together in that order. We symbolize this as S3 =S2 S2 S1. We claim that for n≥3,
1 1 2
n n n n
S =S − S − S − . This holds for n = 3. Suppose it holds for some n≥3. Then we have 1 1 1 2 1 1 2 1 ( ) ( ) ( ) ( ) ( ) n n n n n n n n n n n S t S t S S S t S t S t S S S S + − − − − − − − = = = =
Hence our claim holds for all n≥3. Let a and n b be the respective numbers of n As and Bs in S . Then n a1 =1, b1=0 and, for n≥2, we have an =2an−1+bn−1 and
1
n n
b =a − . These recurrence relations yields the following table.
n 1 2 3 4 5 6 7 8 9 10 n a 1 2 5 12 29 70 169 408 985 2378 n b 0 1 2 5 12 29 70 169 408 985 n n a +b 1 3 7 17 41 99 239 577 1393 3363 Answer: 2378
6. D, E and F are points on the sides BC, CA and AB, respectively, of triangle ABC such that AD, BE and CF are concurrent. The area of triangle ABC is 2016 cm2. If there exists a point P such that both BDPE and AFCP are parallelograms, as shown in the diagram below. Find the area of triangle of DEF, in cm2.
Solution
Extend PE to cut AB at Q. Then BCPQ and DCEQ are also parallelograms. Also, triangles CEP and AEQ are similar. Suppose AE ≠CE . We consider two cases.
Case 1. AE < CE.
Then BD = PE > QE = CD and AF = CP > AQ. Hence Q lies on AF, as shown in the diagram below on the left, so that AF = CP = BQ > BF. Hence BD CE AF 1
DC× EA× FB > ,
which contradicts Ceva’s Theorem.
Case 2. AE > CE.
Then Q lies on BF, as shown in the diagram above on the right. We can prove as in Case 1 that BD CE AF 1
DC× EA× FB < . This also contradicts Ceva’s Theorem. It follows that
AE = CE, so that triangles CEP and AEQ are congruent. Hence Q coincides with F, so that AF = CP = BQ = BF and BD = PE = QE = BD.
Hence D, E and F are the respective midpoints of BC, CA and AB. Thus the area of triangle of DEF is 1 2016 504 4× = cm 2 . Answer: 504 cm2 【Marking Scheme】
Observe that there are 2 cases, 5 marks. Complete the proof for AE < CE, 10 marks.
Complete the proof for AE > CE, 10 marks.
Conclude that D, E and F are the respective midpoints of BC, CA and AB, 10
marks.
Conclude the correct answer, 5 marks.
Just say D, E and F are the respective midpoints of BC, CA and AB and hence get
the correct answer without reasons, 10 marks.
A E D C B Q F P A E D C B Q F P
