MeritPath.com. Problems and Solutions, INMO-2011

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1 Problems and Solutions, INMO Let,, be points on the sides,, respectively of a triangle such that and. Prove that is equilateral. Solution 1: c ka kc b kb a Let ;. Note that +, and hence. Similarly, and. Thus the triangle is similar to. We may take ka, kb and kc, for some positive real constant k. pplying sine rule to triangle, we obtain c sin ka sin Rka, b where R is the circum-radius of. Thus we get Rk sin bc /a. Similarly, we obtain Rk sin ca /b and Rk sin ab /c. We therefore get bc ca ab. 1 a b c If some two sides are equal, say, a b, then ac ca giving a c; we get a b c and is equilateral. Suppose no two sides of are equal. We may assume a is the least. Since 1 is cyclic in a, b, c, we have to consider two cases: a < b < c and a < c < b. ase 1. a < b < c. In this case a < c and hence bc < ab, from 1. Since b > a and c > b, we get bc > ab, which is a contradiction. ase. a < c < b. We may write 1 in the form c a/b a b/c b c/a. Now a < c gives a < c so that b c < a b. This gives b < ac. ut b > a and b > c, so that b > ac, which again leads to a contradiction Thus ase 1 and ase cannot occur. We conclude that a b c. Solution. We write 1 in the form, and start from there. The case of two equal sides is dealt as in Solution 1. We assume no two sides are equal. Using ratio properties in, we obtain a b ab c /ca b c bc a /ab. This may be written as ca bbc a bb cab c. urther simplification gives ab 3 + bc 3 + ca 3 abca + b + c. This may be further written in the form If a < b < c, we write 3 in the form ab b c + bc c a + ca a b ab b c + bc c b + b a + ca a b bc bc ab + cb abc a. Since c > b, c > ab, b > a and bc > a, this is impossible. If a < c < b, we write 3, as in previous case, in the form which again is impossible. MeritPath.com 0 ab cb ca + cc abc a, One can also use inequalities: we can show that ab 3 + bc 3 + ca 3 abca + b + c, and equality holds if and only if a b c. ere are some ways of deriving it:

2 i We can write the inequality in the form b c + c a + a b a + b + c. dding a + b + c both sides, this takes the form ut M-M inequality gives b c + c + c a + a + a + b a + b + c. b b c + c b, c a + a a, a b + b a. ence the inequality follows and equality holds if and only if a b c. ii gain we write the inequality in the form b c + c a + a b a + b + c. We use b/c with weight b, c/a with weight c and a/b with weight a, and apply weighted M-M inequality: b b c + c c a + a a b a + b + c b c b + c a c + a b a which reduces to a + b + c. gain equality holds if and only if a b c. Solution 3. ere is a pure geometric solution given by a student. onsider the triangle, and with, and as bases. The sides, and make equal angles with the bases of respective triangles. If, then it is easy to see that. Now using the triangle, we see that gives. ombining, you get and hence 60.. all a natural number n faithful, if there eist natural numbers a < b < c such that a divides b, b divides c and n a + b + c. i Show that all but a finite number of natural numbers are faithful. ii ind the sum of all natural numbers which are not faithful. Solution 1: Suppose n N is faithful. Let k N and consider kn. Since n a + b + c, with a > b > c, c b and b a, we see that kn ka + kb + kc which shows that kn is faithful. Let p > 5 be a prime. Then p is odd and p p shows that p is faithful. If n N contains a prime factor p > 5, then the above observation shows that n is faithful. This shows that a number which is not faithful must be of the form α 3 β 5 γ. We also observe that , and , so that 4, 3 and 5 are faithful. ence n N is also faithful if it contains a factor of the form α where α 4; a factor of the form 3 β where β ; or a factor of the form 5 γ where γ. Thus the numbers which are not faithful are of the form α 3 β 5 γ, where α 3, β 1 and γ 1. We may enumerate all such numbers: 1,, 3, 4, 5, 6, 8, 10, 1, 15, 0, 4, 30, 40, 60, 10. mong these , , , , , , and It is easy to check that the other numbers cannot be written in the required form. ence the only numbers which are not faithful are Their sum is 65. 1,, 3, 4, 5, 6, 8, 1, 4. Solution : If n a + b + c with a < b < c is faithful, we see that a 1, b and c 4. ence n 7. Thus 1,, 3, 4, 5, 6 are not faithful. s observed earlier, kn is faithful whenever,

3 n is. We also notice that for odd n 7, we can write n 1 ++n 3 so that all odd n 7 are faithful. onsider n, 4n, 8n, where n 7 is odd. y observation, they are all faithful. Let us list a few of them: n : 14, 18,, 6, 30, 34,38, 4, 46, 50, 54, 58, 6,... 4n : 8, 36, 44, 5, 60, 68,... 8n : 56, 7,..., We observe that and hence it is faithful. Thus all multiples of 16 are also faithful. Thus we see that 16, 3, 48, 64,... are faithful. ny even number which is not a multiple of 16 must be either an odd multiple of, or that of 4, or that of 8. ence, the only numbers not covered by this process are 8, 10, 1, 0, 4, 40. Of these, we see that , 0 10, , so that 10,0,40 are faithful. Thus the only numbers which are not faithful are Their sum is 65. 1,, 3, 4, 5, 6, 8, 1, onsider two polynomials P a n n + a n 1 n a 1 + a 0 and Q b n n + b n 1 n b 1 + b 0 with integer coefficients such that a n b n is a prime, a n 1 b n 1 and a n b 0 a 0 b n 0. Suppose there eists a rational number r such that Pr Qr 0. Prove that r is an integer. Solution: Let r u/v where gcdu, v 1. Then we get Subtraction gives a n u n + a n 1 u n 1 v + + a 1 uv n 1 + a 0 v n 0, b n u n + b n 1 u n 1 v + + b 1 uv n 1 + b 0 v n 0. a n b n u n + a n b n u n v + + a 1 b 1 uv n 1 + a 0 b 0 v n 0, since a n 1 b n 1. This shows that v divides a n b n u n and hence it divides a n b n. Since a n b n is a prime, either v 1 or v a n b n. Suppose the latter holds. The relation takes the form u n + a n b n u n v + + a 1 b 1 uv n + a 0 b 0 v n 1 0. ere we have divided through-out by v. If n > 1, this forces v u, which is impossible since gcdv, u 1 v > 1 since it is equal to the prime a n b n. If n 1, then we get two equations: a 1 u + a 0 v 0, b 1 u + b 0 v 0. This forces a 1 b 0 a 0 b 1 0 contradicting a n b 0 a 0 b n 0. Note: The condition a n b 0 a 0 b n 0 is etraneous. The condition a n 1 b n 1 forces that for n 1, we have a 0 b 0. Thus we obtain, after subtraction a 1 b 1 u 0. This implies that u 0 and hence r 0 is an integer. 4. Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium. Solution 1: Suppose four distinct points P, Q, R, Sin that order on the circle among these five are such that PQ RS. Then PQRS is an isosceles trapezium, with PS QR. We use this in our argument. If four of the five points chosen are adjacent, then we are through as observed earlier. In this case four points,,, are such that  Ĉ. See ig 1. 3

4 ig 1. ig. ig 3. Suppose only three of the vertices are adjacent, say,, see ig. Then the remaining two must be among,,,. If these two are adjacent vertices, we can pair them with, or, to get equal arcs. If they are not adjacent, then they must be either, or, or,. In the first two cases, we can pair them with, to get equal arcs. In the last case, we observe that Ĥ Ĉ and is an isosceles trapezium. Suppose only two among the five are adjacent, say,. Then the remaining three are among,,,,. See ig 3. If any two of these are adjacent, we can combine them with, to get equal arcs. If no two among these three vertices are adjacent, then they must be,,. In this case Ĥ and is an isosceles trapezium. inally, if we choose 5 among the 9 vertices of a regular nine-sided polygon, then some two must be adjacent. Thus any choice of 5 among 9 must fall in to one of the above three possibilities. Solution : ere is another solution used by many students. Suppose you join the vertices of the nine-sided regular polygon. You get 9 36 line segments. ll these fall in to 9 sets of parallel lines. Now using any 5 points, you get 5 10 line segments. y pigeon-hole principle, two of these must be parallel. ut, these parallel lines determine a trapezium. 5. Let be a quadrilateral inscribed in a circle Γ. Let,,, be the midpoints of the arcs,,, of the circle Γ. Suppose. Prove that,,, are concurrent. Solution: Let R be the radius of the circle Γ. Observe that 1. ence R sin. Similarly, R sin. ut 180. Thus R cos. We hence get Therefore 4R sin cos R sin R. Similarly, we obtain R. R + +, by Ptolemy s theorem. y the given hypothesis, this gives R +. Thus R + R, using M-M inequality. This implies that 4R. ut and are the chords of Γ, so that R and R. We obtain 4R. It follows that 4R, implying that R. Thus and are two diameters of Γ. Using, we conclude that and are also two diameters of Γ. ence,, and all pass through the centre of Γ. 4

5 6. ind all functions f : R R such that f + yf y f + fy 4 fy, 1 for all, y R, where R denotes the set of all real numbers. Solution 1.: Put y 0; we get f0 4f0 and hence f0 0. Put y: we get 4f 4 f 0 for all. ence for each, either f 0 or f. Suppose f 0. Then we can find 0 0 such that f 0 0. Then f ssume that there eists some y 0 0 such that fy 0 0. Then f 0 + y 0 f 0 y 0 f 0. Now f 0 + y 0 f 0 y 0 0 or f 0 + y 0 f 0 y y 0 0 y 0. If f 0 + y 0 f 0 y 0 0, then f 0 0, a contradiction. ence it must be the latter so that 0 y This reduces to y 0 y Since y 0 0, we get y 0 ± 0. Suppose y 0 0. Put 0 and y 0 in 1; we get f f 10 f 0 + f f0. ut f 0 fy 0 0. Thus we get f f 10 f f Now if LS is equal to 0, we get 0 0, a contradiction. Otherwise LS is equal to which reduces to 4 0. We obtain and this forces again 0 0. ence there is no y 0 such that fy 0. We conclude that f for all. Thus there are two solutions: f 0 for all or f, for all. It is easy to verify that both these satisfy the functional equation. Solution : s earlier, we get f0 0. Putting 0, we will also get fy fy f y 0. s earlier, we may conclude that either fy 0 or fy f y for each y R. Replacing y by y, we may also conclude that f y f y fy 0. If fy 0 and f y 0 for some y, then we must have f y fy 0, a contradiction. ence either fy f y 0 or fy f y for each y. This forces f is an even function. Taking y 1 in 1, we get Replacing y by and by 1, you also get f + 1f 1 f + f1 4 f1. f1 + f1 f1 + f 4f. omparing these two using the even nature of f, we get f c, where c f1. Putting y 1 in 1, you get 4c 4c 0. ence c 0 or 1. We get f 0 for all or f for all. 5