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olympiads.hbcse.tifr.res.in Indian National Mathematical Olympiad INMO Previous Question Papers : Homi Bhabha Centre For Science Education

Organisation : Homi Bhabha Centre For Science Education
Exam : Indian National Mathematical Olympiad (INMO)
Document Type : Previous Question Paper
Location : Indaia
Website : olympiads.hbcse.tifr.res.in

Indian National Mathematical Olympiad Question Paper

Problems and Solutions:
Problem 1:
The in-circle of triangle ABC touches the sides BC, CA and AB in K, L and M respectively. The line through A and parallel to LK meets MK in P and the line through A and parallel to MK meets LK in Q. Show that the line PQ bisects the sides AB and AC of triangle ABC.

Related : HBCSE Pre-Regional Mathematical Olympiad Previous Question Paper : www.pdfquestion.in/9136.html

Solution. :
Let AP,AQ produced meet BC in D,E respectively

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Since MK is parallel to AE, we have \AEK = \MKB. Since BK = BM, both being tangents to the circle from B, \MKB = \BMK. This with the fact that MK is parallel to AE gives us \AEK = \MAE. This shows that MAEK is an isosceles trapezoid. We conclude that MA = KE. Similarly, we can prove that AL = DK. But AM = AL. We get that DK = KE. Since KP is parallel to AE, we get DP = PA and similarly EQ = QA. This implies that PQ is parallel to DE and hence bisects AB,AC when produced.

[The same argument holds even if one or both of P and Q lie outside triangle ABC.]

Problem 2:
Let ABC be a triangle in which no angle is 90?. For any point P in the plane of the triangle, let A1,B1,C1 denote the reflections of P in the sides BC,CA,AB respectively. Prove the following statements: (a) If P is the incentre or an excentre of ABC, then P is the circumcentre of A1B1C1; (b) If P is the circumcentre of ABC, then P is the orthocentre of A1B1C1; (c) If P is the orthocentre of ABC, then P is either the incentre or an excentre of A1B1C1.

Solution:
(a) If P = I is the incentre of triangle ABC, and r its inradius, then it is clear that A1I = B1I = C1I = 2r. It follows that I is the circumcentre of A1B1C1. On the otherhand if P = I1 is the excentre of ABC opposite A and r1 the corresponding ex- radius, then again we see that A1I1 = B1I1 = C1I1 = 2r1. Thus I1 is the circumcentre of A1B1C1.

(b) Let P = O be the circumcentre of ABC. By definition, it follows that OA1 bisects and is bisected by BC and so on. Let D,E, F be the mid-points of BC,CA,AB respectively. Then FE is parallel to BC. But E, F are also mid-points of OB1,OC1 and hence FE is parallel to B1C1 as well. We conclude that BC is parallel to B1C1. Since OA1 is perpendicular to BC, it follows that OA1 is perpendicular to B1C1. Similarly OB1 is perpendicular to C1A1 and OC1 is perpendicular to A1B1. These imply that O is the orthocentre of A1B1C1. (This applies whether O is inside or outside ABC.)

(c) let P = H, the orthocentre of ABC. We consider two possibilities; H falls inside ABC and H falls outside ABC. Suppose H is inside ABC; this happens if ABC is an acute triangle. It is known that A1,B1,C1 lie on the circumcircle of ABC. Thus \C1A1A = \C1CA = 90? – A. Similarly \B1A1A = \B1BA = 90? – A. These show that \C1A1A = \B1A1A. Thus A1A is an internal bisector of \C1A1B1. Similarly we can show that B1 bisects \A1B1C1 and C1C bisects \B1C1A1. Since A1A,B1B,C1C concur at H, we conclude that H is the incentre of A1B1C1.

(or) If D,E, F are the feet of perpendiculars of A,B,C to the sides BC,CA,AB respectively, then we see that EF, FD,DE are respectively parallel to B1C1, C1A1, A1B1. This implies that \C1A1H = \FDH = \ABE = 90? – A, as BDHF is a cyclic quadrilateral. Similarly, we can show that \B1A1H = 90? – A. It follows that A1H is the internal bisector of \C1A1B1. We can proceed as in the earlier case. If H is outside ABC, the same proofs go through again, except that two of A1H, B1H, C1H are external angle bisectors and one of these is an internal angle bisector. Thus H becomes an excentre of triangle A1B1C1.

Problem 3:
For any natural number n, (n = 3), let f(n) denote the number of non-congruent integer-sided triangles with perimeter n (e.g., f(3) = 1, f(4) = 0, f(7) = 2). Show that
(a) f(1999) > f(1996);
(b) f(2000) = f(1997).

Solution:
(a) Let a, b, c be the sides of a triangle with a+b+c = 1996, and each being a positive integer. Then a+1, b+1, c+1 are also sides of a triangle with perimeter 1999 because a < b + c => a + 1 < (b + 1) + (c + 1),

and so on. Moreover (999, 999, 1) form the sides of a triangle with perimeter 1999, which is not obtainable in the form (a+1, b+1, c+1) where a, b, c are the integers and the sides of a triangle with a + b + c = 1996. We conclude that f(1999) > f(1996).

(b) As in the case (a) we conclude that f(2000) = f(1997). On the other hand, if x, y, z are the integer sides of a triangle with x+y+z = 2000, and say x = y = z = 1, then we cannot have z = 1; for otherwise we would get x + y = 1999 forcing x, y to have opposite parity so that x – y = 1 = z violating triangle inequality for x, y, z. Hence x = y = z > 1.

This implies that x – 1 = y – 1 = z – 1 > 0. We already have x < y + z. If x = y + z – 1, then we see that y + z – 1 = x < y + z, showing that y + z – 1 = x. Hence we obtain 2000 = x + y + z = 2x + 1 which is impossible. We conclude that x < y + z – 1.

This shows that x – 1 < (y – 1) + (z – 1) and hence x – 1, y – 1, z – 1 are the sides of a triangle with perimeter 1997.

This gives f(2000) = f(1997). Thus we obtain the desired result

Author : Jency

olympiads.hbcse.tifr.res.in Indian National Mathematical Olympiad INMO Previous Question Papers : Homi Bhabha Centre For Science Education

Indian National Mathematical Olympiad Question Paper

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