TOMATO Objective 21 | ISI Entrance Exam

TIFR 2013 Maths Paper Answer Key

This is TIFR 2013 Maths Paper Answer Key.

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Part A

1. D
2. B
3. C
4. C
5. D
6. C
7. D
8. A
9. C
10. B
11. D
12. C
13. B
14. B
15. D
16. D
17. B
18. C
19. B
20. D

Let C denote the cube $latex [-1, 1]^3 \subset \mathbb{R} $ . How many rotations are there in $latex \mathbb{R}^3 $ which take C to itself?

A. 6; B. 12 C. 18. D. 24

Discussion:

Theorem: A finite subgroup of $latex SO_3 $ is one of the following groups:

$latex C_k $ : the cyclic group of rotations by multiples of $latex \frac {2 \pi } {k} $ about a line, with k arbitrary
$latex D_k $: the Dihedral group of symmetries of a regular k-gon, with k arbitrary
$latex T $ the tetrahedral group of 12 rotational symmetries of a tetrahedron;
$latex O $: the octahedral group of 24 rotational symmetries of a cube or an octahedron
$latex I $: the icosahedral group of 60 rotational symmetries of a dodecahedron or an icosahedron

Part B

21. C
22. A
23.
24. D
25. A
26. B
27. D
28. A
29. A
30. B

(courtesy: Tattwamasi Amrutam)

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Orthocenter on perpendicular bisector | INMO 2013

This is a problem from Indian National Mathematics Olympiad, INMO, 2013 based on Orthocenter on perpendicular bisector. Try out this problem.

Problem: Orthocenter on perpendicular bisector

In an acute angled triangle ABC with AB < AC the circle $latex \Gamma $ touches AB at B and passes through C intersecting AC again at D. Prove that the orthocenter of triangle ABD lies on $latex \Gamma $ if and only if it lies on the perpendicular bisector of BC.

Discussion

Suppose H is the orthocenter of triangle ABD and it lies on the circle $latex \Gamma $. We show that HB = HC (if we can show this then the perpendicular from H on BC will bisect BC).

DF and BE are altitudes of triangle ABD.

First we note that $latex \angle FBH = \angle HCB $ for FB is tangent to the circle and angle made by a chord with a tangent is equivalent to an angle in the alternate segment. In this case the chord is BH.

Again FBDE is cyclic (since $latex \angle BFD = \angle BED = 90^0 $ ). Hence $latex \angle FBH = \angle EDH $ (angle in the same segment FE). .... (ii)

But HDCB is also cyclic (all vertices are on the circle). Hence $latex \angle EDH = \angle HBC $ (exterior angle is equal to the interior opposite angle in a cyclic quadrilateral). .... (iii)

Combining (ii) and (iii) we have $latex \angle HCB = HBC$ implying HB = HC.

Conversely if we have HB = HC, this implies $latex \angle HBC = \angle HCB $ . Also $latex \angle FBD = \angle DCB $ (angles in the alternate segment subtended by chord BD)

Now consider triangles BEC and BFD. We have $latex \angle BEC = \angle BFD = 90^0 $ and $latex \angle ECB = \angle FBD $. Therefore remaining angles BDF and EBC are also equal. But $latex \angle DBC = \angle HCB $ implying $latex \angle BDF = \angle HCB $. Thus HDCB is cyclic. Hence proved.

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Number of 4-tuples (a,b,c,d) of natural numbers

Find the number of 4-tuples (a,b,c,d) of natural numbers with $latex a \le b \le c $ and $latex a! + b! + c! = 3^d $

Discussion: Number of 4-tuples

The basic idea is: factorial function is faster than the exponential function in the long run. Note that all three of a, b, c cannot be larger than 3; then the left side will be divisible by 4 but the right side is not. Hence the possible values of a are 1, 2, and 3.

If a=1, then b and c both cannot be greater than or equal to 3 (then the left-hand side is not divisible by 3). So we have the following cases:

b=1 or b=2

If b =1 then we have $latex 1! + 1! + c! = 3^d $ Surely c! is 1 mod 3 (otherwise the left-hand side is not divisible by 3). Then the possible values of c is 1. And indeed $latex 1! + 1! + 1! = 3^1 $ fits into our equation. Hence (1, 1, 1, 1) is a solution and for b=1 there is no other.

If b=2 then we have $latex 1! + 2! + c! = 3^d $. Surely c! is 0 mod 3. Hence possible values of c are 3, 4, 5, ...

c=3 and c=4 furnish specific solutions as $latex 1! + 2! + 3! = 3^2 $ and $latex 1! + 2! + 4! = 3^3 $ . Hence (1, 2, 3, 2) and (1, 2, 4, 3) are two solutions.

Can c be greater than 4? Surely c cannot be 5 since $latex 1! + 2! + 5! = 123 $ is not a power of 3. From c=6 onward we argue $latex 1! +2! + c! = 3^d $ implies $latex 3(1 + \frac {c!}{3} ) = 3^d $ implies $latex 1 + \frac {c!}{3} = 3^{d-1} $ . Since c is greater than 5, c! will contain atleast two 3's in it's prime factorization (and d will be greater than 4 as 5! = 120 > 81 ) . Hence $latex \frac {c!}{3} $ is divisible by 3.
Thus in the equation $latex 1 + \frac {c!}{3} = 3^{d-1} $ left hand side is 1 mod 3 (that is produces 1 as remainder when divided by 3) and right hand side is 0 mod 3. Hence no solution.

Hence there are exactly 3 solutions.

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Pre-Regional Mathematics Olympiad - 2012 - Problem 17 - Video

Number of 8 digit numbers sum of whose digits is 4

Find the number of 8 digit numbers sum of whose digits is 4.

Discussion:

Suppose the number is $latex a_1 a_2 a_3 ... a_8 $.The possible values of $latex a_1 $ are 1, 2, 3, 4. We consider these four cases.

If $latex a_1 = 4 $ then all other digits are 0 (since sum of digits is 4). Hence there is only 1 such number.

If $latex a_1 = 3 $ then exactly one of the other 7 digits is 1. Hence there are 7 such numbers (depending on where the digit '1' is).

If $latex a_1 = 2 $ then sum of the other seven digits is 2.

Hence we compute the number of non negative integer solutions of $latex a_2 + ... + a_8 = 2 $ .

This equals $latex \binom {6+2}{2} $ = 28

If $latex a_1 = 1 $ then sum of the other seven digits is 3.

Hence we compute the number of non negative integer solutions of $latex a_2 + ... + a_8 $ = 3

This equals $latex \binom {6+3}{3} $ = 84

Hence the answer is 120.

For more problems: Pre-Regional Mathematics Olympiad Problems

Regional Math Olympiad 2013 (RMO 2013)

In this post, there are questions from Regional Math Olympiad 2013. Try out the problems.
Find the number of 8 digit numbers sum of whose digits are 4.
Discussion
Find the number of 4-tuples (a,b,c,d) of natural numbers with $latex a \le b \le c $ and $latex a! + b! + c! = 3^d $
Discussion
In an acute-angled triangle ABC with AB < AC the circle $latex \Gamma $ touches AB at B and passes through C intersecting AC again at D. Prove that the orthocenter of triangle ABD lies on $latex \Gamma $ if and only if it lies on the perpendicular bisector of BC.
A polynomial is called a Fermat Polynomial if it can be written as the sum of squares of two polynomials with integer coefficients. Suppose f(x) is a Fermat Polynomial such that f(0) = 1000. Show that f(x) + 2x is not a Fermat Polynomial.
Let ABC be a triangle which is not right angled. Define a sequence of triangles $latex A_i B_i C_i $ with $ i \ge 0$ as follows. $latex A_0 B_0 C_0 = ABC $ and for $latex i \ge 0 A_{i+1} B_{i+1} C_{i+1} $ are the reflections of the orthocenter of triangle $latex A_i B_i C_i $ in the sides $latex B_i C_i , C_i A_i , A_i B_i $ respectively. Assume that $latex \angle A_n = \angle A_m $ for some distinct natural numbers m, n. Prove that $latex \angle A = 60^o $.
Let $latex n \ge 4 $ be a natural number. Let $latex A_1 , A_ 2 .... A_n $ be a regular polygon and X = { 1, 2, ..., n }. A subset $latex { i_1 , i_2 , ... i_k } $, $latex k \ge 1 $ , $ i_1 < i_2 < ... < i_k $ is called a good subset if the angles of the polygon angles $ A_{i_1} ... A_{i_k}$ when arranged in an increasing order is an arithmetic progression. If n is prime then show that a PROPER good subset of X contains exactly 4 elements.

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