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Problems(6)

Antitours in Mahishmati

Source: India TST 2023 Day 1 P1

7/9/2023
In the fictional country of Mahishmati, there are 5050 cities, including a capital city. Some pairs of cities are connected by two-way flights. Given a city AA, an ordered list of cities C1,,C50C_1,\ldots, C_{50} is called an antitour from AA if
[*] every city (including AA) appears in the list exactly once, and [*] for each k{1,2,,50}k\in \{1,2,\ldots, 50\}, it is impossible to go from AA to CkC_k by a sequence of exactly kk (not necessarily distinct) flights.
Baahubali notices that there is an antitour from AA for any city AA. Further, he can take a sequence of flights, starting from the capital and passing through each city exactly once. Find the least possible total number of antitours from the capital city.
Proposed by Sutanay Bhattacharya
combinatoricsgraph theory
Wait wasn't it the reciprocal in the paper?

Source: India TST 2023 Day 2 P1

7/9/2023
Let Z0\mathbb{Z}_{\ge 0} be the set of non-negative integers and R+\mathbb{R}^+ be the set of positive real numbers. Let f:Z02R+f: \mathbb{Z}_{\ge 0}^2 \rightarrow \mathbb{R}^+ be a function such that f(0,k)=2kf(0, k) = 2^k and f(k,0)=1f(k, 0) = 1 for all integers k0k \ge 0, and f(m,n)=2f(m1,n)f(m,n1)f(m1,n)+f(m,n1)f(m, n) = \frac{2f(m-1, n) \cdot f(m, n-1)}{f(m-1, n)+f(m, n-1)} for all integers m,n1m, n \ge 1. Prove that f(99,99)<1.99f(99, 99)<1.99.
Proposed by Navilarekallu Tejaswi
algebrainequalities
Decimal functions in binary

Source: India TST 2023 Day 3 P1

7/9/2023
Let N\mathbb{N} be the set of all positive integers. Find all functions f:NNf : \mathbb{N} \rightarrow \mathbb{N} such that f(x)+yf(x) + y and f(y)+xf(y) + x have the same number of 11's in their binary representations, for any x,yNx,y \in \mathbb{N}.
number theory
Slightly weird points which are not so weird

Source: India TST 2023 Day 4 P1

7/9/2023
Suppose an acute scalene triangle ABCABC has incentre II and incircle touching BCBC at DD. Let ZZ be the antipode of AA in the circumcircle of ABCABC. Point LL is chosen on the internal angle bisector of BZC\angle BZC such that AL=LIAL = LI. Let MM be the midpoint of arc BZCBZC, and let VV be the midpoint of IDID. Prove that IML=DVM\angle IML = \angle DVM
geometryTST
Odd points give concurrency of perpendicular bisectors

Source: India TST 2023 Practice Test 1 P1

7/9/2023
Let ABCABC be a triangle, and let DD be the foot of the AA-altitude. Points P,QP, Q are chosen on BCBC such that DP=DQ=DADP = DQ = DA. Suppose APAP and AQAQ intersect the circumcircle of ABCABC again at XX and YY. Prove that the perpendicular bisectors of the lines PXPX, QYQY, and BCBC are concurrent.
Proposed by Pranjal Srivastava
geometryperpendicular bisector
Keep on dividing till you can&amp;#039;t no more

Source: India TST 2023 Practice Test 2 P1

7/9/2023
The numbers 1,2,3,4,,391,2,3,4,\ldots , 39 are written on a blackboard. In one step we are allowed to choose two numbers aa and bb on the blackboard such that aa divides bb, and replace aa and bb by the single number ba\tfrac{b}{a}. This process is continued till no number on the board divides any other number. Let SS be the set of numbers which is left on the board at the end. What is the smallest possible value of S|S|?
Proposed by B.J. Venkatachala
number theory