MathBattle : Problems offered for all Leagues
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Muhammad Yusuf
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1. Three cops are trying to catch a gangster on an infinite square grid. All four move along the horizontal and vertical lines of the grid at the same maximal speed. The gangster is caught if he is ever on the same line as a cop. At the start time the location of the gangster is unknown. Is there a strategy that ensures that the cops catch the gangster eventually?
Solution (i) Let the length of the segments of the grid be 1 and the speed be 1. Let us assume that the three cops, (A, B, C), are gathered at point O, the origin of the coordinate system, and the gangster (G) is at a distance of no more than r (r is integer) from O.
Cops Strategy: A runs to the West and B runs to the East while C patrols along OY between (0,−2r) and (0, 2r) resting for 1 minute at each intersection.
Let us show that after r passes C, the x- coordinate of G is between those of A and B. If G even once went to outside of the y-interval [−2r, 2r], then he made at least r vertical moves and thus the x-coordinates of both A and B exceeded (or reached) the x-coordinate of G. On the other hand, if the y-coordinate of G never exceeded 2r, then during each pass of C he must at least once go vertically because if he stayed on the same horizontal line, he would be caught by C.
After we have ensured that the x-coordinate of G is been between those of A and B, let us change the strategy. Now A runs to the North, B run to the South, while C patrols in an East-West direction between the verticals of A and B, staying at each knot for one minute.
Note that G cannot cross these verticals. Further, after some time the y-coordinate of G is localized between those of A and B. Therefore, G is somewhere inside of the rectangle with A and B be as opposite vertices.
Further strategy of catching G is obvious (if he had not been caught so far). We denote this strategy by Σ(r).
(ii) Assume that G is at a distance of R from the origin (but the cops do not know R). They employ the strategy Σ(r).
If G is not caught, then our initial assumption is wrong. Then cops return to point O. Note that the time t(r) that the cops spent trying to catch G according to Σ(r) (and return to O) depends on r only. During this time G will be at distance R + t(r) from O.
Now, let us apply the strategy Σ(r1) with r1 = 2r + t(r); then strategy Σ(r2) with r2 =
3r + t(r) + t(r1) and so on. G will be be caught eventually, since if he is not caught after Σ(rk) it would mean that at the initial time he was at a distance of R ≥ kr from O.
2. A positive non-straight angle with vertex O and a point P inside of it are given. Construct a segment through P creating a triangle of minimal area.
Solution.
Point P must be the midpoint of the segment in question. To see this, let us consider any other segment through P and the triangle created by it. Its area is equal to the area of the triangle containing the segment in question plus and minus the areas of the two triangles created by both segments. One can prove that the triangle to be added has a larger area than the triangle to be subtracted.
Construction: Through point O’ symmetrical to O with respect to P draw two lines parallel to the sides of the angle. Then P is the centre of the created parallelogram.
3. Each inhabitant of the Three Parties Island is either a Truth- Teller or a Liar and a member of exactly one of the three political parties. One day each person was asked whether he belonged to
(a) the First Party
(b) the Second Party
(c) the Third Party
60%, 50%, and 40% of population answered affirmatively on questions (a), (b), and (c), respectively. Who (Truth-Tellers or Liars) constitutes a majority of the Second Party?
Solution
Each Truthteller answered affirmatively to exactly one question while each Liar did so on two questions. Note that the total number of positive answers is 150=40+50+60% (of the population). Therefore, the Liars constitute exactly 50% of the population. The second question was answered positively by either Truthtellers from The Second Party or by Liars who are not from The Second Party. Let Truthtellers from The Second Party constitute x% of the total population; then Liars who are not from The Second Party constitute (50 − x)% of the population. Also, the number of Liars from The Second Party constitute 50%−(50−x)% = x% of the population. Therefore, The Second Party is equally split betweenLiars and Truthtellers.
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