IOQM 2022 Part 1 #IOQM2022 #maths

IOQM 2022 Part 1 #IOQM2022 #maths

IOQM 2022 (Q1 to Q10) A triangle ABC with AC= 20 is inscribed in a circle. A tangent t to circle is drawn through B. The distance of t from A is 25 and that from C is 16. If S denotes the area of the triangle ABC, find the largest integer not exceeding S/20. ๐ผ๐‘› ๐‘Ž ๐‘๐‘Ž๐‘Ÿ๐‘Ž๐‘™๐‘™๐‘’๐‘™๐‘œ๐‘”๐‘Ÿ๐‘Ž๐‘š ๐ด๐ต๐ถ๐ท, ๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘–๐‘›๐‘ก ๐‘ƒ ๐‘œ๐‘› ๐‘Ž ๐‘ ๐‘’๐‘”๐‘š๐‘’๐‘›๐‘ก ๐ด๐ต ๐‘–๐‘  ๐‘ก๐‘Ž๐‘˜๐‘’๐‘› ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐ด๐‘ƒ/๐ด๐ต ๐‘–๐‘  61/2022 ๐‘Ž๐‘›๐‘‘ ๐‘Ž ๐‘๐‘œ๐‘–๐‘›๐‘ก ๐‘„ ๐‘œ๐‘› ๐‘กโ„Ž๐‘’ ๐‘ ๐‘’๐‘”๐‘š๐‘’๐‘›๐‘ก ๐ด๐ท ๐‘–๐‘  ๐‘ก๐‘Ž๐‘˜๐‘’๐‘› ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐ด๐‘„/๐ด๐ท ๐‘–๐‘  61/2065. ๐ผ๐‘“ ๐‘ƒ๐‘„ ๐‘–๐‘›๐‘ก๐‘’๐‘Ÿ๐‘ ๐‘’๐‘๐‘ก๐‘  ๐ด๐ถ ๐‘Ž๐‘ก ๐‘‡, ๐‘“๐‘–๐‘›๐‘‘ ๐ด๐ถ/๐ด๐‘‡ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘›๐‘’๐‘Ž๐‘Ÿ๐‘’๐‘ ๐‘ก ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ. In a trapezoid ABCD, the internal bisector of angle A intersects the base BC (or its extension) at the point E. Inscribed in the triangle ABE is a circle touching the side AB at M and side BE at the point P. Find the angle DAE in degrees, if AB:MP = 2. Starting with a positive integer M written on the board, Alice plays the following game: in each move, if x is the number on the board, she replaces it with 3x + 2. Similarly, starting with a positive integer N written on the board, Bob plays the following game: in each move, if x is the number on the board, he replaces it with 2x + 27. Given that Alice and Bob reach the same number after playing 4 moves each, find the smallest value of M + N. ๐ฟ๐‘’๐‘ก ๐‘š ๐‘๐‘’ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘š๐‘Ž๐‘™๐‘™๐‘’๐‘ ๐‘ก ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ ๐‘ ๐‘ข๐‘โ„Ž ๐‘กโ„Ž๐‘Ž๐‘ก ๐‘š^2 + (๐‘š + 1)^2 + โ€ฆ + (๐‘š + 10)^2 ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘ ๐‘ž๐‘ข๐‘Ž๐‘Ÿ๐‘’ ๐‘œ๐‘“ ๐‘Ž ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ ๐‘›. ๐น๐‘–๐‘›๐‘‘ ๐‘š + ๐‘›. ๐‘ณ๐’†๐’• ๐’‚, ๐’ƒ ๐’ƒ๐’† ๐’‘๐’๐’”๐’Š๐’•๐’Š๐’—๐’† ๐’Š๐’๐’•๐’†๐’ˆ๐’†๐’“๐’” ๐’”๐’‚๐’•๐’Š๐’”๐’‡๐’š๐’Š๐’๐’ˆ ๐’‚^๐Ÿ‘โ€“ ๐’ƒ^๐Ÿ‘ โ€“ ๐’‚๐’ƒ=๐Ÿ๐Ÿ“. ๐‘ญ๐’Š๐’๐’… ๐’•๐’‰๐’† ๐’๐’‚๐’“๐’ˆ๐’†๐’”๐’• ๐’‘๐’๐’”๐’”๐’Š๐’ƒ๐’๐’† ๐’—๐’‚๐’๐’–๐’† ๐’๐’‡ ๐’‚^๐Ÿ + ๐’ƒ^๐Ÿ‘. ๐‘ญ๐’Š๐’๐’… ๐’•๐’‰๐’† ๐’๐’–๐’Ž๐’ƒ๐’†๐’“ ๐’๐’‡ ๐’๐’“๐’…๐’†๐’“๐’†๐’… ๐’‘๐’‚๐’Š๐’“๐’” (๐’‚, ๐’ƒ)๐’”๐’–๐’„๐’‰ ๐’•๐’‰๐’‚๐’• ๐’‚, ๐’ƒ โˆˆ (๐Ÿ๐ŸŽ, ๐Ÿ๐Ÿ, โ€ฆ, ๐Ÿ๐Ÿ—, ๐Ÿ‘๐ŸŽ) ๐’‚๐’๐’… ๐‘ฎ๐‘ช๐‘ซ (๐’‚, ๐’ƒ) + ๐‘ณ๐‘ช๐‘ด (๐’‚, ๐’ƒ) = ๐’‚ + ๐’ƒ. ๐‘†๐‘ข๐‘๐‘๐‘œ๐‘ ๐‘’ ๐‘กโ„Ž๐‘’ ๐‘๐‘Ÿ๐‘–๐‘š๐‘’ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ๐‘  ๐‘ ๐‘Ž๐‘›๐‘‘ ๐‘ž ๐‘ ๐‘Ž๐‘ก๐‘–๐‘ ๐‘“๐‘ฆ ๐‘ž^2 + 3๐‘=197๐‘^2+๐‘ž. ๐‘Š๐‘Ÿ๐‘–๐‘ก๐‘’ ๐‘ž/๐‘ ๐‘Ž๐‘  ๐‘™ ๐‘š/๐‘›, ๐‘คโ„Ž๐‘’๐‘Ÿ๐‘’ ๐‘™, ๐‘š, ๐‘› ๐‘Ž๐‘Ÿ๐‘’ ๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘–๐‘ฃ๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘”๐‘’๐‘Ÿ๐‘ , ๐‘š less than ๐‘› ๐‘Ž๐‘›๐‘‘ ๐บ๐ถ๐ท(๐‘š, ๐‘›) = 1. ๐น๐‘–๐‘›๐‘‘ ๐‘กโ„Ž๐‘’ ๐‘š๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š ๐‘ฃ๐‘Ž๐‘™๐‘ข๐‘’ ๐‘œ๐‘“ ๐‘™ + ๐‘š + ๐‘›. Consider the 10-digit number M = 9876543210. We obtain a new 10-digit number from M according to the following rule: we can choose one or more disjoint pairs of adjacent digits in M and interchange the digits in these chosen pairs, keeping the remaining digits in their own places. For example, from M = 9876543210, by interchanging the 2 underlined pairs, and keeping the others in their places, we get M1 9786453210 . Note that any number of (disjoint) pairs can be interchanged. Find the number of new numbers that can be so obtained from M.