MathDB

2016 BAMO

Part of BAMO Contests

Subcontests

(5)
4
2

Medial hexagon by perpendiculars

In an acute triangle ABCABC let K,L,K,L, and MM be the midpoints of sides AB,BC,AB,BC, and CA,CA, respectively. From each of K,L,K,L, and MM drop two perpendiculars to the other two sides of the triangle; e.g., drop perpendiculars from KK to sides BCBC and CA,CA, etc. The resulting 66 perpendiculars intersect at points Q,S,Q,S, and TT as in the figure to form a hexagon KQLSMTKQLSMT inside triangle ABC.ABC. Prove that the area of this hexagon KQLSMTKQLSMT is half of the area of the original triangle ABC.ABC.
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra; diagram by adihaya*/ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 11.888712276357234, xmax = 17.841346447833423, ymin = 10.61620970860601, ymax = 15.470685507068502; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (12.488234161849352,12.833838721895551), B = (16.50823416184936,15.093838721895553), C = (16.28823416184936,11.353838721895551), K = (14.498234161849355,13.963838721895552), L = (16.39823416184936,13.223838721895552), M = (14.388234161849356,12.093838721895551), D = (13.615830174638527,13.467760858438725), F = (15.75135711740064,11.562938202365055), G = (15.625830174638523,14.597760858438724), H = (16.435061748056253,13.849907687412797), T = (14.02296781802369,12.74356027153236), Q = (16.032967818023693,13.873560271532357), O = (16.325061748056253,11.979907687412794);
draw(A--B--C--cycle, zzttqq); draw((13.426050287639166,13.361068683160477)--(13.532742462917415,13.171288796161116)--(13.722522349916774,13.277980971439364)--D--cycle, qqwuqq); draw((14.054227993863618,12.223925334689998)--(14.133240861538676,12.426796211152979)--(13.930369985075695,12.505809078828037)--(13.851357117400637,12.302938202365056)--cycle, qqwuqq); draw((16.337846386707046,12.19724654447628)--(16.12050752964356,12.210031183127075)--(16.107722890992765,11.992692326063588)--O--cycle, qqwuqq); draw((15.830369985075697,11.765809078828037)--(15.627499108612716,11.844821946503092)--(15.54848624093766,11.641951070040111)--F--cycle, qqwuqq); draw((15.436050287639164,14.491068683160476)--(15.542742462917412,14.301288796161115)--(15.73252234991677,14.407980971439365)--G--cycle, qqwuqq); draw((16.217722890992764,13.86269232606359)--(16.20493825234197,13.645353469000101)--(16.42227710940546,13.63256883034931)--H--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(A--B, zzttqq); draw(B--C, zzttqq); draw(C--A, zzttqq); draw(M--D); draw(K--(13.851357117400637,12.302938202365056)); draw(F--L); draw(L--G); draw(K--H); draw(M--O); /* dots and labels */ dot(A,dotstyle); label("AA", (12.52502834296331,12.93568440300881), NE * labelscalefactor); dot(B,dotstyle); label("BB", (16.548187989892043,15.193580123223922), NE * labelscalefactor); dot(C,dotstyle); label("CC", (16.332661580235147,11.457789022504372), NE * labelscalefactor); dot(K,linewidth(3.pt) + dotstyle); label("KK", (14.536608166427676,14.02357961365791), NE * labelscalefactor); dot(L,linewidth(3.pt) + dotstyle); label("LL", (16.43529320388129,13.28463192340569), NE * labelscalefactor); dot(M,linewidth(3.pt) + dotstyle); label("MM", (14.433976542781535,12.155684063298134), NE * labelscalefactor); dot(D,linewidth(3.pt) + dotstyle); dot((13.851357117400637,12.302938202365056),linewidth(3.pt) + dotstyle); dot(F,linewidth(3.pt) + dotstyle); dot(G,linewidth(3.pt) + dotstyle); dot(H,linewidth(3.pt) + dotstyle); dot((15.922967818023695,12.003560271532354),linewidth(3.pt) + dotstyle); label("SS", (15.96318773510904,12.063315602016607), NE * labelscalefactor); dot(T,linewidth(3.pt) + dotstyle); label("TT", (14.064502697655428,12.802263292268826), NE * labelscalefactor); dot(Q,linewidth(3.pt) + dotstyle); label("QQ", (16.076082521119794,13.931211152376383), NE * labelscalefactor); dot(O,linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

Another year-based olympiad problem

Find a positive integer NN and a1,a2,⋯ ,aNa_1, a_2, \cdots, a_N where ak=1a_k = 1 or ak=−1a_k = -1, for each k=1,2,⋯ ,N,k=1,2,\cdots,N, such that a1⋅13+a2⋅23+a3⋅33⋯+aN⋅N3=20162016a_1 \cdot 1^3 + a_2 \cdot 2^3 + a_3 \cdot 3^3 \cdots + a_N \cdot N^3 = 20162016 or show that this is impossible.
5
2

Chessboard parallelograms

For n>1n>1 consider an n×nn\times n chessboard and place identical pieces at the centers of different squares.
[*] Show that no matter how 2n2n identical pieces are placed on the board, that one can always find 44 pieces among them that are the vertices of a parallelogram. [*] Show that there is a way to place (2n−1)(2n-1) identical chess pieces so that no 44 of them are the vertices of a parallelogram.

The "MOAB" Problem

The corners of a fixed convex (but not necessarily regular) nn-gon are labeled with distinct letters. If an observer stands at a point in the plane of the polygon, but outside the polygon, they see the letters in some order from left to right, and they spell a "word" (that is, a string of letters; it doesn't need to be a word in any language). For example, in the diagram below (where n=4n=4), an observer at point XX would read "BAMOBAMO," while an observer at point YY would read "MOABMOAB."
Diagram to be added soon
Determine, as a formula in terms of nn, the maximum number of distinct nn-letter words which may be read in this manner from a single nn-gon. Do not count words in which some letter is missing because it is directly behind another letter from the viewer's position.
3
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2
1

Weird calculator buttons

A weird calculator has a numerical display and only two buttons, D♯\boxed{D\sharp} and D♭\boxed{D\flat}. The first button doubles the displayed number and then adds 11. The second button doubles the displayed number and then subtracts 11. For example, if the display is showing 55, then pressing the D♯\boxed{D\sharp} produces 1111. If the display shows 55 and we press D♭\boxed{D\flat}, we get 99. If the display shows 55 and we press the sequence D♯\boxed{D\sharp}, D♭\boxed{D\flat}, D♯\boxed{D\sharp}, D♯\boxed{D\sharp}, we get a display of 8787.
[*] Suppose the initial displayed number is 11. Give a sequence of exactly eight button presses that will result in a display of 313313. [*] Suppose the initial displayed number is 11, and we then perform exactly eight button presses. Describe all the numbers that can possibly result? Prove your answer by explaining how all these numbers can be produced and that no other numbers can be produced.
1
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