Difference between revisions of "2013 UMO Problems"
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Consider the following diagram. | Consider the following diagram. | ||
+ | |||
+ | <asy> | ||
+ | path S=polygon(4); | ||
+ | D(S,black+linewidth(.75)); | ||
+ | D(shift(sqrt(2),0)*S,black+linewidth(.75)); | ||
+ | D(shift(2*sqrt(2),0)*S,black+linewidth(.75)); | ||
+ | D(shift(sqrt(2),sqrt(2))*S,black+linewidth(.75)); | ||
+ | D(shift(2*sqrt(2),sqrt(2))*S,black+linewidth(.75)); | ||
+ | D(shift(2*sqrt(2),2*sqrt(2))*S,black+linewidth(.75)); | ||
+ | </asy> | ||
(a) Show that you can retrace the diagram without lifting up your pencil using exactly nine (possibly | (a) Show that you can retrace the diagram without lifting up your pencil using exactly nine (possibly | ||
Line 16: | Line 26: | ||
a straight cut through the sheet (not necessarily parallel to the sides of the page), creating two new | a straight cut through the sheet (not necessarily parallel to the sides of the page), creating two new | ||
sheets. The sheet with smaller area is discarded (either one if the two are equal), and the player | sheets. The sheet with smaller area is discarded (either one if the two are equal), and the player | ||
− | gives the larger sheet to the other player. The first player to receive a sheet of area less than 1 square | + | gives the larger sheet to the other player. The first player to receive a sheet of area less than <math>1</math> square |
centimeter from the opposing player loses. If Alice goes first, describe (with proof) the sizes of paper | centimeter from the opposing player loses. If Alice goes first, describe (with proof) the sizes of paper | ||
for which she has a winning strategy. | for which she has a winning strategy. | ||
Line 32: | Line 42: | ||
Given line <math>\ell_1</math> and distinct points <math>I</math> and <math>X</math> on line <math>\ell_1</math>, draw lines <math>\ell_2</math> and <math>\ell_3</math> through point <math>I</math>, with angles <math>\alpha</math> and <math>\beta</math> as marked in the figure. Also, draw line segment <math>XY</math> at an angle of <math>\gamma</math> from line <math>\ell_1</math> such that it intersects line <math>\ell_2</math> at <math>Y</math>. Establish necessary and sufficient conditions on <math>\alpha</math> , <math>\beta</math> , and <math>\gamma</math> such that a triangle can be drawn with one of its sides as <math>XY</math> with lines <math>\ell_1</math>, <math>\ell_2</math>, and <math>\ell_3</math> as the angle bisectors of that triangle. | Given line <math>\ell_1</math> and distinct points <math>I</math> and <math>X</math> on line <math>\ell_1</math>, draw lines <math>\ell_2</math> and <math>\ell_3</math> through point <math>I</math>, with angles <math>\alpha</math> and <math>\beta</math> as marked in the figure. Also, draw line segment <math>XY</math> at an angle of <math>\gamma</math> from line <math>\ell_1</math> such that it intersects line <math>\ell_2</math> at <math>Y</math>. Establish necessary and sufficient conditions on <math>\alpha</math> , <math>\beta</math> , and <math>\gamma</math> such that a triangle can be drawn with one of its sides as <math>XY</math> with lines <math>\ell_1</math>, <math>\ell_2</math>, and <math>\ell_3</math> as the angle bisectors of that triangle. | ||
+ | |||
+ | <asy> | ||
+ | D((-2,-2)--(2,2),black+linewidth(.75)); | ||
+ | D((-1,2)--(1,-2),black+linewidth(.75)); | ||
+ | D((-2,1)--(2,-1),black+linewidth(.75)); | ||
+ | D((-.7,1.4)--(-1.3,-1.3),black+linewidth(.75)); | ||
+ | MP("\ell_1",(1.5,1.6),N); | ||
+ | MP("\ell_2",(-1.1,1.5),N); | ||
+ | MP("\ell_3",(-1.5,.3),N); | ||
+ | MP("\alpha",(.05,.2),N); | ||
+ | MP("\beta",(-.35,.13),N); | ||
+ | MP("\gamma",(-1.1,-1.1),N); | ||
+ | MP("Y",(-.7,1.4),NE); | ||
+ | MP("X",(-1.3,-1.3),S); | ||
+ | MP("I",(-.1,-.1),S); | ||
+ | dot((-1.3,-1.3));dot((-.7,1.4));dot((0,0)); | ||
+ | </asy> | ||
[[2013 UMO Problems/Problem 4|Solution]] | [[2013 UMO Problems/Problem 4|Solution]] | ||
Line 38: | Line 65: | ||
Cooper and Malone take turns replacing <math>a</math>, <math>b</math>, and <math>c</math> in the equation below with real numbers. | Cooper and Malone take turns replacing <math>a</math>, <math>b</math>, and <math>c</math> in the equation below with real numbers. | ||
− | < | + | <cmath>P(x) = x^3 + ax^2 + bx + c.</cmath> Once a coefficient has been replaced, no one can choose to |
change that coefficient on their turn. The game ends when all three coefficients have been chosen. | change that coefficient on their turn. The game ends when all three coefficients have been chosen. | ||
Malone wins if <math>P(x)</math> has a non-real root and Cooper wins otherwise. If Malone goes first, find the person who has a winning strategy and describe it with proof. | Malone wins if <math>P(x)</math> has a non-real root and Cooper wins otherwise. If Malone goes first, find the person who has a winning strategy and describe it with proof. | ||
Line 50: | Line 77: | ||
prove that your answer holds true for all positive integers <math>n \ge 3</math> (examples of such grids for <math>n = 3</math>, | prove that your answer holds true for all positive integers <math>n \ge 3</math> (examples of such grids for <math>n = 3</math>, | ||
<math>n = 4</math>, <math>n = 5</math>, and <math>n = 6</math> are shown below). | <math>n = 4</math>, <math>n = 5</math>, and <math>n = 6</math> are shown below). | ||
+ | |||
+ | <asy> | ||
+ | path S=rotate(30)*polygon(6); | ||
+ | filldraw(S,green+linewidth(.75)); | ||
+ | filldraw(shift(sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(2*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(3*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | |||
+ | filldraw(shift(6,0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(2*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(3*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(6,0)*shift(3*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(5*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(2*sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(6,0)*shift(3*sqrt(3)/2,9/2)*S,green+linewidth(.75)); | ||
+ | |||
+ | filldraw(shift(14,0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(2*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(3*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(4*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(14,0)*shift(3*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(14,0)*shift(5*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(7*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(14,0)*shift(2*sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(3*sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(3*sqrt(3)/2,9/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(5*sqrt(3)/2,9/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(14,0)*shift(2*sqrt(3),6)*S,green+linewidth(.75)); | ||
+ | |||
+ | |||
+ | filldraw(shift(24,0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(2*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(3*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(4*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(5*sqrt(3),0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(24,0)*shift(3*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(24,0)*shift(5*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(24,0)*shift(7*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(9*sqrt(3)/2,3/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(24,0)*shift(2*sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(24,0)*shift(3*sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(4*sqrt(3),3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(3*sqrt(3)/2,9/2)*S,green+linewidth(.75)); | ||
+ | //filldraw(shift(24,0)*shift(5*sqrt(3)/2,9/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(7*sqrt(3)/2,9/2)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(2*sqrt(3),6)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(3*sqrt(3),6)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(24,0)*shift(5*sqrt(3)/2,15/2)*S,green+linewidth(.75)); | ||
+ | |||
+ | filldraw(shift(38,0)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(38,0)*shift(sqrt(3)/2,3)*S,green+linewidth(.75)); | ||
+ | filldraw(shift(38,0)*shift(-sqrt(3)/2,3)*S,green+linewidth(.75)); | ||
+ | </asy> | ||
[[2013 UMO Problems/Problem 6|Solution]] | [[2013 UMO Problems/Problem 6|Solution]] |
Latest revision as of 02:41, 14 October 2014
Problem 1
Consider the following diagram.
(a) Show that you can retrace the diagram without lifting up your pencil using exactly nine (possibly overlapping) line segments.
(b) Show that you cannot retrace the diagram in the same way using eight or fewer segments.
Problem 2
Alice and Carl play the following game using a square sheet of paper. On each turn, the player makes
a straight cut through the sheet (not necessarily parallel to the sides of the page), creating two new
sheets. The sheet with smaller area is discarded (either one if the two are equal), and the player
gives the larger sheet to the other player. The first player to receive a sheet of area less than square
centimeter from the opposing player loses. If Alice goes first, describe (with proof) the sizes of paper
for which she has a winning strategy.
Problem 3
Find all with
such that the last three digits of
are all equal to the same nonzero
digit.
Problem 4
Given line and distinct points
and
on line
, draw lines
and
through point
, with angles
and
as marked in the figure. Also, draw line segment
at an angle of
from line
such that it intersects line
at
. Establish necessary and sufficient conditions on
,
, and
such that a triangle can be drawn with one of its sides as
with lines
,
, and
as the angle bisectors of that triangle.
Problem 5
Cooper and Malone take turns replacing ,
, and
in the equation below with real numbers.
Once a coefficient has been replaced, no one can choose to
change that coefficient on their turn. The game ends when all three coefficients have been chosen.
Malone wins if
has a non-real root and Cooper wins otherwise. If Malone goes first, find the person who has a winning strategy and describe it with proof.
Problem 6
How many ways can one tile the border of a triangular grid of hexagons of length completely using
only
and
hexagon tiles? Express your answer in terms of a well-known sequence, and
prove that your answer holds true for all positive integers
(examples of such grids for
,
,
, and
are shown below).