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Difference between revisions of "2013 UMO Problems"

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m (→‎Problem 5)
 
(4 intermediate revisions by the same user not shown)
Line 2: Line 2:
  
 
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.
−
<math>P(x) = x^3 + ax^2 + bx + c</math> . Once a coefficient has been replaced, no one can choose to  
+
<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.

[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 overlapping) line segments.

(b) Show that you cannot retrace the diagram in the same way using eight or fewer segments.


Solution

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 $1$ 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.

Solution

Problem 3

Find all $x$ with $1 \le x \le 999$ such that the last three digits of $x^2$ are all equal to the same nonzero digit.

Solution

Problem 4

Given line $\ell_1$ and distinct points $I$ and $X$ on line $\ell_1$, draw lines $\ell_2$ and $\ell_3$ through point $I$, with angles $\alpha$ and $\beta$ as marked in the figure. Also, draw line segment $XY$ at an angle of $\gamma$ from line $\ell_1$ such that it intersects line $\ell_2$ at $Y$. Establish necessary and sufficient conditions on $\alpha$ , $\beta$ , and $\gamma$ such that a triangle can be drawn with one of its sides as $XY$ with lines $\ell_1$, $\ell_2$, and $\ell_3$ 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]

Solution

Problem 5

Cooper and Malone take turns replacing $a$, $b$, and $c$ in the equation below with real numbers. \[P(x) = x^3 + ax^2 + bx + c.\] 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 $P(x)$ 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.

Solution

Problem 6

How many ways can one tile the border of a triangular grid of hexagons of length $n$ completely using only $1 \times 1$ and $1 \times 2$ hexagon tiles? Express your answer in terms of a well-known sequence, and prove that your answer holds true for all positive integers $n \ge 3$ (examples of such grids for $n = 3$, $n = 4$, $n = 5$, and $n = 6$ 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]

Solution

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