MathDB

1

nf(2n + 1) = (2n + 1)(f(n) + n) , f(2n) = 2f(n)

Consider a function f defined on the positive integers that meets the following conditions:

f(1) = 1 \, , \,\, f(2n) = 2f(n) \, , \,\, nf(2n + 1) = (2n + 1)(f(n) + n)

for all

n \ge 1

. a) Prove that

f(n)

is an integer for all

n

. b) Find all positive integers

m

less than

2013

that satisfy the equation

f(m) = 2m

.

6

1

Juan must pay 4 bills and goes to an ATM

Juan must pay

4

bills. He goes to an ATM, but doesn't remember the amount of the bills. Just know that a) Each account is a multiple of

1,000

and is at least

4,000

. b) The accounts total 2

00, 000

. What is the least number of times Juan must use the ATM to make sure he can pay the bills with exact change without any excess money? The cashier has banknotes of

2, 000

,

5, 000

,

10, 000

, and

20,000

. Juan can decide how much money he asks the cashier each time, but you cannot decide how many bills of each type to give to the cashier.

3

1

exists i with sum a_i_>0 if a_1 + a_2 + ... + a_n > 0

Given a finite sequence of real numbers

a_1,a_2,...,a_n

such that

a_1 + a_2 + ... + a_n > 0.

Prove that there is at least one index

i

such that

a_i > 0, a_i + a_{i+1} > 0, ..., a_i + a_{i+1} + ...+ a_n > 0.

2

1

Hannibal and Clarice are still at a barbecue

Hannibal and Clarice are still at a barbecue and there are three anticuchos left, each of which it has

10

pieces. Of the

30

total pieces, there are

29

chicken and one meat, the which is at the bottom of one of the anticuchos. To decide who to stay with the piece of meat, they decide to play the following game: they alternately take out a piece of one of the anticuchos (they can take only the outer pieces) and whoever wins the game manages to remove the piece of meat. Clarice decides if she starts or Hannibal starts. What should she decide?

1

sum of 5-digit numbers with digits 1,2,5

Find the sum of all

5

-digit positive integers that they have only the digits

1, 2

, and

5

, none repeated more than three consecutive times.

5

1

intersection of a conical surface and a plane is an ellipse

A conical surface

C

is cut by a plane

T

as shown in the figure on the back of this sheet. Show that

C \cap T

is an ellipse. You can use as an aid the fact that if you consider the two spheres tangent to

C

and

T

as shown in the figure, they intersect

T

in the bulbs.
[asy] // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); }
// projection of point A onto line BC triple projectionofpointontoline(triple A, triple B, triple C) { return lineintersectplan(B, B - C, A, B - C); }
// calculate area of space triangle with vertices A, B, and C real trianglearea(triple A, triple B, triple C) { return abs(cross(A - C, B - C)/2); }
// calculate incentre of space triangle ABC triple triangleincentre(triple A, triple B, triple C) { return (abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(abs(B - C) + abs(C - A) + abs(A - B)); }
// calculate inradius of space triangle ABC real triangleinradius(triple A, triple B, triple C) { return 2*trianglearea(A,B,C)/(abs(B - C) + abs(C - A) + abs(A - B)); }
// calculate excentre of space triangle ABC triple triangleexcentre(triple A, triple B, triple C) { return (-abs(B - C) * A + abs(C - A) * B + abs(A - B) * C)/(-abs(B - C) + abs(C - A) + abs(A - B)); }
// calculate exradius of space triangle ABC real triangleexradius(triple A, triple B, triple C) { return 2*trianglearea(A,B,C)/(-abs(B - C) + abs(C - A) + abs(A - B)); }
unitsize(2 cm);
pair project (triple A, real t) { return((A.x, A.y*Sin(t) + A.z*Cos(t))); }
real alpha, beta, theta, t; real coneradius = 1, coneheight = 3; real a, b, c; real[] m, r; triple A, B, V; triple ellipsecenter, ellipsex, ellipsey; triple[] F, O, P, R, W; path[] ellipse, spherering;
theta = 15; V = (0,0,-coneheight);
m[1] = sqrt(Cos(theta)^2*coneheight^2 - Sin(theta)^2*coneradius^2)/coneradius; m[2] = -m[1]; alpha = -aTan(Sin(theta)/m[1]); beta = -aTan(Sin(theta)/m[2]) + 180; A = (coneradius*Cos(alpha), coneradius*Sin(alpha), 0); B = (coneradius*Cos(beta), coneradius*Sin(beta), 0);
W[1] = interp(V,(coneradius,0,0),0.6); W[2] = interp(V,(-coneradius,0,0),0.4); O[1] = triangleexcentre(V,W[1],W[2]); O[2] = triangleincentre(V,W[1],W[2]); r[1] = triangleexradius(V,W[1],W[2]); r[2] = triangleinradius(V,W[1],W[2]); F[1] = projectionofpointontoline(O[1],W[1],W[2]); F[2] = projectionofpointontoline(O[2],W[1],W[2]);
P[1] = O[1] - (0,0,r[1]*coneradius/sqrt(coneradius^2 + coneheight^2)); P[2] = O[2] - (0,0,r[2]*coneradius/sqrt(coneradius^2 + coneheight^2)); spherering[11] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta); spherering[12] = shift(project(P[1],theta))*yscale(Sin(theta))*arc((0,0),r[1]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360); spherering[21] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),alpha,beta); spherering[22] = shift(project(P[2],theta))*yscale(Sin(theta))*arc((0,0),r[2]*coneheight/sqrt(coneradius^2 + coneheight^2),beta,alpha + 360);
ellipsecenter = (W[1] + W[2])/2; a = abs(W[1] - ellipsecenter); c = abs(F[1] - ellipsecenter); b = sqrt(a^2 - c^2); ellipsex = (W[1] - W[2])/abs(W[1] - W[2]); ellipsey = (0,1,0);
ellipse[1] = project(ellipsecenter + a*ellipsex, theta);
for (t = 0; t <= 180; t = t + 5) { ellipse[1] = ellipse[1]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta); }
ellipse[2] = project(ellipsecenter - a*ellipsex, theta);
for (t = 180; t <= 360; t = t + 5) { ellipse[2] = ellipse[2]--project(ellipsecenter + a*Cos(t)*ellipsex + b*Sin(t)*ellipsey, theta); }
R[1] = ellipsecenter + 1*ellipsex + ellipsey; R[2] = ellipsecenter - 1.2*ellipsex + ellipsey; R[3] = ellipsecenter - 1*ellipsex - ellipsey; R[4] = ellipsecenter + 1.2*ellipsex - ellipsey;
fill(ellipse[1]--ellipse[2]--cycle, gray(0.9)); draw(yscale(Sin(theta))*Circle((0,0),coneradius)); draw(project(V,theta)--project(A,theta)); draw(project(V,theta)--project(B,theta)); draw(Circle(project(O[1],theta),r[1])); draw(Circle(project(O[2],theta),r[2])); draw(spherering[11], dashed); draw(spherering[12]); draw(spherering[21], dashed); draw(spherering[22]); draw(ellipse[1], dashed); draw(ellipse[2]); draw(project(R[1],theta)--interp(project(R[1],theta),project(R[2],theta),0.13)); draw(interp(project(R[1],theta),project(R[2],theta),0.13)--interp(project(R[1],theta),project(R[2],theta),0.76), dashed); draw(interp(project(R[1],theta),project(R[2],theta),0.76)--project(R[2],theta)); draw(project(R[2],theta)--project(R[3],theta)--project(R[4],theta)--project(R[1],theta));
label("

C

", (-1,0.3)); label("

T

", (1.2,-0.8)); dot(project(F[1],theta)); dot(project(F[2],theta)); //dot("

F_1

", project(F[1],theta)); //dot("

F_2

", project(F[2],theta)); //dot("

O_1

", project(O[1],theta)); //dot("

O_2

", project(O[2],theta)); //dot("

P_1

", project(P[1],theta)); //dot("

V

", project(V,theta)); //dot("

W_1

", project(W[1],theta)); //dot("

W_2

", project(W[2],theta)); [/asy]

2013 Chile National Olympiad

Subcontests

nf(2n + 1) = (2n + 1)(f(n) + n) , f(2n) = 2f(n)

Juan must pay 4 bills and goes to an ATM

exists i with sum a_i_>0 if a_1 + a_2 + ... + a_n > 0

Hannibal and Clarice are still at a barbecue

sum of 5-digit numbers with digits 1,2,5

intersection of a conical surface and a plane is an ellipse

2013 Chile National Olympiad

Subcontests

nf(2n + 1) = (2n + 1)(f(n) + n) , f(2n) = 2f(n)

Juan must pay 4 bills and goes to an ATM

exists i with sum a_i_&gt;0 if a_1 + a_2 + ... + a_n &gt; 0

Hannibal and Clarice are still at a barbecue

sum of 5-digit numbers with digits 1,2,5

intersection of a conical surface and a plane is an ellipse

exists i with sum a_i_>0 if a_1 + a_2 + ... + a_n > 0