Subcontests
(30)Odd Integer Sequence
In the non-decreasing sequence of odd integers {a1,a2,a3,…}={1,3,3,3,5,5,5,5,5,…} each odd positive integer k appears k times. It is a fact that there are integers b, c, and d such that for all positive integers n,
an=b⌊n+c⌋+d,
where ⌊x⌋ denotes the largest integer not exceeding x. The sum b+c+d equals
(A) 0(B) 1(C) 2(D) 3(E) 4 Area Ratios
In triangle ABC, ∡CBA=72∘, E is the midpoint of side AC, and D is a point on side BC such that 2BD=DC; AD and BE intersect at F. The ratio of the area of triangle BDF to the area of quadrilateral FDCE is[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E);
draw(E--B--A--C--B^^A--D);
label("A", A, dir(D--A));
label("B", B, dir(E--B));
label("C", C, dir(0));
label("D", D, SE);
label("E", E, N);
label("F", F, dir(80));[/asy](A) 51(B) 41(C) 31(D) 52(E) none of these A "Cartesian Bug"
A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to (1,0). Then it makes a 90∘ counterclockwise and travels 21 a unit to (1,21). If it continues in this fashion, each time making a 90∘ degree turn counterclockwise and traveling half as far as the previous move, to which of the following points will it come closest?
(A) (32,32)(B) (54,52)(C) (32,54)(D) (32,31)(E) (52,54) Quadrilateral Area
Sides AB,BC,CD and DA of convex polygon ABCD have lengths 3,4,12, and 13, respectively, and ∡CBA is a right angle. The area of the quadrilateral is[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
real r=degrees((12,5)), s=degrees((3,4));
pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s));
draw(A--B--C--D--cycle);
markscalefactor=0.05;
draw(rightanglemark(A,B,C));
pair point=incenter(A,C,D);
label("A", A, dir(point--A));
label("B", B, dir(point--B));
label("C", C, dir(point--C));
label("D", D, dir(point--D));
label("3", A--B, dir(A--B)*dir(-90));
label("4", B--C, dir(B--C)*dir(-90));
label("12", C--D, dir(C--D)*dir(-90));
label("13", D--A, dir(D--A)*dir(-90));[/asy](A) 32(B) 36(C) 39(D) 42(E) 48 Length Chasing
If AB and CD are perpendicular diameters of circle Q, P in AQ, and ∡QPC=60∘, then the length of PQ divided by the length of AQ is
[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0);
draw(P--C--D^^A--B^^Circle(Q,1));
label("A", A, W);
label("B", B, E);
label("C", C, N);
label("D", D, S);
label("P", P, S);
label("Q", Q, SE);
label("60∘", P+0.0.5*dir(30), dir(30));[/asy]
<spanclass=′latex−bold′>(A)</span> 23<spanclass=′latex−bold′>(B)</span> 33<spanclass=′latex−bold′>(C)</span> 22<spanclass=′latex−bold′>(D)</span> 21<spanclass=′latex−bold′>(E)</span> 32 Angle Measure
In the adjoining figure, CDE is an equilateral triangle and ABCD and DEFG are squares. The measure of ∠GDA is(A) 90∘(B) 105∘(C) 120∘(D) 135∘(E) 150∘[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair D=origin, C=D+dir(240), E=D+dir(300), F=E+dir(30), G=D+dir(30), A=D+dir(150), B=C+dir(150);
draw(E--D--G--F--E--C--D--A--B--C);
pair point=(0,0.5);
label("A", A, dir(point--A));
label("B", B, dir(point--B));
label("C", C, dir(point--C));
label("D", D, dir(-15));
label("E", E, dir(point--E));
label("F", F, dir(point--F));
label("G", G, dir(point--G));[/asy]