Subcontests
(30)Counting solutions
The number of real solutions (x,y,z,w) of the simultaneous equations 2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w} is<spanclass=′latex−bold′>(A)</span> 1<spanclass=′latex−bold′>(B)</span> 2<spanclass=′latex−bold′>(C)</span> 4<spanclass=′latex−bold′>(D)</span> 8<spanclass=′latex−bold′>(E)</span> 16 Pentagon Geometry
ABCDE is a regular pentagon. AP, AQ and AR are the perpendiculars dropped from A onto CD, CB extended and DE extended, respectively. Let O be the center of the pentagon. If OP=1, then AO+AQ+AR equals[asy]
size(200);
defaultpen(fontsize(10pt)+linewidth(.8pt));
pair O=origin, A=2*dir(90), B=2*dir(18), C=2*dir(306), D=2*dir(234), E=2*dir(162), P=(C+D)/2, Q=C+3.10*dir(C--B), R=D+3.10*dir(D--E), S=C+4.0*dir(C--B), T=D+4.0*dir(D--E);
draw(A--B--C--D--E--A^^E--T^^B--S^^R--A--Q^^A--P^^rightanglemark(A,Q,S,7)^^rightanglemark(A,R,T,7));
dot(O);
label("O",O,dir(B));
label("1",(O+P)/2,W);
label("A",A,dir(A));
label("B",B,dir(B));
label("C",C,dir(C));
label("D",D,dir(D));
label("E",E,dir(E));
label("P",P,dir(P));
label("Q",Q,dir(Q-A));
label("R",R,dir(R-A));
[/asy]<spanclass=′latex−bold′>(A)</span> 3<spanclass=′latex−bold′>(B)</span> 1+5<spanclass=′latex−bold′>(C)</span> 4<spanclass=′latex−bold′>(D)</span> 2+5<spanclass=′latex−bold′>(E)</span> 5 Ratio of areas
In the adjoining figure, AB is a diameter of the circle, CD is a chord parallel to AB, and AC intersects BD at E, with ∠AED=α. The ratio of the area of △CDE to that of △ABE is[asy]
size(200);
defaultpen(fontsize(10pt)+linewidth(.8pt));
pair A=(-1,0), B=(1,0), E=(0,-.4), C=(.6,-.8), D=(-.6,-.8), E=(0,-.8/(1.6));
draw(unitcircle);
draw(A--B--D--C--A);
draw(Arc(E,.2,155,205));
label("A",A,W);
label("B",B,C);
label("C",C,C);
label("D",D,W);
label("α",E-(.2,0),W);
label("E",E,N);[/asy]<spanclass=′latex−bold′>(A)</span> cos α<spanclass=′latex−bold′>(B)</span> sin α<spanclass=′latex−bold′>(C)</span> cos2α<spanclass=′latex−bold′>(D)</span> sin2α<spanclass=′latex−bold′>(E)</span> 1−sin α Random pick problem
Six distinct integers are picked at random from {1,2,3,…,10}. What is the probability that, among those selected, the second smallest is 3?<spanclass=′latex−bold′>(A)</span> 601<spanclass=′latex−bold′>(B)</span> 61<spanclass=′latex−bold′>(C)</span> 31<spanclass=′latex−bold′>(D)</span> 21<spanclass=′latex−bold′>(E)</span> none of these Trigonometry
In the configuration below, θ is measured in radians, C is the center of the circle, BCD and ACE are line segments and AB is tangent to the circle at A.[asy]
defaultpen(fontsize(10pt)+linewidth(.8pt));
pair A=(0,-1), E=(0,1), C=(0,0), D=dir(10), F=dir(190), B=(-1/sin(10*pi/180))*dir(10);
fill(Arc((0,0),1,10,90)--C--D--cycle,mediumgray);
fill(Arc((0,0),1,190,270)--B--F--cycle,mediumgray);
draw(unitcircle);
draw(A--B--D^^A--E);
label("A",A,S);
label("B",B,W);
label("C",C,SE);
label("θ",C,SW);
label("D",D,NE);
label("E",E,N);
[/asy]A necessary and sufficient condition for the equality of the two shaded areas, given 0<θ<2π, is<spanclass=′latex−bold′>(A)</span> tanθ=θ<spanclass=′latex−bold′>(B)</span> tanθ=2θ<spanclass=′latex−bold′>(C)</span> tanθ=4θ<spanclass=′latex−bold′>(D)</span> tan2θ=θ<spanclass=′latex−bold′>(E)</span> tan2θ=θ Similar triangles
In △ABC, AB=8, BC=7, CA=6 and side BC is extended, as shown in the figure, to a point P so that △PAB is similar to △PCA. The length of PC is[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B);
draw(A--P--C--A--B--C);
label("A", A, W);
label("B", B, E);
label("C", C, NE);
label("P", P, NW);
label("6", 3*dir(A--C), SE);
label("7", B+3*dir(B--C), NE);
label("8", (4,0), S);[/asy]
<spanclass=′latex−bold′>(A)</span> 7<spanclass=′latex−bold′>(B)</span> 8<spanclass=′latex−bold′>(C)</span> 9<spanclass=′latex−bold′>(D)</span> 10<spanclass=′latex−bold′>(E)</span> 11 Line length
In △ABC, AB=13, BC=14 and CA=15. Also, M is the midpoint of side AB and H is the foot of the altitude from A to BC. The length of HM is[asy]
size(200);
defaultpen(linewidth(0.7)+fontsize(10));
pair H=origin, A=(0,6), B=(-4,0), C=(5,0), M=B+3.6*dir(B--A);
draw(B--C--A--B^^M--H--A^^rightanglemark(A,H,C));
label("A", A, NE);
label("B", B, W);
label("C", C, E);
label("H", H, S);
label("M", M, dir(M));
[/asy]<spanclass=′latex−bold′>(A)</span> 6<spanclass=′latex−bold′>(B)</span> 6.5<spanclass=′latex−bold′>(C)</span> 7<spanclass=′latex−bold′>(D)</span> 7.5<spanclass=′latex−bold′>(E)</span> 8 Approximation
The population of the United States in 1980 was 226,504,825. The area of the country is 3,615,122 square miles. The are (5280)2 square feet in one square mile. Which number below best approximates the average number of square feet per person?<spanclass=′latex−bold′>(A)</span> 5,000<spanclass=′latex−bold′>(B)</span> 10,000<spanclass=′latex−bold′>(C)</span> 50,000<spanclass=′latex−bold′>(D)</span> 100,000<spanclass=′latex−bold′>(E)</span> 500,000 Two Identical Blocks of Wood
Using a table of a certain height, two identical blocks of wood are placed as shown in Figure 1. Length r is found to be 32 inches. After rearranging the blocks as in Figure 2, length s is found to be 28 inches. How high is the table?[asy]
size(300);
defaultpen(linewidth(0.8)+fontsize(13pt));
path table = origin--(1,0)--(1,6)--(6,6)--(6,0)--(7,0)--(7,7)--(0,7)--cycle;
path block = origin--(3,0)--(3,1.5)--(0,1.5)--cycle;
path rotblock = origin--(1.5,0)--(1.5,3)--(0,3)--cycle;
draw(table^^shift((14,0))*table);
filldraw(shift((7,0))*block^^shift((5.5,7))*rotblock^^shift((21,0))*rotblock^^shift((18,7))*block,gray);
draw((7.25,1.75)--(8.5,3.5)--(8.5,8)--(7.25,9.75),Arrows(size=5));
draw((21.25,3.25)--(22,3.5)--(22,8)--(21.25,8.25),Arrows(size=5));
unfill((8,5)--(8,6.5)--(9,6.5)--(9,5)--cycle);
unfill((21.5,5)--(21.5,6.5)--(23,6.5)--(23,5)--cycle);
label("r",(8.5,5.75));
label("s",(22,5.75));
[/asy]<spanclass=′latex−bold′>(A)</span>28 inches<spanclass=′latex−bold′>(B)</span>29 inches<spanclass=′latex−bold′>(C)</span>30 inches<spanclass=′latex−bold′>(D)</span>31 inches<spanclass=′latex−bold′>(E)</span>32 inches Easy chapter level geometry
△ABC is a right angle at C and ∠A=20∘. If BD is the bisector of ∠ABC, then ∠BDC=
[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(11pt));
pair A= origin, B = 3 * dir(25), C = (B.x,0);
pair X = bisectorpoint(A,B,C), D = extension(B,X,A,C);
draw(B--A--C--B--D^^rightanglemark(A,C,B,4));
path g = anglemark(A,B,D,14);
path h = anglemark(D,B,C,14);
draw(g);
draw(h);
add(pathticks(g,1,0.11,6,6));
add(pathticks(h,1,0.11,6,6));
label("A",A,W);
label("B",B,NE);
label("C",C,E);
label("D",D,S);
label("20∘",A,8*dir(12.5));
[/asy]<spanclass=′latex−bold′>(A)</span> 40∘<spanclass=′latex−bold′>(B)</span> 45∘<spanclass=′latex−bold′>(C)</span> 50∘<spanclass=′latex−bold′>(D)</span> 55∘<spanclass=′latex−bold′>(E)</span> 60∘