MathDB

2012 AMC 12/AHSME

Part of AMC 12/AHSME

Subcontests

(18)

Unit Cube and Octahedron

A unit cube has vertices P1,P2,P3,P4,P1,P2,P3P_1, P_2, P_3, P_4, P_1', P_2', P_3', and P4P_4'. Vertices P2,P3P_2, P_3, and P4P_4 are adjacent to P1P_1, and for 1i41\leq i\leq 4, vertices PiP_i and PiP_i' are opposite to each other. A regular octahedron has one vertex in each of the segments P1P2,P1P3,P1P4,P1P2,P1P3P_1P_2, P_1P_3, P_1P_4, P_1'P_2', P_1'P_3', and P1P4P_1'P_4'. What is the octahedron's side length? [asy] import three;
size(7.5cm); triple eye = (-4, -8, 3); currentprojection = perspective(eye);
triple[] P = {(1, -1, -1), (-1, -1, -1), (-1, 1, -1), (-1, -1, 1), (1, -1, -1)}; // P[0] = P[4] for convenience triple[] Pp = {-P[0], -P[1], -P[2], -P[3], -P[4]};
// draw octahedron triple pt(int k){ return (3*P[k] + P[1])/4; } triple ptp(int k){ return (3*Pp[k] + Pp[1])/4; } draw(pt(2)--pt(3)--pt(4)--cycle, gray(0.6)); draw(ptp(2)--pt(3)--ptp(4)--cycle, gray(0.6)); draw(ptp(2)--pt(4), gray(0.6)); draw(pt(2)--ptp(4), gray(0.6)); draw(pt(4)--ptp(3)--pt(2), gray(0.6) + linetype("4 4")); draw(ptp(4)--ptp(3)--ptp(2), gray(0.6) + linetype("4 4"));
// draw cube for(int i = 0; i < 4; ++i){ draw(P[1]--P); draw(Pp[1]--Pp); for(int j = 0; j < 4; ++j){ if(i == 1 || j == 1 || i == j) continue; draw(P--Pp[j]); draw(Pp--P[j]); } dot(P); dot(Pp); dot(pt(i)); dot(ptp(i)); }
label("P1P_1", P[1], dir(P[1])); label("P2P_2", P[2], dir(P[2])); label("P3P_3", P[3], dir(-45)); label("P4P_4", P[4], dir(P[4])); label("P1P'_1", Pp[1], dir(Pp[1])); label("P2P'_2", Pp[2], dir(Pp[2])); label("P3P'_3", Pp[3], dir(-100)); label("P4P'_4", Pp[4], dir(Pp[4])); [/asy] <spanclass=latexbold>(A)</span> 324<spanclass=latexbold>(B)</span> 7616<spanclass=latexbold>(C)</span> 52<spanclass=latexbold>(D)</span> 233<spanclass=latexbold>(E)</span> 62 <span class='latex-bold'>(A)</span>\ \frac{3\sqrt{2}}{4}\qquad<span class='latex-bold'>(B)</span>\ \frac{7\sqrt{6}}{16}\qquad<span class='latex-bold'>(C)</span>\ \frac{\sqrt{5}}{2}\qquad<span class='latex-bold'>(D)</span>\ \frac{2\sqrt{3}}{3}\qquad<span class='latex-bold'>(E)</span>\ \frac{\sqrt{6}}{2}

Square inscribed in Equiangular Hexagon

Square AXYZAXYZ is inscribed in equiangular hexagon ABCDEFABCDEF with XX on BC\overline{BC}, YY on DE\overline{DE}, and ZZ on EF\overline{EF}. Suppose that AB=40AB=40, and EF=41(31)EF=41(\sqrt{3}-1). What is the side-length of the square?
[asy] size(200); defaultpen(linewidth(1)); pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; draw(A--B--C--D--E--F--cycle); draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); dot("AA",A,W,linewidth(4)); dot("BB",B,dir(0),linewidth(4)); dot("CC",C,dir(0),linewidth(4)); dot("DD",D,dir(20),linewidth(4)); dot("EE",E,dir(100),linewidth(4)); dot("FF",F,W,linewidth(4)); dot("XX",X,dir(0),linewidth(4)); dot("YY",Y,N,linewidth(4)); dot("ZZ",Z,W,linewidth(4)); [/asy]
<spanclass=latexbold>(A)</span> 293<spanclass=latexbold>(B)</span> 2122+4123<spanclass=latexbold>(C)</span> 203+16 <span class='latex-bold'>(A)</span>\ 29\sqrt{3} \qquad<span class='latex-bold'>(B)</span>\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3}\qquad<span class='latex-bold'>(C)</span>\ 20\sqrt{3}+16 <spanclass=latexbold>(D)</span> 202+133<spanclass=latexbold>(E)</span> 216<span class='latex-bold'>(D)</span>\ 20\sqrt{2}+13\sqrt{3} \qquad<span class='latex-bold'>(E)</span>\ 21\sqrt{6}

Set of Right Triangles

Let S={(x,y):x{0,1,2,3,4},y{0,1,2,3,4,5}S=\{(x,y) : x \in \{0,1,2,3,4\}, y \in \{0,1,2,3,4,5\}, and (x,y)(0,0)}(x,y) \neq (0,0) \}. Let TT be the set of all right triangles whose vertices are in SS. For every right triangle t=ABCt=\triangle ABC with vertices AA, BB, and CC in counter-clockwise order and right angle at AA, let f(t)=tan(CBA)f(t)= \tan (\angle CBA). What is tTf(t)? \displaystyle \prod_{t \in T} f(t) \text{?} [asy] size((120)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((0,1)); dot((0,2)); dot((0,3)); dot((0,4)); dot((0,5)); dot((1,1)); dot((1,2)); dot((1,3)); dot((1,4)); dot((1,5)); dot((2,1)); dot((2,2)); dot((2,3)); dot((2,4)); dot((2,5)); dot((3,1)); dot((3,2)); dot((3,3)); dot((3,4)); dot((3,5)); dot((4,1)); dot((4,2)); dot((4,3)); dot((4,4)); dot((4,5)); label("\circ", (0,0)); label("SS", (-.7,2.5)); [/asy]
<spanclass=latexbold>(A)</span> 1<spanclass=latexbold>(B)</span> 625144<spanclass=latexbold>(C)</span> 12524<spanclass=latexbold>(D)</span> 6<spanclass=latexbold>(E)</span> 62524<span class='latex-bold'>(A)</span>\ 1 \qquad <span class='latex-bold'>(B)</span>\ \frac{625}{144} \qquad <span class='latex-bold'>(C)</span>\ \frac{125}{24} \qquad <span class='latex-bold'>(D)</span>\ 6 \qquad <span class='latex-bold'>(E)</span>\ \frac{625}{24}

Sequences of numbers

Let {ak}k=12011\{a_k\}^{2011}_{k=1} be the sequence of real numbers defined by a_1=0.201,   a_2=(0.2011)^{a_1},  a_3=(0.20101)^{a_2},  a_4=(0.201011)^{a_3}, and more generally ak={(0.201010101k+2 digits)ak1,if k is odd,(0.2010101011k+2 digits)ak1,if k is even. a_k = \begin{cases}(0.\underbrace{20101\cdots0101}_{k+2 \ \text{digits}})^{a_{k-1}}, &\text {if } k \text { is odd,} \\ (0.\underbrace{20101\cdots01011}_{k+2 \ \text{digits}})^{a_{k-1}}, &\text {if } k \text { is even.}\end{cases}
Rearranging the numbers in the sequence {ak}k=12011\{a_k\}^{2011}_{k=1} in decreasing order produces a new sequence {bk}k=12011\{b_k\}^{2011}_{k=1}. What is the sum of all the integers kk, 1k20111\le k \le 2011, such that ak=bka_k = b_k?
<spanclass=latexbold>(A)</span> 671<spanclass=latexbold>(B)</span> 1006<spanclass=latexbold>(C)</span> 1341<spanclass=latexbold>(D)</span> 2011<spanclass=latexbold>(E)</span> 2012 <span class='latex-bold'>(A)</span>\ 671\qquad<span class='latex-bold'>(B)</span>\ 1006\qquad<span class='latex-bold'>(C)</span>\ 1341\qquad<span class='latex-bold'>(D)</span>\ 2011\qquad<span class='latex-bold'>(E)</span>\ 2012