MathDB

22

Part of 2015 AMC 10

Problems(2)

No two Coins Adjacent

Source: 2015 AMC 12A #17/10A #22

2/4/2015
Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
<spanclass=latexbold>(A)</span>47256<spanclass=latexbold>(B)</span>316<spanclass=latexbold>(C)</span>49256<spanclass=latexbold>(D)</span>25128<spanclass=latexbold>(E)</span>51256<span class='latex-bold'>(A) </span>\dfrac{47}{256}\qquad<span class='latex-bold'>(B) </span>\dfrac{3}{16}\qquad<span class='latex-bold'>(C) </span>\dfrac{49}{256}\qquad<span class='latex-bold'>(D) </span>\dfrac{25}{128}\qquad<span class='latex-bold'>(E) </span>\dfrac{51}{256}
probabilityAMCAMC 12countingdistinguishabilityAMC10
Pentagon Ratios

Source: 2015 AMC 10B #22

2/26/2015
In the figure shown below, ABCDEABCDE is a regular pentagon and AG=1AG=1. What is FG+JH+CDFG+JH+CD? [asy] import cse5;pathpen=black;pointpen=black; size(2inch); pair A=dir(90), B=dir(18), C=dir(306), D=dir(234), E=dir(162); D(MP("A",A,A)--MP("B",B,B)--MP("C",C,C)--MP("D",D,D)--MP("E",E,E)--cycle,linewidth(1.5)); D(A--C--E--B--D--cycle); pair F=IP(A--D,B--E), G=IP(B--E,C--A), H=IP(C--A,B--D), I=IP(D--B,E--C), J=IP(C--E,D--A); D(MP("F",F,dir(126))--MP("I",I,dir(270))--MP("G",G,dir(54))--MP("J",J,dir(198))--MP("H",H,dir(342))--cycle); [/asy] <spanclass=latexbold>(A)</span>3<spanclass=latexbold>(B)</span>1245<spanclass=latexbold>(C)</span>5+253<spanclass=latexbold>(D)</span>1+5<spanclass=latexbold>(E)</span>11+11510<span class='latex-bold'>(A) </span> 3 \qquad<span class='latex-bold'>(B) </span> 12-4\sqrt5 \qquad<span class='latex-bold'>(C) </span> \dfrac{5+2\sqrt5}{3} \qquad<span class='latex-bold'>(D) </span> 1+\sqrt5 \qquad<span class='latex-bold'>(E) </span> \dfrac{11+11\sqrt5}{10}
ratioAsymptotetrigonometryUSAMTSgeometryparallelogramangle bisector