olymmath

AMC 12 2024a Problem 18

  • Paper

On top of a rectangular card with sides of length 11 and 2+32+\sqrt{3}, an identical card is placed so that two of their diagonals line up, as shown (AC‾\overline{AC}, in this case).

defaultpen(fontsize(12)+0.85); size(150);
real h=2.25;
pair C=origin,B=(0,h),A=(1,h),D=(1,0),Dp=reflect(A,C)*D,Bp=reflect(A,C)*B;
pair L=extension(A,Dp,B,C),R=extension(Bp,C,A,D);
draw(L--B--A--Dp--C--Bp--A);
draw(C--D--R);
draw(L--C^^R--A,dashed+0.6);
draw(A--C,black+0.6);
dot("CC",C,2*dir(C-R)); dot("AA",A,1.5*dir(A-L)); dot("BB",B,dir(B-R));

Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled BB in the figure?

(A) 6(B) 8(C) 10(D) 12(E) No new vertex will land on B.\textbf{(A) }6 \textbf{(B) }8 \textbf{(C) }10 \textbf{(D) }12 \textbf{(E) }\text{No new vertex will land on }B.

Dataset text AMC 12 2000 to 2025. Quoted with source.

Show the answer

A

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator amc12/2024a/p18 · Tag XZHZ402SF1 · Edit this page on GitHub · Keys: j next, k previous