olymmath

AMC 12 2003b Problem 10

  • Paper

Several figures can be made by attaching two equilateral triangles to the regular pentagon ABCDE in two of the five positions shown. How many non-congruent figures can be constructed in this way?

size(200);
defaultpen(0.9);
real r = 5/dir(54).x, h = 5 tan(54*pi/180);
pair A = (5,0), B = A+10*dir(72), C = (0,r+h), E = (-5,0), D = E+10*dir(108);
draw(A--B--C--D--E--cycle);
label("AA",A+(0,-0.5),SSE);
label("BB",B+(0.5,0),ENE);
label("CC",C+(0,0.5),N);
label("DD",D+(-0.5,0),WNW);
label("EE",E+(0,-0.5),SW);
//
real l = 5*sqrt(3); pair ab = (h+l)*dir(72), bc = (h+l)*dir(54);
pair AB = (ab.y, h-ab.x), BC = (bc.x,h+bc.y), CD = (-bc.x,h+bc.y), DE = (-ab.y, h-ab.x), EA = (0,-l);
draw(A--AB--B^^B--BC--C^^C--CD--D^^D--DE--E^^E--EA--A, dashed);
//
dot(A); dot(B); dot(C); dot(D); dot(E); dot(AB); dot(BC); dot(CD); dot(DE); dot(EA);

(A) 1(B) 2(C) 3(D) 4(E) 5\text {(A) } 1 \text {(B) } 2 \text {(C) } 3 \text {(D) } 4 \text {(E) } 5

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

Show the answer

B

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 amc12/2003b/p10 · Tag H83E2TSBNT · Edit this page on GitHub · Keys: j next, k previous