Difference between revisions of "1983 AHSME Problems/Problem 4"

Revision as of 23:00, 1 September 2017

$[asy] pair A, B, C, D, E, F; A = (0, 1.732); B = (0.5, 0.866); C = (0,0); D = (1, 0); E = (1.5, 0.866); F = (1, 1.732); draw(A--B--C--D--E--F--A); label("A", A, A); label("B", (0.3, 0.866)); label("C", (-0.1, 0)); label("D", D, D); label("E", E, E); label("F", F, F); draw(B--D, dashed+linewidth(0.5)); draw(B--E, dashed+linewidth(0.5)); draw(B--F, dashed+linewidth(0.5)); [/asy]$

Drawing the diagram as described, we create a convex hexagon with all side lengths equal to 1. In this case, it is a natural approach to divide the figure up into four equilateral triangles. The area, A, of one such equilateral triangle is $\frac{\sqrt{3}}{4}$ , which gives a total of $4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}$ , or $\boxed{D}$ .

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Difference between revisions of "1983 AHSME Problems/Problem 4"

Revision as of 23:00, 1 September 2017

@@ Line 8: / Line 8: @@
 F = (1, 1.732);
 draw(A--B--C--D--E--F--A);
-label("A", A, dir(point--A));
+label("A", A, A);
+label("B", (0.3, 0.866));
+label("C", (-0.1, 0));
+label("D", D, D);
+label("E", E, E);
+label("F", F, F);
+draw(B--D, dashed+linewidth(0.5));
+draw(B--E, dashed+linewidth(0.5));
+draw(B--F, dashed+linewidth(0.5));
 </asy>
+Drawing the diagram as described, we create a convex hexagon with all side lengths equal to 1. In this case, it is a natural approach to divide the figure up into four equilateral triangles. The area, A, of one such equilateral triangle is <math>\frac{\sqrt{3}}{4}</math>, which gives a total of <math>4\left(\frac{\sqrt{3}}{4}\right) = \sqrt{3}</math>, or <math>\boxed{D}</math>.