Difference between revisions of "2012 AMC 10B Problems/Problem 14"
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==Solution== | ==Solution== | ||
| − | + | <center><asy> | |
| + | size(8cm); | ||
| + | pair A, B, C, D, E, F, G, H, BF, AF; | ||
| + | A = (0,0); | ||
| + | B = (1,0); | ||
| + | C = (1,1); | ||
| + | D = (0,1); | ||
| + | E = (1/2,0); | ||
| + | H = (1/2,1); | ||
| + | G = (1/2,1/2^(1/2)); | ||
| + | F = (1/2,1-(1/2^(1/2))); | ||
| + | AF = (3^(1/2)/2,1/2); | ||
| + | BF = (1-3^(1/2)/2,1/2); | ||
| + | draw(A--B--C--D--A--AF--D); | ||
| + | draw(C--BF--B); | ||
| + | draw(H--E,linetype("8 8")); | ||
| + | label("$A$",A,SW); | ||
| + | label("$B$",B,SE); | ||
| + | label("$C$",C,NE); | ||
| + | label("$D$",D,NW); | ||
| + | label("$E$",E,S); | ||
| + | label("$H$",H,N); | ||
| + | label("$G$",G+1/20,E); | ||
| + | label("$F$",F-1/20,W); | ||
| + | </asy></center> | ||
Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length <math>\sqrt{3}</math> and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length <math>2\sqrt{3} - 2</math>. The formula for the area of an equilateral triangle of length s is <math>\frac{\sqrt{3}}{4}s^2</math>. It follows that the area of the rhombus is: | Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length <math>\sqrt{3}</math> and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length <math>2\sqrt{3} - 2</math>. The formula for the area of an equilateral triangle of length s is <math>\frac{\sqrt{3}}{4}s^2</math>. It follows that the area of the rhombus is: | ||
| − | <math>2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = 8\sqrt{3} - 12 | + | <math>2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}</math> |
| − | |||
| − | |||
==See Also== | ==See Also== | ||
Revision as of 16:40, 29 December 2017
Problem
Two equilateral triangles are contained in square whose side length is 
. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?
Solution
![[asy] size(8cm); pair A, B, C, D, E, F, G, H, BF, AF; A = (0,0); B = (1,0); C = (1,1); D = (0,1); E = (1/2,0); H = (1/2,1); G = (1/2,1/2^(1/2)); F = (1/2,1-(1/2^(1/2))); AF = (3^(1/2)/2,1/2); BF = (1-3^(1/2)/2,1/2); draw(A--B--C--D--A--AF--D); draw(C--BF--B); draw(H--E,linetype("8 8")); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,S); label("$H$",H,N); label("$G$",G+1/20,E); label("$F$",F-1/20,W); [/asy]](http://latex.artofproblemsolving.com/1/b/4/1b491bea6ff8007acc151d99ff17f5f0ab905857.png)
Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length 
and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length 
. The formula for the area of an equilateral triangle of length s is 
. It follows that the area of the rhombus is:
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
