Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2016 AMC 12A Problems/Problem 17"

(→‎Solution(Coordinates))
(→‎Solution (Coordinates))
 
Line 17: Line 17:
 
Since <math>EG</math> is the diagonal of square <math>EFGH</math>, <math>\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}</math>
 
Since <math>EG</math> is the diagonal of square <math>EFGH</math>, <math>\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}</math>
  
−
==Solution (Coordinates)==
+
==Solution 2 (Coordinates)==
  
 
Note that we can also use coordinates to solve this problem. WLOG, set the side length of square <math>ABCD</math> equal to <math>6</math>.
 
Note that we can also use coordinates to solve this problem. WLOG, set the side length of square <math>ABCD</math> equal to <math>6</math>.

Latest revision as of 15:54, 27 November 2023

Problem 17

Let $ABCD$ be a square. Let $E, F, G$ and $H$ be the centers, respectively, of equilateral triangles with bases $\overline{AB}, \overline{BC}, \overline{CD},$ and $\overline{DA},$ each exterior to the square. What is the ratio of the area of square $EFGH$ to the area of square $ABCD$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{2+\sqrt{3}}{3} \qquad\textbf{(C)}\ \sqrt{2} \qquad\textbf{(D)}\ \frac{\sqrt{2}+\sqrt{3}}{2} \qquad\textbf{(E)}\ \sqrt{3}$

Solution

The center of an equilateral triangle is its centroid, where the three medians meet.

The distance along the median from the centroid to the base is one third the length of the median.

Let the side length of the square be $1$. The height of $\triangle E$ is $\frac{\sqrt{3}}{2},$ so the distance from $E$ to the midpoint of $\overline{AB}$ is $\frac{\sqrt{3}}{2} \cdot \frac{1}{3} = \frac{\sqrt{3}}{6}$

$EG = 2 \cdot \frac{\sqrt{3}}{6}$ (from above) $+ 1$ (side length of the square).

Since $EG$ is the diagonal of square $EFGH$, $\frac{[EFGH]}{ABCD} = \frac{\frac{EG^2}{2}}{1^2} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$

Solution 2 (Coordinates)

Note that we can also use coordinates to solve this problem. WLOG, set the side length of square $ABCD$ equal to $6$.

This makes the coordinates of the square $EFGH$ equal to $(-\sqrt{3}, 3), (3, 6+\sqrt{3}), (6+\sqrt{3}, 3),$ and $(3, -\sqrt{3})$.

Using the first two points, this gives $EF^2 = (3-(-\sqrt{3})^2+(6+\sqrt{3}-3)^2)= (3+\sqrt{3})^2+(3+\sqrt{3})^2 = 24+12 \sqrt{3}$.

Thus, $[EFGH]=24+12\sqrt{3}$.

Because the side length of $ABCD$ is $6$, $[ABCD] = 36$.

Therefore, $\frac{[EFGH]}{[ABCD]} = \frac{24+12\sqrt{3}}{36} = \boxed{\textbf{(B) }\frac{2 + \sqrt{3}}{3}}$

- Pleaseletmewin

See Also

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_17&oldid=206131"