Difference between revisions of "2022 AMC 12B Problems/Problem 10"
(→Solution 2) |
(→Solution 3) |
||
Line 34: | Line 34: | ||
== Solution 3 == | == Solution 3 == | ||
− | [[Image:2022 AMC 12B-10.PNG|thumb| | + | [[Image:2022 AMC 12B-10.PNG|thumb|none|509px|Image caption]] |
We use a coordinates approach. Letting the origin be the center of the hexagon, we can let <math>A = (-1, \sqrt{3}), B = (1, \sqrt{3}), C = (2, 0), D = (1, -\sqrt{3}), E = (-1, -\sqrt{3}), F = (-2, 0).</math> Then, <math>G = (0, \sqrt{3})</math> and <math>H = (0, -\sqrt{3}).</math> | We use a coordinates approach. Letting the origin be the center of the hexagon, we can let <math>A = (-1, \sqrt{3}), B = (1, \sqrt{3}), C = (2, 0), D = (1, -\sqrt{3}), E = (-1, -\sqrt{3}), F = (-2, 0).</math> Then, <math>G = (0, \sqrt{3})</math> and <math>H = (0, -\sqrt{3}).</math> | ||
Line 44: | Line 44: | ||
<cmath>GC^2 = (0-2)^2 + (\sqrt{3}-0)^2 = 7 \rightarrow GC = \sqrt{7}</cmath> | <cmath>GC^2 = (0-2)^2 + (\sqrt{3}-0)^2 = 7 \rightarrow GC = \sqrt{7}</cmath> | ||
− | Thus, the perimeter of <math>GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4\sqrt{7}</math>. | + | Thus, the perimeter of <math>GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)} \ 4\sqrt{7}}</math>. |
~sirswagger21 | ~sirswagger21 |
Revision as of 20:58, 20 November 2022
Problem
Regular hexagon has side length . Let be the midpoint of , and let be the midpoint of . What is the perimeter of ?
Solution 1 (No trig)
Let the center of the hexagon be . , , , , , and are all equilateral triangles with side length . Thus, , and . By symmetry, . Thus, by the Pythagorean theorem, . Because and , . Thus, the solution to our problem is
~mathboy100
Solution 2
Consider triangle . Note that , , and because it is an interior angle of a regular hexagon. (See note for details.)
By the Law of Cosines, we have:
By SAS Congruence, triangles , , , and are congruent, and by CPCTC, quadrilateral is a rhombus. Therefore, the perimeter of is .
Note: The sum of the interior angles of any polygon with sides is given by . Therefore, the sum of the interior angles of a hexagon is , and each interior angle of a regular hexagon measures .
Solution 3
We use a coordinates approach. Letting the origin be the center of the hexagon, we can let Then, and
We use the distance formula four times to get
Thus, the perimeter of .
~sirswagger21
Note: the last part of this solution could have been simplified by noting that
Video Solution 1
~Education, the Study of Everything
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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.