Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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− | == Solution 5 (Answer choices + Pythagorean Theorem | + | == Solution 5 (Answer choices + Pythagorean Theorem Extension) == |
Like the previous solution, note that <math>\triangle{GAF}, \triangle{FEH}, \triangle{HDC},</math> and <math>\triangle{CBG}</math> are all congruent by SAS. It follows that quadrilateral <math>GCHF</math> is a rhombus. | Like the previous solution, note that <math>\triangle{GAF}, \triangle{FEH}, \triangle{HDC},</math> and <math>\triangle{CBG}</math> are all congruent by SAS. It follows that quadrilateral <math>GCHF</math> is a rhombus. | ||
Recall the Pythagorean Theorem, which states <math>a^2+b^2=c^2</math> for all right triangles, where <math>c</math> is the hypotenuse of the triangle. However, by drawing a quick diagram of an obtuse triangle, we can clearly see that <math>a^2+b^2<c^2</math>, in an obtuse triangle. | Recall the Pythagorean Theorem, which states <math>a^2+b^2=c^2</math> for all right triangles, where <math>c</math> is the hypotenuse of the triangle. However, by drawing a quick diagram of an obtuse triangle, we can clearly see that <math>a^2+b^2<c^2</math>, in an obtuse triangle. | ||
− | Returning to the problem, we have: <cmath>AG^2+AF^2<GF^2 \implies 1^2+2^2<GF^2 \implies \sqrt{5}<GF</cmath> | + | Returning to the problem, since <math>ABCDEF</math> is a regular hexagon, <math>\triangle{GAF} is an obtuse triangle. Using the extended Pythagorean theorem, we have: <cmath>AG^2+AF^2<GF^2 \implies 1^2+2^2<GF^2 \implies \sqrt{5}<GF</cmath> |
− | Since <math>GCHF< | + | Since </math>GCHF<math> is a rhombus, the perimeter is </math>4GF<math>. This eliminates all answer choices but </math>D<math> and </math>E<math>, since in all of those options </math>GF<\sqrt{5}<math>. Lastly, </math>E<math> is eliminated due to the triangle inequality, as </math>1+2<math> is not greater than </math>12/4=3<math>. |
− | Hence, the answer is <math>\boxed{\textbf{(D)}\ 4\sqrt7} | + | Hence, the answer is </math>\boxed{\textbf{(D)}\ 4\sqrt7}$. |
~SwordOfJustice | ~SwordOfJustice |
Revision as of 20:50, 15 June 2023
Contents
Problem
Regular hexagon has side length
. Let
be the midpoint of
, and let
be the midpoint of
. What is the perimeter of
?
Diagram
~MRENTHUSIASM
Solution 1
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
Solution 4
Note that triangles and
are all congruent, since they have side lengths of
and
and an included angle of
By the Law of Cosines, Therefore,
-Benedict T (countmath1)
Solution 5 (Answer choices + Pythagorean Theorem Extension)
Like the previous solution, note that and
are all congruent by SAS. It follows that quadrilateral
is a rhombus.
Recall the Pythagorean Theorem, which states for all right triangles, where
is the hypotenuse of the triangle. However, by drawing a quick diagram of an obtuse triangle, we can clearly see that
, in an obtuse triangle.
Returning to the problem, since is a regular hexagon, $\triangle{GAF} is an obtuse triangle. Using the extended Pythagorean theorem, we have: <cmath>AG^2+AF^2<GF^2 \implies 1^2+2^2<GF^2 \implies \sqrt{5}<GF</cmath>
Since$ (Error compiling LaTeX. Unknown error_msg)GCHF4GF
D
E
GF<\sqrt{5}
E
1+2
12/4=3$.
Hence, the answer is$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(D)}\ 4\sqrt7}$.
~SwordOfJustice
Video Solution 1
~Education, the Study of Everything
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=6I3ZNpI7qwE
Video Solution(1-16)
~~Hayabusa1
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.