Difference between revisions of "2022 AMC 12B Problems/Problem 10"
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\textbf{(E)}\ 12</math> | \textbf{(E)}\ 12</math> | ||
− | == | + | ==Diagram== |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
− | + | size(200); | |
+ | pair A, B, C, D, E, F, G, H; | ||
+ | A = dir(120); | ||
+ | B = dir(60); | ||
+ | C = dir(0); | ||
+ | D = dir(-60); | ||
+ | E = dir(-120); | ||
+ | F = dir(180); | ||
+ | G = midpoint(A--B); | ||
+ | H = midpoint(D--E); | ||
+ | filldraw(G--C--H--F--cycle,yellow); | ||
+ | |||
+ | draw(polygon(6)); | ||
+ | dot("$A$",A,1.5*A,linewidth(4)); | ||
+ | dot("$B$",B,1.5*B,linewidth(4)); | ||
+ | dot("$C$",C,1.5*C,linewidth(4)); | ||
+ | dot("$D$",D,1.5*D,linewidth(4)); | ||
+ | dot("$E$",E,1.5*E,linewidth(4)); | ||
+ | dot("$F$",F,1.5*F,linewidth(4)); | ||
+ | dot("$G$",G,1.5*dir(90),linewidth(4)); | ||
+ | dot("$H$",H,1.5*dir(-90),linewidth(4)); | ||
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 1 == | ||
+ | Let the center of the hexagon be <math>O</math>. <math>\triangle AOB</math>, <math>\triangle BOC</math>, <math>\triangle COD</math>, <math>\triangle DOE</math>, <math>\triangle EOF</math>, and <math>\triangle FOA</math> are all equilateral triangles with side length <math>2</math>. Thus, <math>CO = 2</math>, and <math>GO = \sqrt{AO^2 - AG^2} = \sqrt{3}</math>. By symmetry, <math>\angle COG = 90^{\circ}</math>. Thus, by the Pythagorean theorem, <math>CG = \sqrt{2^2 + \sqrt{3}^2} = \sqrt{7}</math>. Because <math>CO = OF</math> and <math>GO = OH</math>, <math>CG = HC = FH = GF = \sqrt{7}</math>. Thus, the solution to our problem is <math>\sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}</math>. | ||
+ | |||
+ | ~mathboy100 | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Consider triangle <math>AFG</math>. Note that <math>AF = 2</math>, <math>AG = 1</math>, and <math>\angle GAF = 120 ^{\circ}</math> because it is an interior angle of a regular hexagon. (See note for details.) | ||
+ | |||
+ | By the [[Law of Cosines]], we have: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\ | FG^2 &= AG^2 + AF^2 - 2 \cdot AG \cdot AF \cdot \cos \angle GAF \\ | ||
FG^2 &= 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\ | FG^2 &= 1^2 + 2^2 - 2 \cdot 1 \cdot 2 \cdot \cos 120 ^{\circ} \\ | ||
− | FG^2 &= | + | FG^2 &= 5 + 4 \cdot \left( \frac 12 \right) \\ |
FG^2 &= 7 \\ | FG^2 &= 7 \\ | ||
FG &= \sqrt 7. | FG &= \sqrt 7. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
+ | By [[SAS Congruence]], triangles <math>AFG</math>, <math>BCG</math>, <math>CDH</math>, and <math>EFH</math> are congruent, and by [[CPCTC]], quadrilateral <math>GCHF</math> is a rhombus. Therefore, the perimeter of <math>GCHF</math> is <math>4 \cdot FG = \boxed{\textbf{(D)}\ 4\sqrt7}</math>. | ||
+ | |||
+ | Note: The sum of the interior angles of any polygon with <math>n</math> sides is given by <math>180 ^{\circ} (n - 2)</math>. Therefore, the sum of the interior angles of a hexagon is <math>720 ^{\circ}</math>, and each interior angle of a regular hexagon measures <math>\frac{720 ^{\circ}}{6} = 120 ^{\circ}</math>. | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | 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 the distance formula four times to get <math>CH, HF, FG, \text{ and } GC.</math> | ||
+ | <cmath>\begin{alignat*}{8} | ||
+ | CH^2 &= (2-0)^2 + (0-(-\sqrt{3}))^2 &&= 7 &&\rightarrow CH &&= \sqrt{7}, \\ | ||
+ | HF^2 &= (0-(-2))^2 + (-\sqrt{3}-0)^2 &&= 7 &&\rightarrow HF &&= \sqrt{7}, \\ | ||
+ | FG^2 &= (-2-0)^2 + (0-\sqrt{3})^2 &&= 7 &&\rightarrow FG &&= \sqrt{7}, \\ | ||
+ | GC^2 &= (0-2)^2 + (\sqrt{3}-0)^2 &&= 7 &&\rightarrow GC &&= \sqrt{7}. | ||
+ | \end{alignat*}</cmath> | ||
+ | Thus, the perimeter of <math>GCHF = CH + HF + FG + GC = \sqrt{7} + \sqrt{7} + \sqrt{7} + \sqrt{7} = \boxed{\textbf{(D)}\ 4\sqrt7}</math>. | ||
+ | |||
+ | ~sirswagger21 | ||
+ | |||
+ | Note: the last part of this solution could have been simplified by noting that <math>CH = HF = FG = GC = \sqrt{7}.</math> | ||
+ | |||
+ | == Solution 4 == | ||
+ | Note that triangles <math>\triangle{GAF}, \triangle{FEH}, \triangle{HDC},</math> and <math>\triangle{CBG}</math> are all congruent, since they have side lengths of <math>1</math> and <math>2</math> and an included angle of <math>120^{\circ}.</math> | ||
+ | |||
+ | By the Law of Cosines, <cmath>FG=\sqrt{1^2+2^2-2\cdot{1}\cdot{-\frac{1}{2}}}=\sqrt{7}.</cmath> Therefore, <math>FG+GC+CH+HF=4\cdot{FG}=\boxed{\textbf{(D)}\ 4\sqrt7}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | == Solution 5 (Answer Choices + Pythagorean Theorem Extension) == | ||
+ | Like the previous solutions, 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 see that <math>a^2+b^2<c^2</math>, in any given obtuse triangle. | ||
+ | |||
+ | Since <math>ABCDEF</math> is a regular hexagon, all of its angles are obtuse. It follows that <math>\triangle{GAF}</math> is an obtuse triangle. Using the extended Pythagorean Theorem for obtuse triangles, we have: <cmath>AG^2+AF^2<GF^2 \implies 1^2+2^2<GF^2 \implies \sqrt{5}<GF</cmath> | ||
+ | |||
+ | 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}</math>. | ||
+ | |||
+ | ~SwordOfJustice | ||
+ | |||
+ | ==Video Solution 1 by mop 2024== | ||
+ | https://youtu.be/ezGvZgBLe8k&t=0s | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ==Video Solution 2 (Under 1 min!)== | ||
+ | https://youtu.be/XUIm4yOwVd0 | ||
+ | |||
+ | ~Education, the Study of Everything | ||
− | + | ==Video Solution 3 by SpreadTheMathLove== | |
+ | https://www.youtube.com/watch?v=6I3ZNpI7qwE | ||
− | == | + | ==Video Solution(1-16)== |
+ | https://youtu.be/SCwQ9jUfr0g | ||
− | + | ~~Hayabusa1 | |
== See Also == | == See Also == |
Latest revision as of 22:16, 27 October 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 solutions, 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 see that
, in any given obtuse triangle.
Since is a regular hexagon, all of its angles are obtuse. It follows that
is an obtuse triangle. Using the extended Pythagorean Theorem for obtuse triangles, we have:
Since is a rhombus, the perimeter is
. This eliminates all answer choices but
and
, since in all of those options
. Lastly,
is eliminated due to the triangle inequality, as
is not greater than
.
Hence, the answer is .
~SwordOfJustice
Video Solution 1 by mop 2024
https://youtu.be/ezGvZgBLe8k&t=0s
~r00tsOfUnity
Video Solution 2 (Under 1 min!)
~Education, the Study of Everything
Video Solution 3 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.