Difference between revisions of "2018 AMC 10B Problems/Problem 24"
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The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN,\triangle NQO,</math> and <math>\triangle ORM</math>). | The desired area (hexagon <math>MPNQOR</math>) consists of an equilateral triangle (<math>\triangle MNO</math>) and three right triangles (<math>\triangle MPN,\triangle NQO,</math> and <math>\triangle ORM</math>). | ||
− | Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio | + | Notice that <math>\overline {AD}</math> (not shown) and <math>\overline {BC}</math> are parallel. <math>\overline {XY}</math> divides transversals <math>\overline {AB}</math> and <math>\overline {CD}</math> into a <math>1:1</math> ratio (This can be shown by similar triangles.). Thus, it must also divide transversal <math>\overline {AC}</math> and transversal <math>\overline {CO}</math> into a <math>1:1</math> ratio. By symmetry, the same applies for <math>\overline {CE}</math> and <math>\overline {EA}</math> as well as <math>\overline {EM}</math> and <math>\overline {AN}.</math> |
− | In <math>\triangle ACE,</math> we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}.</math> Our desired area becomes <cmath>\left(\frac{1}{4}+3 \cdot \frac{1}{8}\right) \cdot \frac{ | + | In <math>\triangle ACE,</math> we see that <math>\frac{[MNO]}{[ACE]} = \frac{1}{4}</math> and <math>\frac{[MPN]}{[ACE]} = \frac{1}{8}.</math> Our desired area becomes <cmath>\left(\frac{1}{4}+3 \cdot \frac{1}{8}\right) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</cmath> |
==Solution 2 (Area Subtraction)== | ==Solution 2 (Area Subtraction)== | ||
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Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on <math>X,</math> is similar to the triangle with a base of <math>YC = \frac12.</math> Using similar triangles, we calculate the base to be <math>\frac{1}{4}</math> and the height to be <math>\frac{\sqrt{3}}{4}</math> giving us an area of <math>\frac{\sqrt{3}}{32}</math> per triangle, and a total area of <math>\frac{3\sqrt{3}}{32}.</math> Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}.</math> Finding the total area, we get <math>6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}.</math> Taking the complement, we get <math>\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</math> | Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on <math>X,</math> is similar to the triangle with a base of <math>YC = \frac12.</math> Using similar triangles, we calculate the base to be <math>\frac{1}{4}</math> and the height to be <math>\frac{\sqrt{3}}{4}</math> giving us an area of <math>\frac{\sqrt{3}}{32}</math> per triangle, and a total area of <math>\frac{3\sqrt{3}}{32}.</math> Adding the two areas together, we get <math>\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}.</math> Finding the total area, we get <math>6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}.</math> Taking the complement, we get <math>\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</math> | ||
− | ==Solution 4 ( | + | ==Solution 4 (Area Subtraction)== |
− | + | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(200); | ||
+ | draw(polygon(6)); | ||
+ | pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; | ||
+ | A = dir(120); | ||
+ | B = dir(60); | ||
+ | C = dir(0); | ||
+ | D = dir(300); | ||
+ | E = dir(240); | ||
+ | F = dir(180); | ||
+ | X = midpoint(A--B); | ||
+ | Y = midpoint(C--D); | ||
+ | Z = midpoint(E--F); | ||
+ | M = intersectionpoint(A--E,X--Z); | ||
+ | N = intersectionpoint(A--C,X--Y); | ||
+ | O = intersectionpoint(C--E,Y--Z); | ||
+ | P = intersectionpoint(A--C,X--Z); | ||
+ | Q = intersectionpoint(C--E,X--Y); | ||
+ | R = intersectionpoint(A--E,Y--Z); | ||
+ | fill(M--P--N--Q--O--R--cycle,mediumgray); | ||
+ | dot("$A$",A,1.5*dir(A),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(B),linewidth(4)); | ||
+ | dot("$C$",C,1.5*dir(C),linewidth(4)); | ||
+ | dot("$D$",D,1.5*dir(D),linewidth(4)); | ||
+ | dot("$E$",E,1.5*dir(E),linewidth(4)); | ||
+ | dot("$F$",F,1.5*dir(F),linewidth(4)); | ||
+ | dot("$X$",X,1.5*dir(X),linewidth(4)); | ||
+ | dot("$Y$",Y,1.5*dir(Y),linewidth(4)); | ||
+ | dot("$Z$",Z,1.5*dir(Z),linewidth(4)); | ||
+ | dot("$M$",M,1.5*dir(165),linewidth(4)); | ||
+ | dot("$N$",N,1.5*dir(45),linewidth(4)); | ||
+ | dot("$O$",O,1.5*dir(-75),linewidth(4)); | ||
+ | dot("$P$",P,1.5*dir(105),linewidth(4)); | ||
+ | dot("$Q$",Q,1.5*dir(-15),linewidth(4)); | ||
+ | dot("$R$",R,1.5*dir(-135),linewidth(4)); | ||
+ | draw(A--C--E--cycle^^X--Y--Z--cycle); | ||
+ | </asy> | ||
+ | |||
+ | We could also subtract <math>\triangle APM,\triangle CQN,</math> and <math>\triangle ERO</math> from <math>\triangle ACE.</math> | ||
+ | |||
+ | Since <math>\angle BAF = 120^{\circ}</math> and <math>\angle BAC = \angle FAE = 30^{\circ},</math> we have <math>\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.</math> | ||
+ | |||
+ | Since <math>AX=BX</math> and <math>FZ=EZ,</math> we have <math>AF \parallel XZ,</math> from which <math>\angle AMX= \angle FAM = 30^{\circ}.</math> | ||
+ | |||
+ | We can show that <math>\triangle APM</math> is <math>30</math>-<math>60</math>-<math>90</math> using a similar method, <math>\triangle CQN</math> and <math>\triangle ERO</math> are also <math>30</math>-<math>60</math>-<math>90.</math> | ||
+ | |||
+ | Since <math>AC=CE=AE=\sqrt{3},</math> we have <math>[ACE]=AC^2 \cdot \frac{\sqrt{3}}{4}=3 \cdot \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3}}{4}.</math> | ||
+ | |||
+ | Since <math>AX= \frac{1}{2}</math> and <math>AP = AX \cdot \frac{\sqrt{3}}{2}= \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4},</math> we have <math>PM = AP \cdot \sqrt{3} = \frac{\sqrt{3}}{4} \cdot \sqrt{3} = \frac{3}{4}.</math> | ||
+ | |||
+ | Note that <cmath>[APM]=[CQN]=[ERO]=\frac{1}{2} \cdot AP \cdot PM = \frac{1}{2} \cdot \frac{\sqrt{3}}{4} \cdot \frac{3}{4} = \frac{3 \sqrt{3}}{32}.</cmath> | ||
+ | Therefore, we get | ||
+ | <cmath>[PNQORM]=[ACE]-[APM]-[CQN]-[ERO]=\frac{3 \sqrt{3}}{4} - 3 \cdot \frac{3 \sqrt{3}}{32} = \frac{24 \sqrt{3}}{32} - \frac{9 \sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==Solution 5 (Partition the Hexagon)== | ==Solution 5 (Partition the Hexagon)== | ||
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dot(M^^N^^O^^P^^Q^^R,linewidth(4)); | dot(M^^N^^O^^P^^Q^^R,linewidth(4)); | ||
</asy> | </asy> | ||
− | Let the brackets denote areas. Note that the desired region contains <math>15</math> of the <math>48</math> small triangles, so the answer is <cmath>\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{ | + | Let the brackets denote areas. Note that the desired region contains <math>15</math> of the <math>48</math> small triangles, so the answer is <cmath>\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</cmath> |
~AlexLikeMath ~MRENTHUSIASM | ~AlexLikeMath ~MRENTHUSIASM | ||
− | ==Solution 6 (Trigonometry)== | + | ==Solution 6 (Partition the Hexagon)== |
+ | Dividing <math>\triangle MNO</math> into two right triangles congruent to <math>\triangle PMN,</math> we see that <math>[MPNQOR]=\dfrac{5}{8}[ACE].</math> Because <math>[ACE] = \dfrac{1}{2}[ABCDEF],</math> we have <math>[MPNQOR]=\dfrac{5}{16}[ABCDEF].</math> From here, you should be able to tell that the answer will have a factor of <math>5,</math> and <math>\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}</math> is the only answer that has a factor of <math>5.</math> However, if you want to actually calculate the area, you would calculate <math>[ABCDEF]</math> to be <math>6 \cdot \dfrac{\sqrt{3}}{2 \cdot 2} = \dfrac{3\sqrt{3}}{2},</math> so <math>[MPNQOR] = \dfrac{5}{16} \cdot \dfrac{3\sqrt{3}}{2} = \frac {15}{32}\sqrt{3}.</math> | ||
+ | |||
+ | ==Solution 7 (Trigonometry)== | ||
Notice, the area of the convex hexagon formed through the intersection of the <math>2</math> triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. | Notice, the area of the convex hexagon formed through the intersection of the <math>2</math> triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. | ||
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^\circ</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^\circ,</math> and thus the side length of this triangle is <math>1+2\left(\frac{1}{2}\cos 60^\circ\right)=1+\frac{1}{2}=\frac{3}{2}.</math> So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}.</math> | To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is <math>120^\circ</math> and the trapezoid is isosceles, we know that the angle opposite is <math>60^\circ,</math> and thus the side length of this triangle is <math>1+2\left(\frac{1}{2}\cos 60^\circ\right)=1+\frac{1}{2}=\frac{3}{2}.</math> So the area of this triangle is <math>\frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16}.</math> | ||
Line 195: | Line 254: | ||
Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</math> | Therefore, the area of the convex hexagon is <math>\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.</math> | ||
− | ==Solution | + | ==Solution 8 (Linear Transformation)== |
If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of <math>ABCDEF</math> then apply it to the old diagram. | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of <math>ABCDEF</math> then apply it to the old diagram. | ||
<asy> | <asy> |
Latest revision as of 03:38, 6 September 2022
- The following problem is from both the 2018 AMC 12B #20 and 2018 AMC 10B #24, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Area Addition)
- 4 Solution 2 (Area Subtraction)
- 5 Solution 3 (Area Subtraction)
- 6 Solution 4 (Area Subtraction)
- 7 Solution 5 (Partition the Hexagon)
- 8 Solution 6 (Partition the Hexagon)
- 9 Solution 7 (Trigonometry)
- 10 Solution 8 (Linear Transformation)
- 11 Video Solution
- 12 See Also
Problem
Let be a regular hexagon with side length
. Denote by
,
, and
the midpoints of sides
,
, and
, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of
and
?
Diagram
~MRENTHUSIASM
Solution 1 (Area Addition)
The desired area (hexagon ) consists of an equilateral triangle (
) and three right triangles (
and
).
Notice that (not shown) and
are parallel.
divides transversals
and
into a
ratio (This can be shown by similar triangles.). Thus, it must also divide transversal
and transversal
into a
ratio. By symmetry, the same applies for
and
as well as
and
In we see that
and
Our desired area becomes
Solution 2 (Area Subtraction)
Instead of directly finding the desired hexagonal area, can be found. It consists of three triangles and the desired hexagon. Given triangle rotational symmetry, the three triangles are congruent. See that
and
are equilateral, so
so
As
is a transversal running through
(use your imagination) and
we have
Then, is a
-
-
triangle. By HL congruence,
Note that
Then, the area of
is
There are three such triangles for a total area of
is
Find the side of
to be
so the area is
The answer is
~BJHHar
Solution 3 (Area Subtraction)
Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of isosceles trapezoids (namely
and
) and
right triangles (namely
and
).
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is and the other base is
(it is halfway in between the side and the longest diagonal, which has length
) with a height of
(by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of
for a total area of
(Alternatively, we could have calculated the area of hexagon
and subtracted the area of
which, as we showed before, had a side length of
).
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on is similar to the triangle with a base of
Using similar triangles, we calculate the base to be
and the height to be
giving us an area of
per triangle, and a total area of
Adding the two areas together, we get
Finding the total area, we get
Taking the complement, we get
Solution 4 (Area Subtraction)
We could also subtract and
from
Since and
we have
Since and
we have
from which
We can show that is
-
-
using a similar method,
and
are also
-
-
Since we have
Since and
we have
Note that
Therefore, we get
Solution 5 (Partition the Hexagon)
We partition hexagon into
congruent
triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains
of the
small triangles, so the answer is
~AlexLikeMath ~MRENTHUSIASM
Solution 6 (Partition the Hexagon)
Dividing into two right triangles congruent to
we see that
Because
we have
From here, you should be able to tell that the answer will have a factor of
and
is the only answer that has a factor of
However, if you want to actually calculate the area, you would calculate
to be
so
Solution 7 (Trigonometry)
Notice, the area of the convex hexagon formed through the intersection of the triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side.
To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is
and the trapezoid is isosceles, we know that the angle opposite is
and thus the side length of this triangle is
So the area of this triangle is
Now let's find the area of the smaller triangles. Notice, cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then
and the sum of the areas is
Therefore, the area of the convex hexagon is
Solution 8 (Linear Transformation)
If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of then apply it to the old diagram.
The isosceles right triangle with a leg length of in the new diagram is
in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract
from the area of
(the red triangles), giving us
However, we need to take the ratio of this area to the area of
which is
Now we know that our answer is
Video Solution
https://www.youtube.com/watch?v=yDbn9Mx2myw
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.