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Difference between revisions of "2018 AIME II Problems/Problem 9"

(→‎Solution 2 (Homothety))
(→‎Supplement)
 
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__TOC__
 
==Problem==
 
==Problem==
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Octagon <math>ABCDEFGH</math> with side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <math>23</math> <math>\times</math> <math>27</math> rectangle with side <math>\overline{AH}</math> on a short side of the rectangle, as shown. Let <math>J</math> be the midpoint of <math>\overline{AH}</math>, and partition the octagon into 7 triangles by drawing segments <math>\overline{JB}</math>, <math>\overline{JC}</math>, <math>\overline{JD}</math>, <math>\overline{JE}</math>, <math>\overline{JF}</math>, and <math>\overline{JG}</math>. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
  
−
Octagon <math>ABCDEFGH</math> with side lengths <math>AB = CD = EF = GH = 10</math> and <math>BC = DE = FG = HA = 11</math> is formed by removing 6-8-10 triangles from the corners of a <math>23</math> <math>\times</math> <math>27</math> rectangle with side <math>\overline{AH}</math> on a short side of the rectangle, as shown. Let <math>J</math> be the midpoint of <math>\overline{AH}</math>, and partition the octagon into 7 triangles by drawing segments <math>\overline{JB}</math>, <math>\overline{JC}</math>, <math>\overline{JD}</math>, <math>\overline{JE}</math>, <math>\overline{JF}</math>, and <math>\overline{JG}</math>. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
+
<asy>
 +
unitsize(6);
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pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0);
 +
pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23),  G = (8, 23), J = (0, 23/2), H = (0, 17);
 +
draw(P--Q--R--SS--cycle);
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draw(J--B);
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draw(J--C);
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draw(J--D);
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draw(J--EE);
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draw(J--F);
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draw(J--G);
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draw(A--B);
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draw(H--G);
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real dark = 0.6;
 +
filldraw(A--B--P--cycle, gray(dark));
 +
filldraw(H--G--Q--cycle, gray(dark));
 +
filldraw(F--EE--R--cycle, gray(dark));
 +
filldraw(D--C--SS--cycle, gray(dark));
 +
dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(EE);
 +
dot(F);
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dot(G);
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dot(H);
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dot(J);
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dot(H);
 +
defaultpen(fontsize(10pt));
 +
real r = 1.3;
 +
label("$A$", A, W*r);
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label("$B$", B, S*r);
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label("$C$", C, S*r);
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label("$D$", D, E*r);
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label("$E$", EE, E*r);
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label("$F$", F, N*r);
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label("$G$", G, N*r);
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label("$H$", H, W*r);
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label("$J$", J, W*r);
 +
</asy>
  
 
==Solution 1 (Massive Shoelace)==
 
==Solution 1 (Massive Shoelace)==
−
We represent Octagon <math>ABCDEFGH</math> in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that <math>A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})</math>. Recall that the centroid is <math>\frac{1}{3}</math> way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point <math>J</math>. Furthermore, we can take advantage of the reflective symmetry across the line parallel to <math>BC</math> going through <math>J</math> by dealing with less coordinates and ommiting the <math>\frac{1}{2}</math> in the shoelace formula.  
+
We represent octagon <math>ABCDEFGH</math> in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that <math>A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})</math>. Recall that the centroid is <math>\frac{1}{3}</math> way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point <math>J</math>. Furthermore, we can take advantage of the reflective symmetry across the line parallel to <math>BC</math> going through <math>J</math> by dealing with less coordinates and ommiting the <math>\frac{1}{2}</math> in the shoelace formula.  
  
 
By doing some basic algebra, we find that the coordinates of the centroids of <math>\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE</math> are <math>\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),</math> and <math>\left(18, -\frac{23}{2}\right)</math>, respectively. We'll have to throw in the projection of the centroid of <math>\bigtriangleup JAB</math> to the line of reflection to apply shoelace, and that point is <math>\left( \frac{8}{3}, -\frac{23}{2}\right)</math>
 
By doing some basic algebra, we find that the coordinates of the centroids of <math>\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE</math> are <math>\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),</math> and <math>\left(18, -\frac{23}{2}\right)</math>, respectively. We'll have to throw in the projection of the centroid of <math>\bigtriangleup JAB</math> to the line of reflection to apply shoelace, and that point is <math>\left( \frac{8}{3}, -\frac{23}{2}\right)</math>
Line 13: Line 54:
 
<math>=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|</math>
 
<math>=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|</math>
 
<math>=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}</math>
 
<math>=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}</math>
 +
 
Solution by ktong
 
Solution by ktong
 +
 +
note: for a slightly simpler calculation, notice that the heptagon can be divided into two trapezoids of equal area and a small triangle.
  
 
==Solution 2 (Homothety)==
 
==Solution 2 (Homothety)==
 
Draw the heptagon whose vertices are the midpoints of octagon <math>ABCDEFGH</math> except <math>J</math>.
 
Draw the heptagon whose vertices are the midpoints of octagon <math>ABCDEFGH</math> except <math>J</math>.
−
Note that <math>J</math> passes through corresponding vertices of the two heptagons. Also, by centroid properties, our ratio between the sidelengths is <math>\frac{2}{3}</math>, and their area ratio is hence <math>\frac{4}{9}</math>. So, we have a homothety.  
+
We have a homothety since:
 +
 
 +
1. <math>J</math> passes through corresponding vertices of the two heptagons.
 +
 
 +
2. By centroid properties, our ratio between the sidelengths is <math>\frac{2}{3},</math> and their area ratio is hence <math>\frac{4}{9}.</math>
 +
 
 +
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is <cmath>= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.</cmath> The area of each triangle is <cmath>=
 +
\frac{1}{2}\cdot 17\cdot 4=34.</cmath>
 +
 
 +
Hence, the area of the large heptagon is <cmath>2\cdot 190+34=414.</cmath> Then, from our homothety, the area of the required heptagon is <cmath>\frac{4}{9}\cdot 414=\boxed{184}.</cmath>
 +
~novus677
 +
 
 +
===Supplement===
 +
Note that we use <math>17.</math> A proof of this is as follows:
 +
*First, note that the shaded triangles are 6-8-10 triangles. Because the homothety is defined on the midpoints of the sides of the octagon, the smaller triangle created by that midpoint and the surrounding rectangle has sides 3-4-5. Thus, the sidelength of the octagon (17) is just 23-2*3=17.
 +
~mathboy282
  
−
We now compute the area of our large heptagon. We can divide this into two congruent trapezoids and a triangle. The area of each trapezoid is <math>\frac{1}{2}(17+23)\cdot \frac{19}{2}=190</math>. The area of each triangle is <math>\frac{1}{2}\cdot 17\cdot 2=34</math>.
+
==Video Solution (Mathematical Dexterity)==
 +
https://www.youtube.com/watch?v=HUwJqixBLUI
  
−
Hence, the area of the large heptagon is <math>2\cdot 190+34=414</math>. Then from our homothety, the area of the required heptagon is <math>frac{4}{9}\cdot 414=\boxed{184}</math>
+
==See Also==
 
{{AIME box|year=2018|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2018|n=II|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:30, 5 January 2024

Problem

Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$, and partition the octagon into 7 triangles by drawing segments $\overline{JB}$, $\overline{JC}$, $\overline{JD}$, $\overline{JE}$, $\overline{JF}$, and $\overline{JG}$. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.

[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23), G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy]

Solution 1 (Massive Shoelace)

We represent octagon $ABCDEFGH$ in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that $A=(0,-6), B=(8, 0), C=(19,0), D=(27, -6), E=(27, -17), F=(19, -23), G=(8, -23), H=(0, -17), J=(0, -\frac{23}{2})$. Recall that the centroid is $\frac{1}{3}$ way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point $J$. Furthermore, we can take advantage of the reflective symmetry across the line parallel to $BC$ going through $J$ by dealing with less coordinates and ommiting the $\frac{1}{2}$ in the shoelace formula.

By doing some basic algebra, we find that the coordinates of the centroids of $\bigtriangleup JAB, \bigtriangleup JBC, \bigtriangleup JCD, \bigtriangleup JDE$ are $\left(\frac{8}{3}, -\frac{35}{6}\right), \left(9, -\frac{23}{6}\right), \left(\frac{46}{3}, -\frac{35}{6}\right),$ and $\left(18, -\frac{23}{2}\right)$, respectively. We'll have to throw in the projection of the centroid of $\bigtriangleup JAB$ to the line of reflection to apply shoelace, and that point is $\left( \frac{8}{3}, -\frac{23}{2}\right)$

Finally, applying Shoelace, we get: $\left|\left(\frac{8}{3}\cdot (-\frac{23}{6})+9\cdot (-\frac{35}{6})+\frac{46}{3}\cdot (-\frac{23}{2})+18\cdot (\frac{-23}{2})+\frac{8}{3}\cdot (-\frac{35}{6})\right) - \left((-\frac{35}{6}\cdot 9) +\\(-\frac{23}{6}\cdot \frac{46}{3})+ (-\frac{35}{6}\cdot 18)+(\frac{-23}{2}\cdot \frac{8}{3})+(-\frac{23}{2}\cdot \frac{8}{3})\right)\right|$ $=\left|\left(-\frac{92}{9}-\frac{105}{2}-\frac{529}{3}-207-\frac{140}{9}\right)-\left(-\frac{105}{2}-\frac{529}{9}-105-\frac{92}{3}-\frac{92}{3}\right)\right|$ $=\left|-\frac{232}{9}-\frac{1373}{6}-207+\frac{529}{9}+\frac{184}{3}+105+\frac{105}{2}\right|$ $=\left|\frac{297}{9}-\frac{690}{6}-102\right|=\left| 33-115-102\right|=\left|-184\right|=\boxed{184}$

Solution by ktong

note: for a slightly simpler calculation, notice that the heptagon can be divided into two trapezoids of equal area and a small triangle.

Solution 2 (Homothety)

Draw the heptagon whose vertices are the midpoints of octagon $ABCDEFGH$ except $J$. We have a homothety since:

1. $J$ passes through corresponding vertices of the two heptagons.

2. By centroid properties, our ratio between the sidelengths is $\frac{2}{3},$ and their area ratio is hence $\frac{4}{9}.$

Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is \[= \frac{1}{2}(17+23)\cdot \frac{19}{2}=190.\] The area of each triangle is \[= \frac{1}{2}\cdot 17\cdot 4=34.\]

Hence, the area of the large heptagon is \[2\cdot 190+34=414.\] Then, from our homothety, the area of the required heptagon is \[\frac{4}{9}\cdot 414=\boxed{184}.\] ~novus677

Supplement

Note that we use $17.$ A proof of this is as follows:

  • First, note that the shaded triangles are 6-8-10 triangles. Because the homothety is defined on the midpoints of the sides of the octagon, the smaller triangle created by that midpoint and the surrounding rectangle has sides 3-4-5. Thus, the sidelength of the octagon (17) is just 23-2*3=17.

~mathboy282

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=HUwJqixBLUI

See Also

2018 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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