Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
m (→‎=Supplement)
 
(7 intermediate revisions by 5 users not shown)
Line 2: Line 2:
 
==Problem==
 
==Problem==
 
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]
+
 
 +
<asy>
 
unitsize(6);
 
unitsize(6);
 
pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0);
 
pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0);
Line 32: Line 33:
 
defaultpen(fontsize(10pt));
 
defaultpen(fontsize(10pt));
 
real r = 1.3;
 
real r = 1.3;
−
label("<math>A</math>", A, W*r);
+
label("$A$", A, W*r);
−
label("<math>B</math>", B, S*r);
+
label("$B$", B, S*r);
−
label("<math>C</math>", C, S*r);
+
label("$C$", C, S*r);
−
label("<math>D</math>", D, E*r);
+
label("$D$", D, E*r);
−
label("<math>E</math>", EE, E*r);
+
label("$E$", EE, E*r);
−
label("<math>F</math>", F, N*r);
+
label("$F$", F, N*r);
−
label("<math>G</math>", G, N*r);
+
label("$G$", G, N*r);
−
label("<math>H</math>", H, W*r);
+
label("$H$", H, W*r);
−
label("<math>J</math>", J, W*r);
+
label("$J$", J, W*r);
−
[/asy]
+
</asy>
  
 
==Solution 1 (Massive Shoelace)==
 
==Solution 1 (Massive Shoelace)==
Line 55: Line 56:
  
 
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)==
Line 69: Line 72:
 
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>
 
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
 
~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
  
 
==Video Solution (Mathematical Dexterity)==
 
==Video Solution (Mathematical Dexterity)==

Latest revision as of 03:49, 27 August 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_9&oldid=225847"