Difference between revisions of "2018 AIME II Problems/Problem 9"
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+ | __TOC__ | ||
==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. | ||
− | + | <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)== | ==Solution 1 (Massive Shoelace)== | ||
− | We represent | + | 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> | ||
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<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>. | ||
+ | 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 | ||
+ | |||
+ | ==Video Solution (Mathematical Dexterity)== | ||
+ | https://www.youtube.com/watch?v=HUwJqixBLUI | ||
+ | |||
+ | ==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 03:49, 27 August 2024
Contents
Problem
Octagon with side lengths and is formed by removing 6-8-10 triangles from the corners of a rectangle with side on a short side of the rectangle, as shown. Let be the midpoint of , and partition the octagon into 7 triangles by drawing segments , , , , , and . Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
Solution 1 (Massive Shoelace)
We represent octagon in the coordinate plane with the upper left corner of the rectangle being the origin. Then it follows that . Recall that the centroid is way up each median in the triangle. Thus, we can find the centroids easily by finding the midpoint of the side opposite of point . Furthermore, we can take advantage of the reflective symmetry across the line parallel to going through by dealing with less coordinates and ommiting the in the shoelace formula.
By doing some basic algebra, we find that the coordinates of the centroids of are and , respectively. We'll have to throw in the projection of the centroid of to the line of reflection to apply shoelace, and that point is
Finally, applying Shoelace, we get:
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 except . We have a homothety since:
1. passes through corresponding vertices of the two heptagons.
2. By centroid properties, our ratio between the sidelengths is and their area ratio is hence
Compute the area of the large heptagon by dividing into two congruent trapezoids and a triangle. The area of each trapezoid is The area of each triangle is
Hence, the area of the large heptagon is Then, from our homothety, the area of the required heptagon is ~novus677
Supplement
Note that we use 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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