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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "2016 AIME II Problems/Problem 7"

m (→‎Solution)
(→‎Solution)
 
(11 intermediate revisions by 4 users not shown)
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
−
Letting <math>AI=a</math> and <math>IB=b</math>, we have <math>IJ^{2}=a^{2}+b^{2} \geq 1008</math> by Cauchy-Schwarz inequality. Also, since <math>EFGH||ABCD</math>, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and <math>2</math> adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since <math>2016=12^{2} \cdot 14</math>, we have the maximum area is <math>2016 \cdot \dfrac{11}{12} = 1848</math> (the areas of the squares from largest to smallest are <math>12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14</math> forming a geometric progression).  
+
Letting <math>AI=a</math> and <math>IB=b</math>, we have <cmath>IJ^{2}=a^{2}+b^{2} \geq 1008</cmath> by [[AM-GM inequality]]. Also, since <math>EFGH||ABCD</math>, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and <math>2</math> adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since <cmath>2016=12^{2} \cdot 14</cmath> we have the maximum area is <cmath>2016 \cdot \dfrac{11}{12} = 1848</cmath> (the areas of the squares from largest to smallest are <math>12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14</math> forming a geometric progression).  
  
  
−
The minimum area is <math>1008</math> (every square is half the area of the square whose sides its vertices touch), so the desired answer is <math>1848-1008=\boxed{840}</math>.
+
The minimum area is <math>1008</math> (every square is half the area of the square whose sides its vertices touch), so the desired answer is <cmath>1848-1008=\boxed{840}</cmath>
  
−
Solution by Shaddoll (edited by ppiittaattoo)
+
<asy>
 +
pair A,B,C,D,E,F,G,H,I,J,K,L;
 +
A=(0,0);
 +
B=(2016,0);
 +
C=(2016,2016);
 +
D=(0,2016);
 +
I=(1008,0);
 +
J=(2016,1008);
 +
K=(1008,2016);
 +
L=(0,1008);
 +
E=(504,504);
 +
F=(1512,504);
 +
G=(1512,1512);
 +
H=(504,1512);
 +
draw(A--B--C--D--A);
 +
draw(I--J--K--L--I);
 +
draw(E--F--G--H--E);
 +
label("$A$",A,SW);
 +
label("$B$",B,SE);
 +
label("$C$",C,NE);
 +
label("$D$",D,NW);
 +
label("$E$",E,SW);
 +
label("$F$",F,SE);
 +
label("$G$",G,NE);
 +
label("$H$",H,NW);
 +
label("$I$",I,S);
 +
label("$J$",J,NE);
 +
label("$K$",K,N);
 +
label("$L$",L,NW);
 +
</asy>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2016|n=II|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 23:10, 1 September 2021

Problem

Squares $ABCD$ and $EFGH$ have a common center and $\overline{AB} || \overline{EF}$. The area of $ABCD$ is 2016, and the area of $EFGH$ is a smaller positive integer. Square $IJKL$ is constructed so that each of its vertices lies on a side of $ABCD$ and each vertex of $EFGH$ lies on a side of $IJKL$. Find the difference between the largest and smallest positive integer values for the area of $IJKL$.

Solution

Letting $AI=a$ and $IB=b$, we have \[IJ^{2}=a^{2}+b^{2} \geq 1008\] by AM-GM inequality. Also, since $EFGH||ABCD$, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and $2$ adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since \[2016=12^{2} \cdot 14\] we have the maximum area is \[2016 \cdot \dfrac{11}{12} = 1848\] (the areas of the squares from largest to smallest are $12^{2} \cdot 14, 11 \cdot 12 \cdot 14, 11^{2} \cdot 14$ forming a geometric progression).


The minimum area is $1008$ (every square is half the area of the square whose sides its vertices touch), so the desired answer is \[1848-1008=\boxed{840}\]

[asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(2016,0); C=(2016,2016); D=(0,2016); I=(1008,0); J=(2016,1008); K=(1008,2016); L=(0,1008); E=(504,504); F=(1512,504); G=(1512,1512); H=(504,1512); draw(A--B--C--D--A); draw(I--J--K--L--I); draw(E--F--G--H--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,SW); label("$F$",F,SE); label("$G$",G,NE); label("$H$",H,NW); label("$I$",I,S); label("$J$",J,NE); label("$K$",K,N); label("$L$",L,NW); [/asy]

See also

2016 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AIME_II_Problems/Problem_7&oldid=161346"