Difference between revisions of "2016 AIME II Problems/Problem 7"
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Solution by Shaddoll (edited by ppiittaattoo) | 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); | ||
+ | </asy> | ||
+ | *Drawing not finished, someone please finish it. | ||
== 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}} |
Revision as of 11:40, 21 February 2019
Problem
Squares and
have a common center and
. The area of
is 2016, and the area of
is a smaller positive integer. Square
is constructed so that each of its vertices lies on a side of
and each vertex of
lies on a side of
. Find the difference between the largest and smallest positive integer values for the area of
.
Solution
Letting and
, we have
by Cauchy-Schwarz inequality. Also, since
, the angles that each square cuts another are equal, so all the triangles are formed by a vertex of a larger square and
adjacent vertices of a smaller square are similar. Therefore, the areas form a geometric progression, so since
, we have the maximum area is
(the areas of the squares from largest to smallest are
forming a geometric progression).
The minimum area is (every square is half the area of the square whose sides its vertices touch), so the desired answer is
.
Solution by Shaddoll (edited by ppiittaattoo)
- Drawing not finished, someone please finish it.
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 |