Difference between revisions of "2011 AMC 12B Problems/Problem 12"
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After careful inspection of the regular octagon, we find that all of the sides of the regular octagon must be <math>1</math>. | After careful inspection of the regular octagon, we find that all of the sides of the regular octagon must be <math>1</math>. | ||
− | This gives the total area to be <math>4 * | + | This gives the total area to be <math>4 * triangle area</math> + <math>4 * rectangle area</math> + <math>square area</math> = <math>4(1/4) + 4(sqrt(1/2)) + 1</math>, which simplifies to <math>2(1+sqrt(2))</math>. |
− | The percent area of the center square is the area of the square divided by the total area of the regular octagon. This gives <math> | + | The percent area of the center square is the area of the square divided by the total area of the regular octagon. This gives <math>1/(2(1+sqrt(2))).</math> Multiplying by the conjugate gives (A). |
== See also == | == See also == | ||
{{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}} | {{AMC12 box|year=2011|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:10, 31 August 2024
Contents
Problem
A dart board is a regular octagon divided into regions as shown below. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
Solution
Let's assume that the side length of the octagon is . The area of the center square is just
. The triangles are all
triangles, with a side length ratio of
. The area of each of the
identical triangles is
, so the total area of all of the triangles is also
. Now, we must find the area of all of the 4 identical rectangles. One of the side lengths is
and the other side length is
, so the area of all of the rectangles is
. The ratio of the area of the square to the area of the octagon is
. Cancelling
from the fraction, the ratio becomes
. Multiplying the numerator and the denominator each by
will cancel out the radical, so the fraction is now
Solution
Instead of doing mindless algebra, let's set the side lengths of the square to .
After careful inspection of the regular octagon, we find that all of the sides of the regular octagon must be .
This gives the total area to be +
+
=
, which simplifies to
.
The percent area of the center square is the area of the square divided by the total area of the regular octagon. This gives Multiplying by the conjugate gives (A).
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.