Difference between revisions of "2002 AMC 10B Problems/Problem 17"
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Then the area of <math>ADG</math> equals <math>\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}</math>. | Then the area of <math>ADG</math> equals <math>\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}</math>. | ||
+ | |||
+ | Solution 2 (alcmus solution) | ||
+ | If the mouse is at <math>(x, y) = (x, 18 - 5x)</math>, then the square of the distance from the mouse to the cheese is\[ | ||
+ | (x - 12)^2 + (8 - 5x)^2 = | ||
+ | 26(x^2 - 4x + 8) = 26((x - 2)^2 + 4). | ||
+ | \]The value of this expression is smallest when <math>x = 2</math>, so the mouse is closest to the cheese at the point <math>(2, 8)</math>, and <math>a+b=2+8 = \boxed{10}</math>. | ||
== See Also == | == See Also == |
Revision as of 13:26, 21 May 2021
Problem
A regular octagon has sides of length two. Find the area of
.
Solution
The area of the triangle can be computed as
. We will now find
and
.
Clearly, is a right isosceles triangle with hypotenuse of length
, hence
.
The same holds for triangle
and its leg
. The length of
is equal to
.
Hence
, and
.
Then the area of equals
.
Solution 2 (alcmus solution)
If the mouse is at , then the square of the distance from the mouse to the cheese is\[
(x - 12)^2 + (8 - 5x)^2 =
26(x^2 - 4x + 8) = 26((x - 2)^2 + 4).
\]The value of this expression is smallest when
, so the mouse is closest to the cheese at the point
, and
.
See Also
2002 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
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