Difference between revisions of "2006 AMC 10B Problems/Problem 24"
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== Solution == | == Solution == | ||
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+ | [[File:2006AMC10B24.png]] | ||
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+ | If that doesn't load go to https://www.imgur.com/a/7aphGaa | ||
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+ | Sorry for the wrong point names, I didn't know how to change them. | ||
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Since a [[tangent line]] is [[perpendicular]] to the [[radius]] containing the point of tangency, <math>\angle OAD = \angle PDA = 90^\circ</math>. | Since a [[tangent line]] is [[perpendicular]] to the [[radius]] containing the point of tangency, <math>\angle OAD = \angle PDA = 90^\circ</math>. | ||
Revision as of 23:37, 30 December 2020
Problem
Circles with centers and
have radii
and
, respectively, and are externally tangent. Points
and
on the circle with center
and points
and
on the circle with center
are such that
and
are common external tangents to the circles. What is the area of the concave hexagon
?
Solution
If that doesn't load go to https://www.imgur.com/a/7aphGaa
Sorry for the wrong point names, I didn't know how to change them.
Since a tangent line is perpendicular to the radius containing the point of tangency, .
Construct a perpendicular to that goes through point
. Label the point of intersection
.
Clearly is a rectangle, so
and
. By the Pythagorean Theorem,
.
The area of is
. The area of
is
, so the area of quadrilateral
is
. Using similar steps, the area of quadrilateral
is also
. Therefore, the area of hexagon
is
.
See also
2006 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.